text
stringlengths 81
47k
| source
stringlengths 59
147
|
|---|---|
Question: <p>If the Frattini subgroup is trivial,for a group <em>P</em>, then <em>P</em> is elementary abelian. Where, Frattini subgroup is the intersection of all maximal subgroups. Please prove the above statement.</p>
Answer: <p>Let $G$ be a $p$-group. Then every maximal subgroup is normal. </p>
<p>Notice that $G/\Phi(G)=G/\bigcap_{i=1}^{n} M_i$ can be embedded into $G/M_1\times G/M_2... G/M_n$. (Show this !)</p>
<p>Since the target group is elementary abelian (Why?), $G/\Phi(G)$ is elementary abelian.</p>
|
https://math.stackexchange.com/questions/2197158/frattini-subgroup-is-trivial-hen-the-group-is-elementary-abelian
|
Question: <p>I have to prove that if $G$ and $G'$ are two finites group of same cardinal, then they are isomorphic.</p>
<p>Actually, it looks obvious. Suppose $G=\{g_1,...,g_n\}$ and $G'=\{h_1,...,h_n\}$. Does the homomorphism $g_i\longmapsto h_i$ work ? </p>
Answer: <p>This is not true in general., For example $\Bbb Z/4\Bbb Z$ is not isomorphic to $\Bbb Z/2\Bbb Z\times \Bbb Z/2\Bbb Z$.</p>
<p>You probably also need to assume both groups are cyclic. In that case you can write one as $\{a,a^2,\dots,a^n\}$ and the other as $\{b,b^2,\dots,b^n\}$. Then map $a\mapsto b$ to get the isomorphism.</p>
|
https://math.stackexchange.com/questions/1649636/if-g-and-g-are-two-finite-group-of-same-cardinal-then-g-cong-g
|
Question: <blockquote>
<p>If <span class="math-container">$G$</span> is perfect, show <span class="math-container">$G/N$</span> is also perfect (for a <span class="math-container">$N\trianglelefteq G$</span>)</p>
</blockquote>
<p>I have some proof, but I don't think it right.</p>
<h3>Proof?</h3>
<p>Choose a <span class="math-container">$Ng\in G/N$</span> then since <span class="math-container">$g\in G$</span> and <span class="math-container">$G$</span> is perfect, then there must exists a <span class="math-container">$h,k\in G$</span> such that <span class="math-container">$[h,k]=g$</span>.</p>
<p>Now consider <span class="math-container">$[Nh, Nk] = (Nh^{-1})(Nk^{-1})(Nh)(Nk) = N (h^{-1}k^{-1}hk) = Ng$</span>. <span class="math-container">$\square$</span></p>
<h3>However</h3>
<p>The proper definition of a perfect group is</p>
<blockquote>
<p><span class="math-container">$G$</span> is perfect if it's commutator subgroup equals <span class="math-container">$G$</span>.</p>
</blockquote>
<p>Or <span class="math-container">$[G,G]=G$</span>, where the commutator subgroup is defined as</p>
<blockquote>
<p><span class="math-container">$[G,G] = \langle \{[g,h]: g,h\in G\}\rangle$</span></p>
</blockquote>
<p>Now I'm a bit confused with this definition. This implies that <span class="math-container">$[G,G]$</span> is the group generated by the commutators <span class="math-container">$[g,h]$</span>. I guess this implies that if not all elements of <span class="math-container">$[G,G]$</span> can be written as <span class="math-container">$[g,h]$</span>, but for example some can be written as a product of commutators.</p>
<p>This would imply that the step "<em>then there must exists a <span class="math-container">$h,k\in G$</span> such that <span class="math-container">$[h,k]=g$</span></em>" is faulty.</p>
<p>Is this really an issue? And what is the best way to fix this issue?</p>
Answer: <p>If $G$ is perfect, that is $G=G'$, then $(G/N)'=G'N/N=GN/N=G/N$.</p>
|
https://math.stackexchange.com/questions/2573644/if-g-is-perfect-show-g-n-is-also-perfect-for-a-n-trianglelefteq-g
|
Question: <p>I'm studying dimension of the group now. But I have some trouble with it.</p>
<p>Consider the dimension of 2 by 2 orthogonal group, <span class="math-container">$O(2)=\{A \in M(2,2,R): AA^T=I\}$</span></p>
<p>Let <span class="math-container">$A = \begin{bmatrix}
a & b \\
c & d
\end{bmatrix}$</span>.</p>
<p>Then by some calculation, we get three equations <span class="math-container">$a^2+b^2=1 , c^2+d^2=1, ac+bd=0$</span>.</p>
<p>So, can I say that 'there was four free variable, and three equations, now it's dimesion is one although equations are not linear equation'?</p>
<p>Thank you</p>
Answer: <p>Yes and no.
You need to show that your system of equations does not include "repetition".</p>
<p>Suppose your conditions <span class="math-container">$A \in G \iff \phi(A)=0$</span>, where <span class="math-container">$\phi : M_n(\mathbb{R}) \rightarrow \mathbb{R}^p$</span> (i.e. "there are <span class="math-container">$p$</span> condition equations).
In order to be able to conclude that the dimension of <span class="math-container">$G$</span> as a Lie group is <span class="math-container">$n^2 - p$</span>, you need to show that <span class="math-container">$\phi$</span> is locally surjective. This translates to a condition on the rank of the differential of <span class="math-container">$\phi$</span>.</p>
<p>More specifically, your conditions could be rephrased as <span class="math-container">$\phi(A) = AA^T-I = 0$</span>, which is <span class="math-container">$n^2$</span> equations (in your case, n=2). However, only <span class="math-container">$n(n+1)/2$</span> of those are "free equations", namely the rank of the differential of <span class="math-container">$\phi$</span> is only <span class="math-container">$n(n+1)/2$</span>, and the dimension of <span class="math-container">$O(n)$</span> is <span class="math-container">$n(n-1)/2$</span>.</p>
|
https://math.stackexchange.com/questions/4168724/find-the-dimension-of-the-system-with-equations-and-variables
|
Question: <p>this is my first question so let me know if I broke any rules. I am writing my dissertation on the zero divisor conjecture in group rings, and I am struggling to find any examples I could put in my chapter examining the classes of groups mentioned above. This might be a dumb question, but I have searched for hours on google scholar and through the books at my university library, but I never find anything like</p>
<p><span class="math-container">$$G = \langle \text{generating elements} \mid \text{relation of elements} \rangle.$$</span></p>
<p>Ideally, I'm looking for a group that satisfies the unique product condition, but is not orderable, and an orderable group that is not free. If anyone knows any literature that could help me, that would be greatly appreciated too!</p>
Answer: <ol>
<li><p>There are many examples of non-free right-orderable groups, with explicit (finite) presentations for instance, surface groups, Braid groups, torsion-free 1-relator groups
<span class="math-container">$$
\langle x_1,...,x_n| w\rangle,
$$</span>
i.e. groups where <span class="math-container">$w$</span> is a cyclically reduced word which is not a proper power of another word. </p></li>
<li><p>Examples of groups with the UP property which are not right-orderable (RO) are harder to find, one such example is given by Dunfield in his appendix to</p></li>
</ol>
<p>S. Kionke, J. Raimbault, <a href="https://arxiv.org/pdf/1411.6449.pdf" rel="nofollow noreferrer">On geometric aspects of diffuse groups</a>.
With an appendix by Nathan Dunfield.
Doc. Math. 21 (2016), 873–915.</p>
<p>Specifically, he gives an example (with an explicit presentation) of a <strong>diffuse group</strong> which is not RO. On the other hand, <strong>diffuse groups</strong> are proven to have the UP property in </p>
<p>B. Bowditch, A variation on the unique product property.
J. London Math. Soc. (2) 62 (2000), no. 3, 813–826.</p>
<ol start="3">
<li>Just for completeness, here is the definition of the UP (Unique Product) Property:</li>
</ol>
<p>A group <span class="math-container">$G$</span> is said to have the <strong>unique product property</strong> if for every two finite non-empty subsets <span class="math-container">$A, B \subset G$</span> there is an element in the product <span class="math-container">$x \in A \cdot B$</span> for which the representation in the form <span class="math-container">$x = ab$</span> with <span class="math-container">$a \in A$</span> and <span class="math-container">$b \in B$</span> is unique. (I.e., if <span class="math-container">$x=a_1b_1=a_2b_2, a_i\in A, b_i\in B, i=1,2$</span>, then <span class="math-container">$a_1=a_2, b_1=b_2$</span>.) </p>
<p>The interest in such groups comes from the (rather easy) fact that very group satisfying the UP property has no zero divisors in the group ring <span class="math-container">${\mathbb Z}G$</span>. Such <span class="math-container">$G$</span> is necessarily torsion-free. For a while, it was unknown if every torsion-free group has the UP property. The first examples were constructed in </p>
<p>E. Rips and Y. Segev, Torsion-free group without unique product property, J. Algebra, 108 (1987), no. 1, 116–126.</p>
|
https://math.stackexchange.com/questions/3544023/looking-for-examples-of-free-groups-ordered-groups-and-unique-product-groups
|
Question: <p>Does there exist nontrivial homomorphism between symmetric group $S_n$ to set of integer $\mathbb Z$?</p>
<p>If yes, how many?</p>
<p>If not, then why?</p>
Answer: <p>An element of finite order is by a homomorphism always mapped to an element of finite order. This is proven quite directly from the defining properties of homomorphisms. So given an $s\in S_n$ and a homomorhism $h:S_n\to \Bbb Z$, what can $h(s)$ possibly be? Does this argument change for $D_n$ instead of $S_n$?</p>
|
https://math.stackexchange.com/questions/2660136/homomorphism-between-s-n-to-mathbb-z
|
Question: <p>Let $*$ be defined on $\mathbb R^3$ such that $(x,y,t)*(x',y',t'):=( x+x',y+y',t+t'+\dfrac 12(xy'-x 'y ) )$ , I can show that $(\mathbb R^3 , *)$ is a group ; I want to find the center of this group , Please help . </p>
Answer: <p>Suppose $(x',y',t')$ is in the center. Then $(x,y,t)*(x',y',t')=(x',y',t')*(x,y,t)$ for all $(x,y,t)\in\mathbb{R}^3$. This translates to $$(x+x',y+y',t+t'+\frac{1}{2}(xy'-x'y))=(x'+x,y'+y, t'+t+\frac{1}{2}(x'y-xy'))$$
Since addition is commutative, the condition simplifies to $xy'-x'y=x'y-xy'$ for all $x,y$. This rearranges to $2xy'=2x'y$. The only way this holds for all $x,y$ is if $x'=y'=0$.</p>
|
https://math.stackexchange.com/questions/1474870/let-mathbb-r3-be-a-group-with-operation-x-y-tx-y-t-xx-y
|
Question: <p>If you have a group $G$ and a proper subgroup $H$ inside of the group. Then is $H$ a proper subgroup of the quotient group $G/H$? </p>
Answer: <p>In order to be a subgroup of a group, one must first be a subset. However, $H$ is not a subgroup of $G/H$ for any normal subgroup $H$ because $G/H$ contains as its elements cosets in $G$. For instance $H$ is the identity of $G/H$ because $H$ is the class of the identity $e\in G$. The identity in $H$ is $e$ though and so if it was a subgroup then it would share the same identity as $G/H$. This is not the case.</p>
<p>It is possible for $H$ to be <em>isomorphic</em> to a proper subgroup of $G/H$ and this is a more interesting question. The easiest example is if $H$ is the trivial subgroup generated by the identity. In this case $G/\langle e\rangle=\{\{g\}\mid g\in G\}$ and so the subgroup $\{\{e\}\}\in G/\langle e\rangle$ is isomorphic to the subgroup $\{e\}$ in $G$. (note the double brackets in the notation used for the trivial subgroup in $G/\langle e\rangle$. It's important).</p>
|
https://math.stackexchange.com/questions/676442/quotient-a-group-by-a-proper-subgroup
|
Question: <p>I am trying to find an example of a homomorphism from $GL_n(\mathbb{R}) \rightarrow \mathbb{R^+}$ with kernel $K = \{A\in GL_n(\mathbb{R})| \det(A) = \pm 1\}$ however it is eluding me. I can't seem to make it a homomorphism.</p>
Answer: <p>If $\mathbb{R^+}$ is the multiplicative group of positive reals,
then use this function: $f(A)=|\det A|$. </p>
<p>If $\mathbb{R^+}$ is the additive group of the all reals,
then use this function: $f(A)=\log |\det A|$. </p>
|
https://math.stackexchange.com/questions/1539892/example-of-a-homomorphism-from-gl-n-mathbbr-rightarrow-mathbbr-with
|
Question: <p>I am trying to prove (a) - (e) but am struggling with how to start.</p>
<p>$\bf{Question:}$ Let $H,K < G$ and consider the map $f: H \times K \rightarrow G$ given by $f(h,k) = hk$. The image of $f$ we will denote $HK$.</p>
<p>(a) Show that $f$ is injective iff $H \cap K = \{e\}$</p>
<p>(b) Suppose $H$ normalizes $K$. Show that $HK$ is a subgroup of $G$ and that $<H \cup K> = HK$. Further, show that $K \unlhd HK$.</p>
<p>(c) Let $HK$ be the semidirect product of $H,K$ and let $q: HK \rightarrow (HK)/K$ be the quotient map. Directly show that the restriction $q\restriction_H : H \rightarrow (HK)/K$ is an isomorphism.</p>
<p>(d) Show that $H,K$ commute: $hk = kh$ whenever $h \in H, k \in K$.</p>
<p>(e) Show that the map $f$ is an isomorphism onto its image (it's a bijection by part (a); you need to show it is a group homomorphism).</p>
<hr>
<p>All I have so far for (a) is assuming $f$ is injective, then $f(h,k) = f(h',k') \implies hk = h'k'\implies h = h' \& \ k = k'$.</p>
<p>For (b), if $H$ normalizes $K$ then $\forall h \in H, \forall k \in K$ we have $k^{-1}hk = h$.</p>
<p>Not sure where to go from here.</p>
Answer: <p>For a)</p>
<p>$\Rightarrow$: consider $h\in H\cap K$ and $k\in K$, then $hk=e(hk)$, that is $f(h,k)=f(e,hk)$. Using $f$ injective we conclude $h=e$</p>
<p>$\Leftarrow$: consider $h,h'\in H,$ $k,k' \in K$, such that $hk=h'k'$. From this we can show $hh'^{-1}=kk'^{-1}$. Since $hh'^{-1}\in H$ and $kk'^{-1}\in K$ we see $hh'^{-1}, kk'^{-1}\in H\cap K=\{e\}$. This implies $h=h', k=k'$</p>
<hr>
<p>For b) </p>
<p>First, a small note. Your notion of normalizer is wrong. $H$ normalizes $K$ iif for all $h\in H$, $hK=Kh$.</p>
<p>To see $HK$ is subgroup of $G$, let $g_1=h_1k_1,g_2=h_2k_2\in HK$. Then
\begin{equation}
g_1g_2^{-1}=(h_1k_1)(k_2^{-1}h_2^{-1})=h_1(k_1k_2^{-1})h_2^{-1}
\end{equation}Since $H$ normalizes $K$, there is $k_3\in K$ s.t. $(k_1k_2^{-1})h_2^{-1}=h_2^{-1}k_3$. Then
\begin{equation}
g_1g_2^{-1}=h_1(k_1k_2^{-1})h_2^{-1}=(h_1h_2^{-1})k_3 \in HK
\end{equation}
Let's prove $<H\cup K>=HK$. It's easy to see $H\cup K\subseteq HK$ (just note $h=he$). Now, if $G'$ is a subgroup of $G$ such that contains $H\cup K$ then, for any $h\in H, k\in K$ we have $hk\in G'$, that is, $HK\subseteq G'$.</p>
<p>Let's prove $K$ is a normal subgroup of $HK$. Let $h'\in H,k,k'\in K$. Since $H$ normalizes $K$, there is $k''\in K$ s.t. $h'(k'kk'^{-1})=k''h'$, then
\begin{equation}
(h'k')k(k'^{-1}h'^{-1})=(k'')h'h'^{-1}=k''\in K
\end{equation}</p>
<hr>
<p>For c)</p>
<p>To see $q|_H$ is a group homomorphism, let $g\in h_1h_2K$, then there is some $k\in K$ such that $g=h_1h_2k=(h_1e)(h_2k)\in (h_1K)(h_2K)$. Conversely, $g\in (h_1K)(h_2K)$, then there are $k_1, k_2\in K$ such that $g=(h_1k_1)(h_2k_2)=h_1(k_1h_2)k_2$. Since $K$ is a normal subgroup of $HK$, then $h_2 K=Kh_2$, which implies there is $k_3\in K$ such that $k_1h_2=h_2k_3$, so we conclude
\begin{equation}
g=h_1(h_2k_3)k_2 \in h_1h_2K
\end{equation}To see $q|_H$ injective, let $h_1,h_2\in H$ such that $h_1K=h_2K$. There is some $k\in K$ for which
\begin{equation}
h_1e=h_2k \Rightarrow h_1h_2^{-1}=k\in H\cap K=\{e\} \Rightarrow h_1=h_2
\end{equation}To see $q|_H$ surjective, let $aK$ in $HK/K$ with $a=hk\in HK$. We see $aK=hK$. Indeed, $z\in aK$, then $z=h(kk')$ for $k'\in K$, that is, $z\in hK$. Conversely, $z\in hK$, then there is $k''\in K$ such that $z=hk''=hkk^{-1}k''=ak^{-1}k''\in aK$</p>
<hr>
<p>Since (d) is not necessarily true, as pointed out in the comments, (e) is not necessarily true either.</p>
|
https://math.stackexchange.com/questions/1482008/understanding-semidirect-products-in-group-theory-through-exercises
|
Question: <blockquote>
<p>Prove that the composition of two group homomorphisms is a group homomorphism.</p>
</blockquote>
<p>Let $f:G \longrightarrow G'$ and $g:G' \longrightarrow G''$ be two group homomorphisms.</p>
<p>Let $x$ and $y$ be two arbitrary elements of $G$. Then,</p>
<p>\begin{eqnarray}
(g \circ f)(x \cdot y) &=& g(f(x \cdot y)) \\
&=& g(f(x) \cdot f(y)) \\
&=& g(f(x)) \cdot g(f(y)) \\
&=& (g \circ f)(x) \cdot (g \circ f)(y)
\end{eqnarray}</p>
<p>This completes the proof - but which property of a group is used in the second step?
\begin{eqnarray}
g(f(x \cdot y)) &=& g(f(x) \cdot f(y)) \\
\end{eqnarray}</p>
<p>Is it a cauchy function? Is there another property that lets a group operation $ \cdot $ be pushed outside of a function?</p>
Answer: <p>A <em>group homomorphism</em> is a map from a group $G$ to another group $G'$ that <em>preserves</em> the "group structure" (homomorphism are in general "structur-preserving maps"). This structure is the multiplication for multiplicative groups (addition for addivite groups). A homomorphism $f$ has then the defining property that for any $g,h \in G$:</p>
<p>$$f(g\cdot h) = f(g) \cdot' f(h)$$</p>
<p>where $\cdot$ denotes multiplication in $G$ and $\cdot'$ multiplication in $G'$.</p>
|
https://math.stackexchange.com/questions/2130210/proving-the-composition-of-two-group-homomorphisms-is-a-group-homomorphism
|
Question: <p>Let $f:G \rightarrow G'$ be a homomorphism and let $H$ be the kernel of $G$. Suppose $G$ is finite. Show ord$(G)=$ord$(f(G)) \cdot $ord$(H)$.</p>
<p>What I want to do is to construct a bijection, $\Phi$ from $G$\ $H$ (the factor group) to $f(G)$.</p>
<p>This should then tell me, I think, that ord$(G$\ $H)$ = ord$(f(G))$, which I can then use because ord$(G)=$ord$(G$\ $H)\cdot $ord$(H)$.</p>
<p>My problem is I'm not certain how to define the bijection. I think I want it to be something like $\Phi : aH \mapsto a$ or $\Phi : aH \mapsto f(a)$ such that $a \in G$. Which one should I choose and then when chosen should I just show it's well-defined and a bijection?</p>
Answer: <p>The elements of the quotient $G/H$ ($H$ is a normal subgroup) are cosets $gH$ for some $g\in H$. You can define the bijection this way:
\begin{align*}\bar f\colon G/H &\longrightarrow f(G)\\gH&\longmapsto f(g)
\end{align*}
This map is well-defined (i.e. the image does not depend on the representative in the class) since $ gH=f^{-1}\bigl(f(g)\bigr)$. It's not hard to check it is a group homomorphism, and it is bijective by construction.</p>
|
https://math.stackexchange.com/questions/1327197/show-that-the-order-of-g-order-fg-times-order-kerg
|
Question: <p>Suppose that G is a finite, nonabelian group with odd order. Show s is surjective, and hence bijective.</p>
<p>I have been told to look at the effects of the squaring map, $s\colon G\to G$, defined by $s(g)=g^2$ on the elements of cyclic groups $\langle g\rangle$ of $G$.</p>
<p>I'm stumped. Could anyone give me a nudge in the right direction or (being hopeful) a full solution?</p>
<p>Thanks a lot.</p>
Answer: <p>On each cyclic subgroup $C$ of $G$, the $s$ map is a homomorphism. Prove that $s$ is injective by considering $\ker s$ in $C$. It is here that you use that $G$ of odd order.</p>
|
https://math.stackexchange.com/questions/57913/suppose-that-g-is-a-finite-nonabelian-group-with-odd-order-show-s-is-surjectiv
|
Question: <p>I know that $|Z_{24}|=24=2^3. 3$, So we can use the equation $t=1+k\,p$ to find the number of sylow groups for each $p=2,3$. Therefore we have $1$ or $4$ sylow $3$-subgroups and $1$ or $3$ sylow $2$-subgroups.</p>
<p>Is $<(12)>$ the Sylow $2$-subgroup and $<(8)>$ the sylow $3$-subgroup?</p>
<p>Is that right?</p>
<p>Are there any others?</p>
<p>Thanks.</p>
Answer: <p>Since $\mathbb{Z}_{24}$ is abelian, all subgroups are normal; all Sylow $p$-subgroups for a fixed $p$ are conjugate. Combining these two facts, you can see that there will be a unique Sylow $p$-subgroup for each $p$. </p>
<p>Keep in mind that your Sylow $p$-subgroup should have order $p^{k}$, where $k$ is the largest integer such that $p^{k} \mid |G|$. Your Sylow $3$-subgroup is correct.</p>
|
https://math.stackexchange.com/questions/1744801/what-are-the-sylow-subgroups-of-z-24
|
Question: <p>I am working on a question that asks me to list all abelian groups of order $1188$.</p>
<p>I have a list that I will put below that I obtained using the classification of finite abelian groups.</p>
<p>My question is could someone verify whether my answer is correct and is in fact a list of all such groups.</p>
<p>Is there some kind of way to check answers like this online or do you have to manually compute them to verify?</p>
<p>Anyway here is the list:</p>
<blockquote>
<p>$$C_{1188}$$</p>
<p>$$C_{3} \times C_{396}$$</p>
<p>$$C_{2} \times C_{594}$$</p>
<p>$$C_{6} \times C_{198}$$</p>
<p>$$C_{3} \times C_{3} \times C_{132}$$</p>
<p>$$C_{3} \times C_{6} \times C_{66}$$</p>
</blockquote>
<p>Much obliged!</p>
Answer: <p>For any prime factorisation $p_1^{a_1}...p_n^{a_n}$ , the number of abelian groups of this order are $\pi(a1)*...*\pi (a_n)$ where pi denotes partitions...apply it here to be absolutely sure.1*2*3=6 so you can do this in general for any order</p>
|
https://math.stackexchange.com/questions/2293862/list-of-all-abelian-groups-of-order-1188
|
Question: <p>Are there any groups (not necessarily finite) that have a non-zero even number of elements of order two?</p>
<p>My attempt: since any finite group containing an element of order 2 must be of even order, hence the number of elements of order 2 must be odd, it suffices to find groups of infinite order.</p>
<p>I browsed through those familiar infinite groups but non of them seem to fit in the construction. How to construct such a group?</p>
Answer:
|
https://math.stackexchange.com/questions/4184824/groups-that-have-a-non-zero-even-number-of-elements-of-order-two
|
Question: <p>Let <span class="math-container">$G$</span> be a finitely generated group. For a finite subset <span class="math-container">$S \subset G$</span> which generates <span class="math-container">$G$</span>,</p>
<p>define <span class="math-container">$D_{G,S(n)}$</span> = {<span class="math-container">$s_1 s_2 ...s_r \vert s_i \in S \cup S^{-1} , r \le n $</span>}</p>
<p><span class="math-container">$l_{G,S} (k)$</span> = min {n <span class="math-container">$\vert$</span> <span class="math-container">$\# D_{G,S(n)}$</span> <span class="math-container">$ \geq k $</span>}</p>
<p>For any finite subsets <span class="math-container">$S,T\subset G$</span>generating <span class="math-container">$G$</span>,</p>
<p>show that <span class="math-container">$\limsup\limits_{k\rightarrow \infty}$</span> <span class="math-container">${l_{G,S} (k)}$</span> / <span class="math-container">${l_{G,T} (k)}$</span> <span class="math-container">$<$</span> <span class="math-container">$ \infty$</span></p>
<p><span class="math-container">$\liminf\limits_{k\rightarrow \infty}$</span> <span class="math-container">$l_{G,S} (k)$</span> / <span class="math-container">${l_{G,T} (k)}$</span> <span class="math-container">$>$</span> <span class="math-container">$0$</span></p>
<p>i cannot understand how to compute this. Can you give any hint or solution ?</p>
Answer:
|
https://math.stackexchange.com/questions/4189011/compute-limsup-limits-k-rightarrow-infty-l-g-s-k-l-g-t-k
|
Question: <p><strong>Do group operators that commute with fixed positive powers necessarily commute with all integer powers as well?</strong></p>
<p>Let's take a group with operators <span class="math-container">$(G, \Omega)$</span> and relax the requirements on each <span class="math-container">$\omega$</span> in <span class="math-container">$\Omega$</span> so that they are only required commute with raising group elements to powers, namely</p>
<p><span class="math-container">$$ (g^n)^\omega = (g^\omega)^n \; \text{for all $n \in \mathbb{Z}_{\ge 1}$} \tag{E1} $$</span></p>
<p>I'm calling this property <em>power associativity</em> in the title for brevity, and it does bear a resemblance to power associative binary operations, but I might be using the term wrong.</p>
<p>Raising group elements to a fixed power (<span class="math-container">$g \mapsto g^k$</span> for some fixed <span class="math-container">$k$</span> in <span class="math-container">$\mathbb{Z}_{\ge 1}$</span>) and conjugating by a fixed group element (<span class="math-container">$g \mapsto h^{-1}gh$</span> for some fixed <span class="math-container">$h$</span>) are both examples of a power associative operator. This is my motivation for this question, since raising group elements to fixed powers are not operators in general <a href="https://en.wikipedia.org/wiki/Group_with_operators" rel="nofollow noreferrer">under the usual definition</a>.</p>
<p>Anyway, I'm curious whether insisting on commuting with powers for positive powers in (E1) is enough to give me (G9) below. I think that commuting with inverses and all positive integers is enough to give me (G9).</p>
<p><span class="math-container">$$ (g^a)^\omega = (g^\omega)^a \;\; \text{for all $a \in \mathbb{Z}$} \tag{G9} $$</span></p>
Answer:
|
https://math.stackexchange.com/questions/4193994/groups-with-operators-omega-ni-omega-that-commute-with-positive-powers-doe
|
Question: <p>I understand that since $a^6$ belongs to $H$, then $a^{60}=e$. But I am not sure what are the possibilities for the order of $a$?</p>
Answer: <p>We have seen already that the order of <span class="math-container">$a$</span> must divide 60. In fact all divisors of 60 are possible orders of <span class="math-container">$a$</span>.</p>
<p>To see this let <span class="math-container">$G=\langle x \rangle$</span> be cyclic of order 60 and <span class="math-container">$H=\langle x^6\rangle$</span>. Then all powers <span class="math-container">$a$</span> of <span class="math-container">$x$</span> satisfy <span class="math-container">$a^6\in H$</span>. Also for any divisor <span class="math-container">$k$</span> of 60, <span class="math-container">$x^{60/k}$</span> has order <span class="math-container">$k$</span>.</p>
|
https://math.stackexchange.com/questions/2903641/suppose-that-h-is-a-subgroup-of-a-group-g-and-h-10-if-a-belongs-to-g
|
Question: <p>Let $E$ be a non empty set. How to prove that there exists $\star:E\times E\rightarrow E$ for which $(E,\star)$ is a group?</p>
Answer: <p>If $E$ is finite, you can just identify $E$ with $\Bbb Z/n\Bbb Z$, where $n=|E|$.</p>
<p>If $E$ is countably infinite, you can similarly identify it with $\Bbb Z$ (or $\Bbb Q$ for that matter).
More generally, for <em>any</em> infinite cardinality of $E$, you can consider the set of groups with carrier sets $\subseteq E$, ordered by inclusion. Using Zorn's lemma, there is a maximal such group $M$. Conclude that $|M|>|E\setminus M|$ (and hence $|M|=|E|$, as desired) because otherwise you could build $M\hookrightarrow M\oplus\Bbb Z/2\Bbb Z\subseteq E$, contradicting maximality. </p>
|
https://math.stackexchange.com/questions/2909462/non-empty-set-and-group
|
Question: <p>Is this true that <span class="math-container">$N_G(H) \subseteq N_G(H \cap K)$</span> if not what is the counter-example? I am confused in this question?</p>
<p>Clearly <span class="math-container">$g \in N_G(H)$</span> then <span class="math-container">$g(H \cap K)g^{-1}\subseteq H$</span></p>
Answer: <p>It's not true. Consider the 2-Sylow subgroup <span class="math-container">$V\trianglelefteq A_4$</span>. Let <span class="math-container">$G=A_4$</span>, <span class="math-container">$H=V$</span> and let <span class="math-container">$K$</span> be any 2 element subgroup of <span class="math-container">$V$</span>. Then <span class="math-container">$N_G(H)=G$</span> since <span class="math-container">$V\trianglelefteq G$</span>, but <span class="math-container">$N_G(H\cap K)=N_G(K)=V$</span>, which is smaller. (<span class="math-container">$G/V$</span> permutes those three 2 element subgroups cyclically.)</p>
|
https://math.stackexchange.com/questions/2929568/is-this-true-that-n-gh-subseteq-n-gh-cap-k-if-not-what-is-the-counter-ex
|
Question: <p><em>Let <span class="math-container">$G$</span> be a finite group and <span class="math-container">$J=\langle g_1,\cdots g_k\rangle$</span> be a sequence of group elements. For any <span class="math-container">$\delta \ge 1$</span>, <span class="math-container">$J$</span> is said to be a cube generating sequence for <span class="math-container">$G$</span> with closeness parameter <span class="math-container">$\delta$</span>, if the probability distribution <span class="math-container">$D_J$</span> on <span class="math-container">$G$</span> given by <span class="math-container">$g_1^{\epsilon_1}\cdots g_k^{\epsilon_k},$</span> where each <span class="math-container">$\epsilon_i$</span> is independent and uniformally distributed in {0,1}, is <span class="math-container">$\delta$</span>-close to the uniform distribution in the <span class="math-container">$L_2$</span> norm.</em></p>
<p><strong>Question:</strong> What is the difference between ordinary generating set and generating set given abobve?</p>
Answer: <p>The main difference is that any generating set will not give you something "close" to the uniform distribution. Consider the group <span class="math-container">$\mathbb{Z}^2$</span> with standard generators <span class="math-container">$a,b$</span>. Choose as a generating set <span class="math-container">$b,b^{-1},a^{-1}$</span> and <span class="math-container">$a,a^2,a^3,...,a^k$</span> for some big <span class="math-container">$k$</span>.</p>
<p>This generating set will not be "cube generating" according to your terminology. More precisely, for any given <span class="math-container">$\delta$</span>, you can choose <span class="math-container">$k$</span> so that the generating set is not <span class="math-container">$\delta$</span>-cube generating.</p>
<p>The reason is that you will have more chance to pick an element with a huge power of <span class="math-container">$a$</span> than any element in the group.</p>
<p>If you are familiar with free groups, this is even clearer in the free group generated by <span class="math-container">$a$</span> and <span class="math-container">$b$</span> with the same generating set.</p>
<p>Now, the second difference is that it is not clear for me that a "cube generating" sequence is actually a generating set, since you only require some closeness to the uniform distribution. It seems that a <span class="math-container">$\delta$</span>-cube generating sequence for <span class="math-container">$G$</span> is a <span class="math-container">$\delta'$</span>-cube generating sequence for <span class="math-container">$G\times F$</span>, where <span class="math-container">$F$</span> is a finite group. </p>
|
https://math.stackexchange.com/questions/2946994/difficulty-in-understanding-the-definition-of-this-sequence
|
Question: <p>I'm trying to see if I can find a bijection between two groups that are infinite of which one in the subset of the other. If I find the inverse <span class="math-container">$\phi^{-1}(x)=\frac{1}{5}x$</span> since it doesn't work for <span class="math-container">$x \in \mathbb{Z}$</span> (because I will have values in <span class="math-container">$\mathbb{Q}$</span>) then there isn't an isomorphism right?</p>
<p>Or have I approached the problem incorrectly?</p>
<p>Thanks for your help.</p>
Answer: <p>The subgroups of <span class="math-container">$\Bbb{Z}$</span> are given by <span class="math-container">$n\Bbb{Z}$</span>. For <span class="math-container">$n\neq 0$</span>, <span class="math-container">$n\Bbb{Z}$</span> is an infinite cyclic group with generator <span class="math-container">$n$</span>, and hence isomorphic to <span class="math-container">$\Bbb{Z}$</span>. You do not necessarily need to find an explicit isomorphism (assuming you had cyclic groups in class already). So in fact <span class="math-container">$5\Bbb{Z}\cong \Bbb{Z}$</span>, in contrast to your title.</p>
|
https://math.stackexchange.com/questions/2959686/proving-something-isnt-isomorphic-varphi-5-mathbbz-rightarrow-mathbbz
|
Question: <p>Let <span class="math-container">$G_1, G_2$</span> and <span class="math-container">$H$</span> be finite groups such that <span class="math-container">$|G_1|=|G_2|$</span>. Assume that <span class="math-container">$|L(G_1)|<|L(G_2)|$</span>. Is there a way to show that <span class="math-container">$|L(G_1\times H)|<|L(G_2\times H)|$</span>?</p>
<p>I am especially interested if this would hold if all 3 groups (<span class="math-container">$G_1$</span>, <span class="math-container">$G_2$</span> and <span class="math-container">$H$</span>) are finite abelian <span class="math-container">$p$</span>-groups. I know that the number of subgroups of a direct product of groups may be found via Goursat's Lemma. However, counting the number of subgroups in such a case seems difficult (even if we restrict the study to abelian <span class="math-container">$p$</span>-groups). Any suggestion is appreciated. </p>
Answer:
|
https://math.stackexchange.com/questions/2968181/inequality-involving-the-number-of-subgroups-of-a-direct-product-of-groups
|
Question: <p>Do we just use the subgroup criteria on this? Finding that it's closed, has inverse and identity within the subgroup?</p>
<p>But I still don't how does that prove the fact that's <span class="math-container">$H = \{2^n:𝑛 \in\mathbb{Z}\}$</span> is a subset of <span class="math-container">$\mathbb{Q}\setminus\{0\}$</span>, what if it's not a subset?</p>
Answer: <p>Since <span class="math-container">$2^n$</span> is a non-zero rational number for every <span class="math-container">$n \in \mathbb{Z}$</span>, indeed <span class="math-container">$H \subset \mathbb{Q} \setminus \left\{0\right\}$</span>. Then as you said, <span class="math-container">$H$</span> is a subgroup by the subgroup criterion (where the group operation on <span class="math-container">$H$</span> is multiplication).</p>
|
https://math.stackexchange.com/questions/2993916/show-that-the-set-h-2n-in-mathbbz-is-a-subgroup-of-for-mathbbq
|
Question: <p>Past year exam question:</p>
<p>Let <span class="math-container">$G$</span> be a group and <span class="math-container">$H$</span> a subgroup of <span class="math-container">$G$</span>. Suppose there exist some <span class="math-container">$g \in G$</span>, <span class="math-container">$g \notin H$</span>, such that <span class="math-container">$gH=Hg$</span>, and <span class="math-container">$[G:H]$</span> is a prime p. Prove that <span class="math-container">$H$</span> is normal.</p>
<p>Sketch:
We have <span class="math-container">$H, gH, g^2H, ... g^{p-1}H$</span> to be the cosets of <span class="math-container">$H$</span>. Then for any <span class="math-container">$x \in G$</span>, <span class="math-container">$x=g^kh$</span> for some <span class="math-container">$h \in H$</span> and <span class="math-container">$k$</span> integer. Then <span class="math-container">$x^{-1}=h^{-1}g^{-k}$</span>.</p>
<p>Then <span class="math-container">$xHx^{-1} = g^khHh^{-1}g^{-k} = ... = H$</span>. Hence <span class="math-container">$H$</span> is normal in <span class="math-container">$G$</span>.</p>
<p>I am not sure about my proof, because I don't know how to show that <span class="math-container">$g^kH$</span> can generate all the cosets.</p>
<p>Is there a way to show that this is the case? Or if not, how to go about proving this question?</p>
<p>Thanks for the help!</p>
Answer: <p>A very straighforward proof can be obtained by remarking the following:</p>
<ol>
<li>If <span class="math-container">$K \leqslant G$</span> is a subgroup such that the index <span class="math-container">$(G \colon K)=p$</span> is prime, then <span class="math-container">$K$</span> is a <em>maximal</em> subgroup of <span class="math-container">$G$</span>. Indeed, given the situation <span class="math-container">$K \leqslant E \leqslant G$</span>, we have <span class="math-container">$(G\colon K)=(G \colon E)(E \colon K)$</span>, which means that <span class="math-container">$(E\colon K)$</span> and <span class="math-container">$(G\colon E)$</span> are both natural and furthermore divide <span class="math-container">$p$</span>. Therefore, we either have <span class="math-container">$(E \colon K)=1$</span> in which case <span class="math-container">$E=K$</span> or <span class="math-container">$(E\colon K)=p$</span> in which case we derive <span class="math-container">$(G \colon E)=1$</span> and hence <span class="math-container">$G=E$</span>.</li>
<li>In the particular case of your problem, the normaliser <span class="math-container">$H <\mathrm{N}_G(H)$</span> contains an element in <span class="math-container">$G \setminus H$</span> by hypothesis and is thus compelled to be the improper subgroup -- in other words <span class="math-container">$\mathrm{N}_G(H)=G$</span> -- by the above observation. This means that <span class="math-container">$H \trianglelefteq G$</span> is normal.</li>
</ol>
|
https://math.stackexchange.com/questions/2999699/let-h-a-subgroup-of-g-exists-g-in-g-g-notin-h-such-that-gh-hg-a
|
Question: <p>Let <span class="math-container">$G$</span> be a finite group and <span class="math-container">$K$</span> a normal subgroup of <span class="math-container">$G$</span>. Suppose <span class="math-container">$|K|^2 \nmid |G|$</span> and <span class="math-container">$K$</span> is simple, Prove <span class="math-container">$G$</span> does not have any subgroup isomorphic to <span class="math-container">$K$</span> except <span class="math-container">$K$</span>. Show <span class="math-container">$K$</span> is a characteristic subgroup.</p>
Answer: <p>Hint: if <span class="math-container">$H$</span> and <span class="math-container">$K$</span> are normal subgroups then <span class="math-container">$HK$</span> is a normal subgroup and <span class="math-container">$|HK|=\frac{|H| \cdot |K|}{|H \cap K|}$</span>.</p>
|
https://math.stackexchange.com/questions/3004107/let-g-be-a-finite-group-and-k-a-normal-subgroup-of-g-suppose-k2-nmid
|
Question: <p>“Prove that the additive group of real numbers doesn’t have any proper subgroup with finite index.”
I want to know how to prove this.</p>
Answer: <p>Assume that <span class="math-container">$G\leq\mathbb{R}$</span> is a subgroup with <span class="math-container">$[\mathbb{R}:G]=n\in\mathbb{Z}_{>0}$</span>. Then for any <span class="math-container">$x\in\mathbb{R}$</span> we have <span class="math-container">$n\cdot x\in G$</span> (why?). In particular, for any <span class="math-container">$x\in\mathbb{R}$</span>, we have <span class="math-container">$x = n\cdot(\frac{x}{n})\in G$</span> so <span class="math-container">$\mathbb{R}=G$</span>.</p>
|
https://math.stackexchange.com/questions/3005921/proof-of-group-theory
|
Question: <p>Consider a group and its subgroup:
<span class="math-container">$$\,K\,<\,G\,\;.$$</span>
An element <span class="math-container">$g\in G$</span> generates a bijection of <span class="math-container">$\,G\,$</span> onto itself,
<span class="math-container">\begin{eqnarray}
\hat{\cal{L}}_g\;\colon\quad G\longrightarrow G\;,\qquad
x\;\longrightarrow\;\hat{
\cal
{L}}_g\,x\;=\;g\;x\;\,,
\end{eqnarray}</span>
and a bijection of the coset space onto itself,
<span class="math-container">\begin{eqnarray}
\hat{L}_g\;\colon\quad G/K\longrightarrow G/K\;,\qquad
x\,K\;\longrightarrow\;\hat{L}_g\,(x\,K)\;=\;g\;x\;K\;\,.
\end{eqnarray}</span></p>
<p>The set of operators <span class="math-container">$\,{\mathtt{L}}\,=\,\{\,\hat{\mathit{L}}_g\;\big{|}\;g\in G\,\}\,$</span> can be interpreted as a map
<span class="math-container">\begin{eqnarray}
{\mathtt{L}}\,:\quad G \longrightarrow \operatorname{Aut_{set}}(G/K)\,\;,
\end{eqnarray}</span>
where <span class="math-container">$\,\operatorname{Aut_{set}}(G/K)\,$</span> denotes the set of all bijections of <span class="math-container">$\,G/K\,$</span> onto itself.</p>
<p>Is this map into or onto?</p>
<p>In other words, can any bijection of <span class="math-container">$\,G/K\,$</span> onto itself be written as <span class="math-container">$\,L_g\,$</span>, with <span class="math-container">$\,g\in G\,$</span>?</p>
Answer:
|
https://math.stackexchange.com/questions/3008670/a-simple-question-on-groups-and-quotient-spaces
|
Question: <p>Is there a compact matrix group <span class="math-container">$G\subseteq GL(n,\mathbb{R})$</span> such that <span class="math-container">$|G|$</span> is countable infinite?</p>
Answer: <p>So the answer is negative. In general there is no countably infinite and compact Hausdorff group. See this:</p>
<p><a href="https://mathoverflow.net/questions/4939/is-there-a-compact-group-of-countably-infinite-cardinality">https://mathoverflow.net/questions/4939/is-there-a-compact-group-of-countably-infinite-cardinality</a></p>
<p>To add a bit to the answer (so that it isn't completely naked ;) ): every compact subgroup of <span class="math-container">$GL_n(\mathbb{R})$</span> is obviously contained in the preimage <span class="math-container">$\det^{-1}(\{-1,1\})$</span> which is the largest compact subgroup of <span class="math-container">$\mathbb{R}^*$</span>. But we can do better. The orthogonal group <span class="math-container">$O(n)$</span> is a maximal subgroup of <span class="math-container">$GL_n(\mathbb{R})$</span> among compact subgroups (see <a href="https://math.stackexchange.com/questions/202850/prove-that-on-is-a-maximal-compact-subgroup-of-gln-mathbb-r">this</a>). Therefore every compact subgroup of <span class="math-container">$GL_n(\mathbb{R})$</span> is a subconjugate of <span class="math-container">$O(n)$</span> which is a consequence of the Cartan-Iwasawa-Malcev theorem.</p>
<p>In particular if you wish to analyze compact subgroups of <span class="math-container">$GL_n(\mathbb{R})$</span> then it is enough to look at subgroups of <span class="math-container">$O(n)$</span>.</p>
|
https://math.stackexchange.com/questions/3023788/is-there-such-a-group
|
Question: <blockquote>
<p>Suppose centraliser of group of element has order 4 . Then what information can we deduct about center </p>
</blockquote>
<p>I know that <span class="math-container">$Z(G)\subset Z(x)$</span> where <span class="math-container">$|Z(x)|=4$</span></p>
<p>Any group of order 4 is abelian <span class="math-container">$|Z(G)|=1,2,4$</span>.</p>
<p>Now I got some examples of with above property </p>
<p>Like In <span class="math-container">$D_4$</span> or <span class="math-container">$Q_8$</span> <span class="math-container">$|Z(G)|=2$</span></p>
<p>In <span class="math-container">$Z_4$</span> ,<span class="math-container">$|Z(G)|=4$</span></p>
<p>But I could not able to find example with <span class="math-container">$|Z(G)|=1|
$</span> and <span class="math-container">$|Z(x)=4|$</span></p>
<p>What else info can we deduce?</p>
<p>Any Help will be appreciated</p>
<p>Edit: </p>
<p>As Derek says <span class="math-container">$A_4$</span> has <span class="math-container">$Z(G)=1$</span> there is (12)(43) which has 4 centraliser element</p>
Answer:
|
https://math.stackexchange.com/questions/3028409/how-to-deduce-information-about-center-of-group-if-only-order-of-centraliser-is
|
Question: <blockquote>
<p>Let <span class="math-container">$G$</span> a group and let <span class="math-container">$A,B$</span> subgroups of <span class="math-container">$G$</span>. Let a map <span class="math-container">$f:A/A\cap B\to G/B$</span> by <span class="math-container">$(A\cap B)a\mapsto Ba$</span>. Prove that <span class="math-container">$f$</span> is well defined (not depends on representitives).</p>
</blockquote>
<p>I know that I need to prove that if <span class="math-container">$(A\cap B)a=(A\cap B)a'$</span> then <span class="math-container">$Ba=Ba'$</span>. I don't understand why I can't argue that if <span class="math-container">$(A\cap B)a=(A\cap B)a'$</span> then <span class="math-container">$f((A\cap B)a)=f((A\cap B)a')$</span> and thus <span class="math-container">$Ba=Ba'$</span>.</p>
<p>I have another solution that I don't understand why is true and the previous one doesn't:</p>
<p>Let <span class="math-container">$(A\cap B)a=(A\cap B)a'$</span>. Then <span class="math-container">$a-a'\in A\cap B$</span>. Thus <span class="math-container">$a-a'\in B$</span>, thus <span class="math-container">$aB=a'B$</span>.</p>
Answer: <p>As pointed out in the comments, one can't argue that</p>
<p><span class="math-container">$(A \cap B)a_1 = (A \cap B)a_2 \Longrightarrow Ba_1 = f((A \cap B)a_1) = f((A \cap B)a_2) = Ba_2, \tag 1$</span> </p>
<p>since <span class="math-container">$f$</span> is defined in terms of whatever <span class="math-container">$b$</span> occurs in the <em>exact</em> expression <span class="math-container">$A \cap B)b$</span>; such a definition, on the face of it and without further analysis, yields a <span class="math-container">$b$</span>-dependent <span class="math-container">$f$</span>, not a <em>coset</em>-dependent <span class="math-container">$f$</span>; one needs to show that if, as <em>sets</em>,</p>
<p><span class="math-container">$(A \cap B)a_1 = (A \cap B)a_2, \tag 2$</span></p>
<p>then</p>
<p><span class="math-container">$Ba_1 = Ba_2, \tag 3$</span></p>
<p>again as <em>sets</em>; that is, <span class="math-container">$f$</span> depends only on the the coset <span class="math-container">$(A \cap B)b$</span>, not on it's particular represetative <span class="math-container">$b$</span>.</p>
<p>To properly prove the desired assertion, it helps to realize that, for a
any subgroup <span class="math-container">$C$</span> of any group <span class="math-container">$G$</span>, </p>
<p><span class="math-container">$Cg_1 = Cg_2, \; g_1, g_2 \in G \Longleftrightarrow g_2 g_1^{-1} \in C; \tag 4$</span></p>
<p>for if</p>
<p><span class="math-container">$Cg_1 = Cg_2, \tag 5$</span></p>
<p>and <span class="math-container">$e \in G$</span> denotes the identity element, then since <span class="math-container">$e \in C$</span> (because <span class="math-container">$C$</span> is a subgroup of <span class="math-container">$G$</span>), </p>
<p><span class="math-container">$\exists c \in C, \; cg_1 = eg_2 = g_2 \Longrightarrow g_2g_1^{-1} = c \in C; \tag 6$</span></p>
<p>likewise,</p>
<p><span class="math-container">$g_2g_1^{-1} \in C \Longrightarrow g_2g_1^{-1} = c \in C \Longrightarrow g_2 = cg_1 \Longrightarrow Cg_2 = C(cg_1) = (Cc)g_1 = Cg_1, \tag 7$</span></p>
<p>since</p>
<p><span class="math-container">$c \in C \Longrightarrow Cc = C. \tag 8$</span></p>
<p>If this notion is applied to the present problem, we may argue that</p>
<p><span class="math-container">$(A \cap B)a_1 = (A \cap B)a_2 \Longrightarrow a_2a_1^{-1} \in A \cap B \Longrightarrow a_2 a_1^{-1} \in B \Longrightarrow Ba_1 = Ba_2, \tag 8$</span></p>
<p>and we see the mapping</p>
<p><span class="math-container">$(A \cap B)b \mapsto Bb \tag 9$</span></p>
<p>depends only on the coset <span class="math-container">$(A \cap B)b$</span> and not its particular representative <span class="math-container">$b$</span>; thus <span class="math-container">$f$</span> is, in fact, well-defined.</p>
|
https://math.stackexchange.com/questions/3077227/prove-that-a-map-is-defined-properly
|
Question: <p><span class="math-container">$G$</span> is a Group of order <span class="math-container">$pq$</span>, if <span class="math-container">$G$</span> has exactly one subgroup of order <span class="math-container">$p$</span> and another with order <span class="math-container">$q$</span>, then <span class="math-container">$G$</span> is cyclic (<span class="math-container">$p,q$</span> are prime)</p>
<p>Here's what I have so far, I'm not sure if it is correct.</p>
<p>Let <span class="math-container">$H, K$</span> be subgroups of <span class="math-container">$G$</span><br/>
<span class="math-container">$|H| = p$</span> <br/>
<span class="math-container">$|K| = q$</span> <br/></p>
<p>We know that by Lagrange's Theorem: <br/>
<span class="math-container">$\forall a \in H : |a|=p$</span> where <span class="math-container">$a \neq e$</span> <span class="math-container">$\Rightarrow H = <a>$</span> <br/>
Similarly: <span class="math-container">$\forall a \in K : K = <a>$</span> <br/>
We also know that: <span class="math-container">$\forall a \in G : |a| | pq \Rightarrow |a| = p$</span> or <span class="math-container">$|a| = q$</span> <br/></p>
<p><span class="math-container">$|a| = p \Rightarrow a \in H$</span> since there is exactly one subgroup of order <span class="math-container">$p$</span> <br/>
<span class="math-container">$|a| = q \Rightarrow a \in K$</span> for similar reasons</p>
<p>This means every element in <span class="math-container">$G$</span> is either in <span class="math-container">$H$</span> or <span class="math-container">$K$</span>, which are both cyclic. That means <span class="math-container">$G$</span> is cyclic.</p>
<p>What am I missing here?<br/>
Have I made any mistakes and/or is the proof incomplete?</p>
Answer: <p>Possible orders of a non-trivial element of <span class="math-container">$G $</span> are <span class="math-container">$p $</span>, <span class="math-container">$q $</span> or <span class="math-container">$pq $</span>. If we show the existence of an element of order <span class="math-container">$pq$</span>, then we are done. </p>
<p>If <span class="math-container">$g\in G $</span> is of order <span class="math-container">$p $</span>, then it must be in the subgroup <span class="math-container">$H $</span>(why?). Similarly if <span class="math-container">$g $</span> has order <span class="math-container">$q $</span>, it must be in <span class="math-container">$K $</span>. Now the number of elements of <span class="math-container">$G $</span> which are neither in <span class="math-container">$H $</span> nor in <span class="math-container">$K $</span> is equal to <span class="math-container">$$\begin{align} pq-(p+q-1) &=p (q-1)-(q-1) \\
&=(p-1)(q-1)\\
&\gt 0.
\end{align}$$</span> </p>
<p>So what can you conclude now? </p>
|
https://math.stackexchange.com/questions/3141832/g-is-a-group-of-order-pq-if-g-has-exactly-one-subgroup-of-order-p-and-a
|
Question: <p>Let <span class="math-container">$M$</span> be a non prime number and <span class="math-container">$G$</span> be the set of non-zero integers modulo <span class="math-container">$M$</span>, under multiplication modulo <span class="math-container">$M$</span>.
Show this is not a group.</p>
<p>My attempt:
Since <span class="math-container">$M$</span> is non prime so there exists integers <span class="math-container">$r,s>1$</span> such that <span class="math-container">$rs=M$</span>,
that is <span class="math-container">$rs=0\mod M$</span>, so there exists integers in <span class="math-container">$G$</span> where closure doesn't hold. Is this correct?</p>
<p>Also can a left coset equal right coset if group is not abelian?
Thanks in advance.</p>
Answer: <p>There is more than one definition of a group, which I guess is where the confusion in the comments lie. (A similar confusion came up here before, in the comments to <a href="https://math.stackexchange.com/q/2654537/10513">this question</a>.) The first definition I learned as an undergrad, which I guess is the one the OP is working with, is roughly as follows (this is an amended version of the Wikipedia definition):</p>
<p>A <em>group</em> is a set, <span class="math-container">$G$</span>, together with an operation <span class="math-container">$\circ$</span> which may be applied to pairs of elements in <span class="math-container">$G$</span>. To qualify as a group, the set and operation, <span class="math-container">$(G, \circ)$</span>, must satisfy four requirements known as the group axioms:</p>
<ul>
<li><strong>Closure.</strong> For all <span class="math-container">$a, b\in G$</span> their product <span class="math-container">$a \circ b$</span> is an element of <span class="math-container">$G$</span>.</li>
<li><strong>Associativity.</strong> For all <span class="math-container">$a, b, c \in G$</span> we have <span class="math-container">$(a \circ b) \circ c = a \circ (b \circ c)$</span>.</li>
<li><strong>Identity element.</strong> There exists an element <span class="math-container">$e \in G$</span> such that, for every element <span class="math-container">$a \in G$</span>, the equation <span class="math-container">$e \circ a = a \circ e = a$</span> holds.</li>
<li><strong>Inverse element.</strong> For each <span class="math-container">$a \in G$</span>, there exists an element <span class="math-container">$b \in G$</span> such that <span class="math-container">$a \circ b = b \circ a = e$</span>, where <span class="math-container">$e$</span> is an (the) identity element.</li>
</ul>
<p>As you point out, <span class="math-container">$G$</span> contains two (equivalence classes) of integers such that their product is not in <span class="math-container">$G$</span>. For example, if <span class="math-container">$m=6$</span> then <span class="math-container">$G=\{1,2,3,4,5\}$</span> but <span class="math-container">$2\times3=6\not\in G$</span>. Hence, <span class="math-container">$G$</span> is not a group as it fails the "closure" axiom.</p>
<p>(In conclusion: your proof is fine.)</p>
<hr>
<p>You also ask:</p>
<blockquote>
<p>Can a left coset equal right coset if group is not abelian?</p>
</blockquote>
<p>Yes. A subgroup <span class="math-container">$H$</span> is <em><a href="https://en.wikipedia.org/wiki/Normal_subgroup" rel="nofollow noreferrer">normal</a></em> precisely when its left and right cosets are equal. Non-abelian groups can have normal subgroups. For example, <span class="math-container">$S_3$</span> is non-abelian, and if <span class="math-container">$H=\{1, (123)(132)\}$</span> then the left cosets are <span class="math-container">$H$</span> and <span class="math-container">$(12)H$</span>, while the right cosets are <span class="math-container">$H$</span> and <span class="math-container">$H(12)$</span>.</p>
<p>In the above example, <span class="math-container">$H$</span> is abelian but <span class="math-container">$G=S_3$</span> is not. There are other examples where both <span class="math-container">$H$</span> and <span class="math-container">$G$</span> are non-abelian. For example, if you have come across the alternating group <span class="math-container">$A_n$</span>, then <span class="math-container">$A_n$</span> is a normal subgroup of <span class="math-container">$S_n$</span> (so left cosets=right cosets), and <span class="math-container">$A_n$</span> is non-abelian when <span class="math-container">$n>3$</span>. (In the above example we actually have <span class="math-container">$H=A_3$</span>, and this is cyclic so abelian.)</p>
|
https://math.stackexchange.com/questions/3145241/modulo-groups-and-non-prime-numbers
|
Question: <p>It's easy to show that <span class="math-container">$\mathbb{Q}$</span> and <span class="math-container">$\mathbb{Q}\times\mathbb{Q}$</span> are not ring isomorphic as the first one has no zero divisors where as the second one has zero divisors. But I can't find any solution in case of group isomorphism.</p>
Answer: <p>In <span class="math-container">$\mathbb{Q}$</span>, as a group under addition, any two elements are commensurable; that is, for any <span class="math-container">$p,q$</span>, there is some <span class="math-container">$r$</span> such that both <span class="math-container">$p$</span> and <span class="math-container">$q$</span> are multiples of <span class="math-container">$r$</span>. Consequently, every finitely generated subgroup is cyclic.</p>
<p>The same is not true of <span class="math-container">$\mathbb{Q}\times\mathbb{Q}$</span>; <span class="math-container">$(1,0)$</span> and <span class="math-container">$(0,1)$</span> are incommensurable.</p>
|
https://math.stackexchange.com/questions/3151523/how-do-i-show-that-mathbbq-and-mathbbq-times-mathbbq-are-not-group-i
|
Question: <p>I know that if <span class="math-container">$z\in Z(G)$</span>, the centre of group <span class="math-container">$G$</span> then it is true that <span class="math-container">$cl(z)=\{z\}$</span> where <span class="math-container">$cl(g)$</span> is the conjugacy class that contains element <span class="math-container">$g\in G$</span>. </p>
<p>But what if <span class="math-container">$cl(g)=\{g\}$</span> can we imply that <span class="math-container">$g\in Z(G)?$</span></p>
<p>My attempt:
<span class="math-container">$$cl(g)=\{h\in G|\exists k\in G\text{ such that }h=k^{-1}gk\}=\{g\}$$</span>
so there exists a <span class="math-container">$k$</span> in <span class="math-container">$G$</span> such that <span class="math-container">$g=k^{-1}gk$</span>, but might not be the case for all <span class="math-container">$k$</span> in <span class="math-container">$G$</span> hence the above statement is false.</p>
<p>Thanks.</p>
Answer: <p>We have <span class="math-container">$\operatorname{cl}(g)=\{g\}$</span>, so if <span class="math-container">$h\in G$</span>, then <span class="math-container">$h^{-1}gh\in \operatorname{cl}(g)$</span> implies <span class="math-container">$h^{-1}gh=g$</span>, <em>i.e.</em>, <span class="math-container">$$gh=hg.$$</span> But <span class="math-container">$h$</span> was arbitrary in <span class="math-container">$G$</span>. Hence <span class="math-container">$g\in Z(G)$</span>.</p>
|
https://math.stackexchange.com/questions/3151972/if-clg-g-can-we-imply-that-g-in-zg
|
Question: <blockquote>
<p><span class="math-container">$\phi \in \operatorname{Aut}(\Bbb Z_{50})$</span> via <span class="math-container">$\phi(11) = 3$</span> Then <span class="math-container">$\phi(x) = $</span>? For any <span class="math-container">$x \in \Bbb Z_{50}$</span></p>
</blockquote>
<p>The answer is <span class="math-container">$23x$</span> but I'm not quite sure how to figure that out. Here's what I did:</p>
<p><span class="math-container">$\phi(11) = 3 \Rightarrow$</span> The function maps <span class="math-container">$11x \rightarrow 3x$</span> <br/>
Adding <span class="math-container">$(11)^{-1} = 39$</span> (mod 50) on both sides gives us <br/>
<span class="math-container">$x \rightarrow 42x$</span></p>
<p>What am I not doing correct?</p>
Answer: <p>We have
<span class="math-container">$$1\equiv -9\cdot11\pmod{50}\ .$$</span>
Therefore,
<span class="math-container">$$\phi(x)\equiv\phi(-9x\cdot11)\equiv-9x\cdot\phi(11)\equiv-27x\equiv23x\pmod{50}\ .$$</span>
In the second equivalence we used that <span class="math-container">$\phi(nx) = \phi(x)+\cdots+\phi(x) = n\phi(x)$</span>.</p>
<hr>
<p>Adding <span class="math-container">$39\equiv-11\pmod{50}$</span> to both sides just tells you that
<span class="math-container">$$\phi(11x)-11\equiv3x+39\ ,$$</span>
which is no help at all.</p>
|
https://math.stackexchange.com/questions/3153385/phi-in-operatornameaut-bbb-z-50-via-phi11-3-then-phix
|
Question: <blockquote>
<p>Let <span class="math-container">$G$</span> be a group. <span class="math-container">$|G|=21$</span> and <span class="math-container">$|Z(G)| \neq 1$</span> <span class="math-container">$\Rightarrow$</span> <span class="math-container">$|Z(G)| = $</span> ?</p>
</blockquote>
<p>We know that <span class="math-container">$Z(G)$</span> is a subgroup of <span class="math-container">$G$</span>. <br/>
By Lagrange's Theorem <span class="math-container">$|Z(G)|$</span> divides <span class="math-container">$|G|$</span> <br/>
<span class="math-container">$\Rightarrow |Z(G)| = 3,7,21$</span></p>
<p>What else do we have to do here to show that <span class="math-container">$|Z(G)| = 21$</span>? </p>
Answer: <blockquote>
<p><strong>Lemma</strong>: Let <span class="math-container">$G$</span> be a group, and denote <span class="math-container">$Z(G)$</span> its center. If <span class="math-container">$G/Z(G)$</span> is cyclic then <span class="math-container">$G$</span> is abelian.</p>
</blockquote>
<p>Going back to the problem we have that <span class="math-container">$G/Z(G)=1 ~\text{or} ~3 ~\text{or}~ 7$</span>. In each case since the order of <span class="math-container">$G/Z(G)$</span> is prime, we can conclude that <span class="math-container">$G/Z(G)$</span> is cyclic. By the Lemma <span class="math-container">$G$</span> is abelian hence <span class="math-container">$Z(G)=G$</span>.</p>
<p><strong>Proof of the Lemma:</strong> Suppose that <span class="math-container">$G/Z(G)= <a>$</span> for some <span class="math-container">$a\in G$</span>. Then, take <span class="math-container">$c,d \in G$</span>.</p>
<p>From the hypothesis we have,
<span class="math-container">$c=a^{k_1} z_1$</span> and <span class="math-container">$d=a^{k_2}z_2$</span> where <span class="math-container">$z_i\in Z(G)$</span> and <span class="math-container">$k_i\in \mathbb{N}$</span>. So,</p>
<p><span class="math-container">\begin{align}
cd&=a^{k_1}z_1 a^{k_2} z_2 \\
&=a^{k_1}a^{k_2}z_1 z_2\\
&=a^{k_2}a^{k_1}z_2 z_1\\
&=a^{k_2}z_2a^{k_1} z_1\\
&=dc
\end{align}</span></p>
|
https://math.stackexchange.com/questions/3153429/let-g-be-a-group-g-21-and-zg-neq-1-rightarrow-zg
|
Question: <blockquote>
<p>Let <span class="math-container">$G$</span> be a group with order 12. Which of the following claims are false?</p>
</blockquote>
<p>Does Lagrange's Theorem imply that there could exist a subgroup with order 6? I'm not sure where to begin. The question was taken from a past test with multiple choice answers, and the question asked which of the following claims were false: <br/></p>
<p>• G must have an element of order 2. <br/>
• G must have a subgroup of order 6. <br/>
• G must have a subgroup of order 2. <br/>
• None of the above, they are all true.</p>
Answer: <p>The first option is true because of Cauchy's theorem. This will also imply the existence of subgroup of order <span class="math-container">$2$</span>(consider the subgroup generated by element of order <span class="math-container">$2$</span>). Second option is false because <span class="math-container">$A_4$</span> is a group of order <span class="math-container">$12$</span> which has <a href="https://math.stackexchange.com/questions/582658/a-4-has-no-subgroup-of-order-6">no subgroup of order <span class="math-container">$6$</span></a>.</p>
|
https://math.stackexchange.com/questions/3153581/let-g-be-a-group-with-order-12-which-of-the-following-claims-are-false
|
Question: <p>Let
<span class="math-container">\begin{align*}
G=\langle s, t \mid s^4=1, s^2 = t^3 \rangle
\end{align*}</span>
and
<span class="math-container">\begin{align*}
G'=\langle S, T \mid S^2=1, T^3=1 \rangle
\end{align*}</span>
be two groups. Are these groups isomorphic to each other? Thank you very much.</p>
Answer: <p>No because <span class="math-container">$G$</span> has an element of order <span class="math-container">$4$</span> whereas <span class="math-container">$G'$</span> has no such element. </p>
|
https://math.stackexchange.com/questions/3161957/are-the-two-groups-g-g-isomorphic
|
Question: <blockquote>
<p><span class="math-container">$H$</span> is a subgroup of <span class="math-container">$G$</span>, prove <span class="math-container">$|gHg^{-1}| = |H|$</span>, <span class="math-container">$\forall g \in G$</span></p>
</blockquote>
<p>Here's what I know:</p>
<p>If <span class="math-container">$H$</span> is a normal subgroup of <span class="math-container">$G$</span> then the statement is trivial.</p>
<p>If <span class="math-container">$H$</span> is not a normal subgroup of <span class="math-container">$G$</span> then we know <span class="math-container">$|H| | |G|$</span> by Lagrange's Theorem (although I'm not sure if this is helpful).</p>
<p>Where else can I go from here?</p>
<p>Is my approach correct?</p>
Answer: <p>The map <span class="math-container">$$\begin{align}\phi_g: H&\to gHg^{-1}\\ h&\mapsto ghg^{-1}\end{align}$$</span> is a bijection with inverse <span class="math-container">$$\begin{align}\phi_g^{-1}: gHg^{-1}&\to H\\ \eta&\mapsto g^{-1}\eta g.\end{align}$$</span> It is called <em>conjugation by <span class="math-container">$g$</span></em>.</p>
|
https://math.stackexchange.com/questions/3194039/h-is-a-subgroup-of-g-prove-ghg-1-h-forall-g-in-g
|
Question: <blockquote>
<p><span class="math-container">$H$</span> is a subgroup of <span class="math-container">$G$</span>, <span class="math-container">$N$</span> is a normal subgroup of <span class="math-container">$G$</span>, <span class="math-container">$\gcd(|H|, |G/N|) = 1 \Rightarrow$</span> <span class="math-container">$H$</span> is a subgroup of <span class="math-container">$N$</span></p>
</blockquote>
<p>I'll attempt to prove the contrapositive: <br/>
<span class="math-container">$H$</span> isn't a subgroup of <span class="math-container">$N \Rightarrow \gcd(|H|, |G/N|) \neq 1$</span></p>
<p>We know that <span class="math-container">$|H|$</span> doesn't divide <span class="math-container">$|N|$</span> by Lagrange's Theorem. And we know that <span class="math-container">$|G/N| = \frac{|G|}{|N|} \Rightarrow |N| | |G|$</span></p>
<p>How do we show the <span class="math-container">$\gcd$</span> doesn't equal <span class="math-container">$1$</span> from this information? Do we assume the <span class="math-container">$\gcd$</span> is <span class="math-container">$1$</span>? Or is there something more direct?</p>
Answer: <p>Consider the quotient map <span class="math-container">$p:G\rightarrow G/N$</span>, <span class="math-container">$p(H)$</span> is a subgroup of <span class="math-container">$G/N$</span> its cardinal divides the cardinal of <span class="math-container">$G/N$</span> (Lagrange) and the cardinal of <span class="math-container">$H$</span>, since <span class="math-container">$|H|=|p(H)||Ker(p_{\mid H})|$</span> so it divides <span class="math-container">$gcd(|H|,|G/N|)$</span> which is <span class="math-container">$1$</span>. We deduce that <span class="math-container">$p(H)=1$</span> and <span class="math-container">$H\subset N$</span>.</p>
|
https://math.stackexchange.com/questions/3195123/h-is-a-subgroup-of-g-n-is-a-normal-subgroup-of-g-gcdh-g-n-1
|
Question: <p>I want to use this and the fact that <span class="math-container">$H_n \cap H_m = \{1\}$</span> which I've already proved to show that <span class="math-container">$G \cong H_n \times H_m$</span></p>
<p>Since this are subgroups, and are the only subgroups like this, <span class="math-container">$H_n$</span> and <span class="math-container">$H_m$</span> are normal. Then <span class="math-container">$H_n H_m \triangleleft G$</span>. Then, I show that <span class="math-container">$|H_n H_m| \geq nm$</span> by saying the following:</p>
<p>I have <span class="math-container">$n$</span> possible elections of <span class="math-container">$h_n \in H_n$</span> and <span class="math-container">$m$</span> possible elections of <span class="math-container">$h_m \in H_m$</span>. Then if <span class="math-container">$|H_n H_m| < nm$</span>, then it should be some <span class="math-container">$h_{n_1}, h_{n_2} \in H_n$</span> and <span class="math-container">$h_{m_1}, h_{m_2} \in H_m$</span> such that <span class="math-container">$h_{n_1}h_{m_1} = h_{n_2}h_{m_2}$</span>, but this cannot be because <span class="math-container">$H_n \cap H_m = \{1\}$</span>. Then <span class="math-container">$|H_nH_m| \geq nm$</span>. Trivially, <span class="math-container">$|H_nH_m| \leq nm$</span>, then <span class="math-container">$|H_nH_m|=nm$</span> so <span class="math-container">$G = H_nH_m$</span>. </p>
<p>Is this correct?</p>
Answer: <p>Your argument is correct but can be written more clearly as follows.</p>
<p>Consider the function <span class="math-container">$\phi: H_n \times H_m \to G$</span> given by <span class="math-container">$(u,v) \mapsto uv$</span>.</p>
<p>Then <span class="math-container">$\phi$</span> is injective because <span class="math-container">$H_n \cap H_m = \{1\}$</span>. Indeed, <span class="math-container">$u_1 v_1 = u_2 v_2$</span> implies <span class="math-container">$u_2^{-1} u_1 = v_2 v_1^{-1} \in H_n \cap H_m = \{1\}$</span>.</p>
<p>Therefore, <span class="math-container">$\phi$</span> is surjective because both sets have the same size, <span class="math-container">$mn$</span> elements. </p>
<p>Since the image of <span class="math-container">$\phi$</span> is <span class="math-container">$H_n H_m$</span>, it is equal to <span class="math-container">$G$</span>.</p>
|
https://math.stackexchange.com/questions/3221041/is-h-nh-m-g-if-g-nm-and-h-n-h-m-are-the-only-subgroups-of-g-of-orde
|
Question: <p>Let <span class="math-container">$H$</span> be an infinite cyclic subgroup of a group <span class="math-container">$G$</span>. </p>
<p>Is the quotient group <span class="math-container">$G/C_G (H)$</span> of order two?</p>
Answer: <p>As pointed out in the comments, this is not true. Any abelian group is a counterexample, since <span class="math-container">$C_G(H) = G$</span>. However, we can find conditions on <span class="math-container">$H$</span> where this <strong>becomes</strong> true!</p>
<hr>
<p>First, note that there is a homomorphism from <span class="math-container">$N_G(H)$</span> to <span class="math-container">$\operatorname{Aut}(H)$</span> by</p>
<p><span class="math-container">$$g\mapsto (h\mapsto ghg^{-1})$$</span></p>
<p>that has kernel <span class="math-container">$C_G(H)$</span>. <strong>Since <span class="math-container">$H$</span> is infinite cyclic</strong>, this means that <span class="math-container">$\operatorname{Aut}(H)\cong \Bbb{Z}/2\Bbb{Z}$</span>. Thus, the first isomorphism theorem tells us that <span class="math-container">$N_G(H)/C_G(H)$</span> has order one or order two. </p>
<p>Now, let <span class="math-container">$G$</span> act on its set of subgroups by conjugation. The stabilizer of <span class="math-container">$H$</span> here is exactly <span class="math-container">$N_G(H)$</span>, and the orbit of <span class="math-container">$H$</span>, which we’ll denote <span class="math-container">$\mathcal{O}_H$</span>, is the set of conjugate subgroups of <span class="math-container">$H$</span> in <span class="math-container">$G$</span>. The orbit-stabilizer theorem says that the cosets of <span class="math-container">$N_G(H)$</span> are in bijection with <span class="math-container">$\mathcal{O}_H$</span>.</p>
<p>Putting these two facts together, this means that the size of <span class="math-container">$G/C_G(H)$</span> is at most twice the size of <span class="math-container">$G/N_G(H)$</span> (which is equal to the size of <span class="math-container">$\mathcal{O}_H$</span>). In order for the size of <span class="math-container">$G/C_G(H)$</span> to be <span class="math-container">$2$</span>, there are two options. First, either</p>
<ul>
<li><span class="math-container">$|\mathcal{O}_H|=1$</span>, i.e. <span class="math-container">$H$</span> is normal</li>
<li><span class="math-container">$N_G(H) \ne C_G(H)$</span></li>
</ul>
<p>or </p>
<ul>
<li><span class="math-container">$|\mathcal{O}_H|=2$</span></li>
<li><span class="math-container">$N_G(H) = C_G(H)$</span>.</li>
</ul>
<p>Thus, the size of <span class="math-container">$G/C_G(H)$</span> is two if either of the above situations occur.</p>
|
https://math.stackexchange.com/questions/3222863/about-centralizer-of-an-infinite-cyclic-subgroup
|
Question: <p>I am not sure exactly how to phrase this problem so I appologise if it is not clear, also this is somewhat long but I wanted to explain exactly where I was with the problem. If you have any questions feel free to ask.</p>
<hr>
<p><strong>Description of Problem</strong></p>
<p>Given a set of variables <span class="math-container">$\{x,y,z,...\}$</span> and a variable <span class="math-container">$o$</span> is it possible to define a finite product of these variables and their inverses <span class="math-container">$\sigma(x,y,z,...,x^{-1},y^{-1},z^{-1},...,o,o^{-1})$</span> (i.e. a finite sequence made up of these variables and their inverses) such that;</p>
<p>1) For any group <span class="math-container">$G$</span> and any assignment of values from <span class="math-container">$G$</span> to <span class="math-container">$\{x,y,z,...\}$</span> there exists a unique element <span class="math-container">$g$</span> of <span class="math-container">$G$</span> such that if <span class="math-container">$o$</span> is set to <span class="math-container">$g$</span>;</p>
<p><span class="math-container">$\sigma(x,y,z,...,x^{-1},y^{-1},z^{-1},...,o,o^{-1})=1$</span></p>
<p>and</p>
<p>2) There does not exist a finite product <span class="math-container">$\gamma(x,y,z,...,x^{-1},y^{-1},z^{-1},...)$</span> such that;</p>
<p><span class="math-container">$o=\gamma(x,y,z,...,x^{-1},y^{-1},z^{-1},...)$</span> for all groups <span class="math-container">$G$</span></p>
<hr>
<p><strong>Rough explaination as to why I am asking here</strong></p>
<p>My intuition is no but I am unsure how to prove this. There are clearly examples of equations like these solvable in all groups (i.e. <span class="math-container">$xo=1$</span>) but these have algebraic solutions (in that example <span class="math-container">$o=x^{-1}$</span>) and there are examples of these equations which are solvable in wide classes of groups (i.e. <span class="math-container">$o^{n!+1}x=1$</span> is solvable in any group of order less than <span class="math-container">$n$</span> with <span class="math-container">$o=x^{-1}$</span>) but these are not solvable in all groups. In addition some equations are solvable in all groups but not uniquely (i.e. <span class="math-container">$o^2=1$</span> has many solutions in groups with elements of order 2 but can always be solved with <span class="math-container">$o=1$</span>)</p>
<hr>
<p><strong>Progress on proof (or proof of falsehood)</strong></p>
<p>It can be shown that <span class="math-container">$\sigma$</span> must contain exactly <span class="math-container">$\pm1$</span> total occurences of <span class="math-container">$o$</span> (where <span class="math-container">$o^{-1}$</span> counts as <span class="math-container">$-1$</span> occurence of o) using the following argument.</p>
<p>If <span class="math-container">$G$</span> is abelian then <span class="math-container">$\sigma$</span> can be written as <span class="math-container">$Ao^n$</span> for some <span class="math-container">$A$</span> which is a product of the other variables. For this to be solvable <span class="math-container">$o^n=A^{-1}$</span> must be solvable in every abelian group. If <span class="math-container">$A=1$</span> <span class="math-container">$o$</span> is not uniquely defined for groups of order <span class="math-container">$|n|$</span>. If <span class="math-container">$A\neq1$</span> and <span class="math-container">$|n|\neq1$</span> then <span class="math-container">$o$</span> is not defined for groups of order <span class="math-container">$|n|$</span> or <span class="math-container">$n=0$</span> and so <span class="math-container">$o$</span> is not unique. Therefore <span class="math-container">$|n|=1$</span> and so the total number of occurences of <span class="math-container">$o$</span> in <span class="math-container">$\sigma$</span> must be <span class="math-container">$\pm1$</span>.</p>
<p>In addition it is clear that there must be an odd number of occurences of <span class="math-container">$o$</span> greater than <span class="math-container">$1$</span> (this time counting <span class="math-container">$o^{-1}$</span> as <span class="math-container">$1$</span> occurence). This follows as otherwise there is a clear definition of <span class="math-container">$\gamma$</span> (if there is <span class="math-container">$1$</span> occurence) or the observation above is violated (if there are an even number of occurences).</p>
<p>This is where I am and I am not sure how to proceed. Appologies again for this being overly long. Any information or advice would be appreciated.</p>
Answer: <p>Okay so I have no answered my question. If this doesn't seem like an answer it is probably due to me being unable to clearly describe the question I wanted to ask so sorry. Also I'm not sure how best to explain the answer.</p>
<hr>
<p>It can be shown that no such equation exists by contradiction. First assume there exists <span class="math-container">$\sigma$</span> as is required by the problem. Let <span class="math-container">$n$</span> be the number of variables it takes (excluding <span class="math-container">$o$</span> and only counting <span class="math-container">$x$</span> not <span class="math-container">$x^{-1}$</span> for example). Take the group <span class="math-container">$F_n$</span> (the free group with <span class="math-container">$n$</span> generators) and assign to each of the variables that <span class="math-container">$\sigma$</span> takes <span class="math-container">$x_1,x_2,...$</span> a different one of the generators <span class="math-container">$g_1,g_2,...$</span>. Find the value of <span class="math-container">$o$</span> corresponding to this situation. By definition of <span class="math-container">$F_n$</span>, <span class="math-container">$o$</span> can be written as a finite product of <span class="math-container">$g_1,g_2,...$</span>. Write <span class="math-container">$\gamma$</span> as this representaion of <span class="math-container">$o$</span> but where the generators are replaced by the corresponding variables. Therefore by definition of the free group in all groups <span class="math-container">$o=\gamma$</span> is a solution.</p>
<hr>
<p>Sorry if this wasn't clear. Also thanks to Santana Afton for mentioning using the free group in there answer.</p>
|
https://math.stackexchange.com/questions/3240948/are-there-equations-which-have-solutions-in-all-groups-but-which-are-not-algebra
|
Question: <p>I want to find an example of a homomorphism <span class="math-container">$f : \mathbb{Z}^2 \ast \mathbb{Z}^2 \to \mathbb{Z}^2$</span> such that <span class="math-container">$\ker f$</span> is free and not finitely generated.</p>
<p>My idea is to define <span class="math-container">$f$</span> on each copy of <span class="math-container">$\mathbb{Z}^2$</span> as the identity map, then use the universal property of free products to get a map <span class="math-container">$f : \mathbb{Z}^2 \ast \mathbb{Z}^2 \to \mathbb{Z}^2$</span>.</p>
<p>The kernel is free since conjugates of <span class="math-container">$\mathbb{Z}^2$</span> map isomorphically to their images: <span class="math-container">$g \mathbb{Z}^2 g^{-1}$</span> maps to <span class="math-container">$\mathbb{Z}^2$</span> isomorphically since in the image, we can move the <span class="math-container">$g^{-1}$</span> past the elements of <span class="math-container">$\mathbb{Z}^2$</span> and cancel with <span class="math-container">$g$</span>. Then the kernel intersects trivially with all conjugates, so acts freely on the Bass-Serre tree of <span class="math-container">$\mathbb{Z}^2 \ast \mathbb{Z}^2$</span> (which has vertex stabilisers given by conjugates of <span class="math-container">$\mathbb{Z}^2$</span>), so <span class="math-container">$\ker f$</span> is free.</p>
<p>I can't think of any finite generating set for <span class="math-container">$\ker f$</span>, but I am having a tough time thinking about why it <em>can't</em> be finitely generated. Certainly there are finitely generated subgroups with not finitely generated subgroups (e.g. the rank 2 free group <span class="math-container">$F_2$</span> has the countably infinite rank free group <span class="math-container">$F_\infty$</span> as a subgroup), so thinking about subgroups won't help. Maybe constructing an isomorphism to <span class="math-container">$F_\infty$</span>?</p>
Answer: <p>It's a classical and not hard theorem that if a finitely generated group <span class="math-container">$G$</span> has a finitely generated subgroup <span class="math-container">$N$</span> such that <span class="math-container">$N$</span> is normal and both <span class="math-container">$N$</span> and <span class="math-container">$G/N$</span> are infinite, then <span class="math-container">$G$</span> has a single end.</p>
<p>Hence, in your case, the kernel being infinite with infinite quotient, it is infinitely generated.</p>
<p>There's also a more explicit approach. Namely, consider the action of <span class="math-container">$G\ast G$</span> on its Bass-Serre tree (which has valency <span class="math-container">$|G|$</span>). Let <span class="math-container">$N$</span> be the kernel of the homomorphism onto <span class="math-container">$G$</span> that is identity on each factor. It acts freely, hence is free. It's not hard to see that <span class="math-container">$N$</span> has the same vertex orbits (namely 2). But all directed edges emanating from a single vertex are in distinct orbits. Hence, <span class="math-container">$N$</span> is infinitely generated as soon as <span class="math-container">$G$</span> is infinite.</p>
|
https://math.stackexchange.com/questions/3241462/proving-that-there-is-a-homomorphism-f-mathbbz2-ast-mathbbz2-to-ma
|
Question: <p>Wikipedia has a basic <a href="https://en.wikipedia.org/wiki/*-algebra#*-ring" rel="nofollow noreferrer">reference on *-rings in the *-algebra article</a>, which defines a *-ring as a ring equipped with a <code>*</code> operator which is an antiautomorphism and an involution.</p>
<p>However, I am not interested in rings equipped with such an operator, but Groups equipped with one. I tried searching for "*-group" to find references, but wasn't able to find much (I also tried the groupified versions of the alternative names Wikipedia suggested for *-rings: "involutive ring, involutory ring, and ring with involution", with no luck).</p>
<p>I am wondering if such a structure as a "*-group" exists in the literature, and if so, what name does it go by?</p>
<p>Is there some reference that could give a basic introduction to the subject? (Or maybe the structure is trivial, which might explain so little has been written about it?)</p>
<p>(For the context of why I am interested, I am thinking about generalizations of a <a href="https://www.chiark.greenend.org.uk/~sgtatham/algorithms/cbtree.html" rel="nofollow noreferrer">Counted B-Tree</a> for non-commutative sums, where segments can be added together, subtracted off from each other, and "flipped" to reverse the order).</p>
Answer:
|
https://math.stackexchange.com/questions/3259366/groups-or-group-with-involution
|
Question: <p>In this <a href="https://en.wikipedia.org/wiki/Lagrange%27s_theorem_(group_theory)" rel="nofollow noreferrer">page</a> of wikipedia there is a disproving of the converse of Lagrange's theorem. I would like to see a more simple (or short) disproving of Lagrange's theorem.</p>
Answer: <p>I prefer the following proof:</p>
<p>Let <span class="math-container">$G$</span> be a finite group, <span class="math-container">$N \subset G$</span> a normal subgroup and <span class="math-container">$g \in G$</span>. Then we obviously have that <span class="math-container">$\text{ord}(\overline{g})$</span> is a divisor of <span class="math-container">$\text{ord}(g)$</span>, where <span class="math-container">$\overline{g}$</span> is the image of <span class="math-container">$g$</span> under the quotient map <span class="math-container">$G \rightarrow G/N$</span>. Now assume that <span class="math-container">$[G:N] = 2$</span> and that <span class="math-container">$\text{ord}(g)$</span> has odd order. Then we get <span class="math-container">$\text{ord}(\overline{g}) = 1$</span> by Lagrange's theorem, such that <span class="math-container">$g \in N$</span>. We just showed that every element of odd order needs to be in <span class="math-container">$N$</span>. Therefore no group having more than <span class="math-container">$50$</span>% elements of odd order can have a subgroup of index <span class="math-container">$2$</span>. Since <span class="math-container">$A_4$</span> has <span class="math-container">$9$</span> elements of odd order it cannot have a subgroup of order <span class="math-container">$6$</span>.</p>
|
https://math.stackexchange.com/questions/3266823/disproving-the-converse-of-lagranges-theorem
|
Question: <p>Suppose that <span class="math-container">$(A,+)$</span> is an abelian group and that <span class="math-container">$A=B \cup C$</span>. Define for any <span class="math-container">$X \subseteq A$</span> the following</p>
<p><span class="math-container">$$\Delta X = \{ x_1-x_2 ; x_1 , x_2 \in X \}$$</span></p>
<p>Show that, if <span class="math-container">$B$</span> and <span class="math-container">$C$</span> have non-empty intersection, then either <span class="math-container">$A=\Delta B$</span> or <span class="math-container">$A=\Delta C$</span>.</p>
<p>This problem has challenged me. My approach is elementary, I am taking <span class="math-container">$x \in B \cap C$</span> and trying to write any <span class="math-container">$a \in A$</span> as <span class="math-container">$a + kx - kx$</span> where <span class="math-container">$k$</span> is some integer such that <span class="math-container">$a-kx \in B$</span> (or <span class="math-container">$C$</span>) and <span class="math-container">$kx$</span> also in <span class="math-container">$B$</span> (or <span class="math-container">$C$</span>). The problem is, this may not happen, I can have (almost) every multiple of <span class="math-container">$x$</span> in <span class="math-container">$C \setminus B$</span> and every <span class="math-container">$a+kx$</span> in <span class="math-container">$B \setminus C$</span> for example.</p>
<p>I have tried some contradiction arguments, but there are inevitably too many cases to consider (as far as I have tried). Can anyone offer some insight? Elementary hints (or solutions) are prefered (no quotients).</p>
Answer: <p>Note that the hypotheses <span class="math-container">$A = B \cup C$</span> and <span class="math-container">$B \cap C \ne \emptyset$</span> are unaffected if we replace <span class="math-container">$B, C$</span> by <span class="math-container">$-a + B, - a + C$</span>, for some <span class="math-container">$a \in A$</span>, while <span class="math-container">$\Delta B, \Delta C$</span> are unchanged. Choosing <span class="math-container">$a \in B \cap C$</span>, then, we may assume <span class="math-container">$0 \in B \cap C$</span>. This yields <span class="math-container">$B \subseteq \Delta B, C \subseteq \Delta C$</span>. </p>
<p>If <span class="math-container">$\Delta B \ne A$</span>, choose <span class="math-container">$a \in A \setminus \Delta B$</span>. Then for all <span class="math-container">$b\in B$</span>, we have <span class="math-container">$a + b \notin B$</span>, so that <span class="math-container">$a + b \in C$</span>. Note in particular that <span class="math-container">$a \in C$</span>, either because <span class="math-container">$a \notin \Delta B \supseteq B$</span>, or directly because <span class="math-container">$0 \in B$</span>. </p>
<p>But then for all <span class="math-container">$b \in B$</span> we have <span class="math-container">$b = (a + b) - a \in \Delta C$</span>, so that <span class="math-container">$B \subseteq \Delta C$</span>. Thus <span class="math-container">$A = B \cup C \subseteq \Delta C$</span>. </p>
|
https://math.stackexchange.com/questions/3274585/prove-that-either-a-delta-b-or-a-delta-c
|
Question: <p>Let <span class="math-container">$(G,*)$</span> be a group. And let <span class="math-container">$ x$</span> be a element of odd order of <span class="math-container">$G$</span> , then prove or disprove that , there is a element <span class="math-container">$y$</span> in <span class="math-container">$G$</span> such that , <span class="math-container">$y^2 = x$</span></p>
<p>Please provide some hint, i am not able to show any contradicting examples nor able to prove it.</p>
Answer: <p>Suppose <span class="math-container">$x$</span> has order <span class="math-container">$2n+1$</span>. Then <span class="math-container">$x^{n+1}$</span> can serve as <span class="math-container">$y$</span>, since <span class="math-container">$y^2=x^{2n+2}=x^{2n+1}x=x$</span>. Therefore the statement is true.</p>
|
https://math.stackexchange.com/questions/3278714/prove-or-disprove-that-there-is-a-element-y-in-g-such-that-y2-x
|
Question: <p>I am trying to learn group theory from the textbook Algebra by Michael Artin 2nd edition. I am looking for finite group and subgroup examples that are useful for developing an intuition about group theory. So far, I have been using the symmetric group <span class="math-container">$S_3$</span>. </p>
<p>Preferably, I am looking for an example of a finite group and a subgroup that I can use to grasp the concepts of the book. A list would be nice. Or just direction towards a database of group examples could work too.</p>
Answer: <p>Of course any answer to such a broad question has to be incomplete, but you might find the following helpful:</p>
<hr>
<p>At the very beginning I think that (as you say) <span class="math-container">$S_3$</span> is a good example as an easy-to-grasp nonabelian group. And abelian groups aren't entirely uninteresting either - you should understand why <span class="math-container">$\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z}\cong\mathbb{Z}/6\mathbb{Z}$</span> but <span class="math-container">$\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}\not\cong\mathbb{Z}/4\mathbb{Z}$</span>. This sets the stage for the <a href="https://en.wikipedia.org/wiki/Abelian_group#Classification" rel="nofollow noreferrer">classification of finite(ly generated) abelian groups</a>, and ultimately of finitely generated modules over a PID, but long before then is just a key point to understand - and the latter also serves as a good example of how the "obvious" statement <span class="math-container">$$A/B=C\implies A=B\times C$$</span> fails miserably in the context of groups, quotient groups, and direct products.</p>
<p>What about a bit further on - e.g. when we start talking about <a href="https://en.wikipedia.org/wiki/Normal_subgroup" rel="nofollow noreferrer"><strong><em>normal</em> subgroups</strong></a>? In my opinion, <a href="https://en.wikipedia.org/wiki/Alternating_group" rel="nofollow noreferrer"><span class="math-container">$A_4$</span> and <span class="math-container">$A_5$</span></a> are quite good examples. Each is non-abelian, and <span class="math-container">$A_5$</span> is simple while <span class="math-container">$A_4$</span> is not - and that's already a good thing to have examples of. And proving that <span class="math-container">$A_5$</span> is the smallest non-abelian simple group is a good problem. And of course, eventually you'll care about solvable groups, and then <span class="math-container">$A_5$</span> is crucial.</p>
<p>There are also examples of groups which early on will just seem weird, but are really valuable. For example, a small <a href="https://en.wikipedia.org/wiki/Semidirect_product" rel="nofollow noreferrer">semidirect product</a> like <span class="math-container">$D_3$</span> <em>(yes, that's just <span class="math-container">$S_3$</span>, but thinking about it as a dihedral group is a bit better right now)</em>. It's easy enough to write out how this group works <em>(without defining semidirect products - just defining this specific group)</em>, and compare it with the corresponding direct product <span class="math-container">$\mathbb{Z}/3\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}\cong\mathbb{Z}/6\mathbb{Z}$</span>; while it will then be a while (probably) before you care about semidirect products, it will be useful to have in your head early on the idea that there are more complicated ways to combine groups than just the direct product.</p>
<p>And then there are infinite groups which are still "finite-flavored" - I'm thinking e.g. of matrix groups like <a href="https://en.wikipedia.org/wiki/General_linear_group" rel="nofollow noreferrer"><span class="math-container">$GL_n(\mathbb{R})$</span></a>, the group of all invertible <span class="math-container">$n$</span>-by-<span class="math-container">$n$</span> matrices with real number entries. This isn't such a great example if you're not already familiar with some linear algebra, but if you are it's really quite good to get your hands on it early.</p>
|
https://math.stackexchange.com/questions/3289459/what-examples-to-use-when-learning-group-theory
|
Question: <p>Let <span class="math-container">$G$</span> be the smallest group of <span class="math-container">$2 \times 2$</span> matrices whose entries are complex numbers which contains both matrices<br>
<span class="math-container">$$\begin{bmatrix}0&1\\-1&0\end{bmatrix} \quad \text{ and } \quad \begin{bmatrix}i&0\\0&-i\end{bmatrix},$$</span>
where the complex number <span class="math-container">$i$</span> satisfies <span class="math-container">$i^2 = −1$</span> and where the binary operation is matrix multiplication. Determine all conjugacy classes of <span class="math-container">$G$</span>.</p>
<p>Any ideas?</p>
<p>I had the idea of multiplying the matrices together, and finding the inverses of the matrices for the elements of the group, but feel like this wouldn't be helpful, and may make things more difficult for me
matrices.</p>
Answer: <p><em>Hint:</em>
Those two matrices commute, because the second one is <span class="math-container">$iI$</span>.
Therefore, the group they generate is an abelian group.
What are the conjugacy classes of an abelian group?</p>
|
https://math.stackexchange.com/questions/3304910/group-of-2-times-2-matrices-with-complex-numbers
|
Question: <p>Let <span class="math-container">$H$</span> be a subgroup of finite index in group <span class="math-container">$G$</span> and let <span class="math-container">$g \in G$</span>. </p>
<p><strong>Question:</strong> Is it true, that <span class="math-container">$C_H(g)$</span> has finite index in <span class="math-container">$C_G(g)$</span> for all <span class="math-container">$g \in G$</span>? </p>
<p>Here <span class="math-container">$C_H(g) := \{ h \in H\mid gh = hg\}$</span> denotes the centralizer of <span class="math-container">$g$</span> in <span class="math-container">$H$</span>. </p>
<p>I know that the statement is true for <span class="math-container">$g \in H$</span>. In this case it is a byproduct of the theory of ranks of projective modules over group rings. For, there one shows that for each conjugacy class <span class="math-container">$[h]$</span> of <span class="math-container">$H$</span> there is an integer <span class="math-container">$n_h$</span> such that for each projective <span class="math-container">$\mathbb{Z}G$</span>-module <span class="math-container">$P$</span> the Hattori-Stallings rank satisfies: </p>
<p><span class="math-container">$$R_H(P) = \sum_h n_h R_G(P)(h)\cdot [h]$$</span>
where <span class="math-container">$h$</span> runs over a system of representatives of the conjugacy classes of <span class="math-container">$H$</span>. Moreover, one can show <span class="math-container">$n_h = (C_G(h):C_H(h))$</span> (cf. Kenneth S. Brown: Cohomology of Groups, IX, Prop. 4.1 and Exercise 2). </p>
Answer:
|
https://math.stackexchange.com/questions/3315817/finite-index-of-centralizer-c-hg-in-c-gg
|
Question: <p>If <span class="math-container">$H$</span> and <span class="math-container">$K$</span> are subgroups of a group <span class="math-container">$G,*$</span>, how can one proof that:
<span class="math-container">$$HK = \operatorname{grp}\{H \cup K\} \iff HK \text{ is a subgroup of } G$$</span> </p>
<p>I've seen similar proofs on the internet, but there the subgroups <span class="math-container">$H$</span> and <span class="math-container">$K$</span> were normal subgroups. I think it is also possible to prove with subgroups, but can't figure out how to prove it without the condition of normal subgroups.</p>
Answer: <p>If <span class="math-container">$HK$</span> is the subgroup generated by <span class="math-container">$H\cup K$</span>, then it is of course a subgroup.</p>
<p>On the other hand, if <span class="math-container">$HK$</span> is a subgroup of <span class="math-container">$G$</span>, then it is clearly the smallest subgroup containing both <span class="math-container">$H$</span> and <span class="math-container">$K$</span> (as any such subgroup must contain all <span class="math-container">$hk$</span> with <span class="math-container">$h\in H$</span>, <span class="math-container">$k\in K$</span>).</p>
|
https://math.stackexchange.com/questions/3321901/proof-for-hk-operatornamegrp-h-cup-k-iff-hk-text-is-a-subgroup-of
|
Question: <p>Let G be a group. If order of the element in the group <span class="math-container">$G$</span> is prime (<span class="math-container">$G$</span> is not a <span class="math-container">$p$</span>-group, order of the elements are different prime). Is it true that every element of order <span class="math-container">$p$</span> are in the same conjugacy class of <span class="math-container">$G$</span>?</p>
Answer: <p>No. Not all elements of order 5 in the alternating group <span class="math-container">$A_5$</span> are conjugate.</p>
|
https://math.stackexchange.com/questions/3329586/let-g-be-a-group-if-order-of-the-element-in-the-group-g-is-prime
|
Question: <blockquote>
<p>Let <span class="math-container">$G$</span> be group, <span class="math-container">$S \subset G$</span>, <span class="math-container">$g \in G $</span>. Prove:
<span class="math-container">$$gC_{G}(S)(g)^{-1} = C_{G}(gS(g)^{-1}), \forall g\in G.$$</span></p>
</blockquote>
<p>I have managed to prove that <span class="math-container">$gC_{G}(S)(g)^{-1} \subseteq C_{G}(gS(g)^{-1})$</span>, but I have troubles with this <span class="math-container">$ C_{G}(gS(g)^{-1}) \subseteq gC_{G}(S)(g)^{-1} $</span>. I have tried to prove it by definition but I got stuck.
P.S. <span class="math-container">$C_{G}(S) = \{ g \in G \colon gsg^{-1} = s, \forall s \in S \}$</span>.</p>
Answer: <p>Let <span class="math-container">$x \in C_G(gSg^{-1})$</span>, then <span class="math-container">$$\forall s \in S, g \in G: g^{-1}xgs = g^{-1}xgsg^{-1}g.$$</span> Since <span class="math-container">$x$</span> commutes with <span class="math-container">$gsg^{-1}$</span> this is <span class="math-container">$$\underbrace{g^{-1}g}_{=e}sg^{-1}xg = sg^{-1}xg.$$</span> Hence for every <span class="math-container">$s \in S, g \in G$</span>, <span class="math-container">$g^{-1}xg$</span> commutes with <span class="math-container">$s$</span> i.e. <span class="math-container">$g^{-1}xg \in C_G(S)$</span> and thus <span class="math-container">$x \in g^{-1}C_G(S)g$</span>.</p>
|
https://math.stackexchange.com/questions/3333011/let-g-be-group-s-subset-g-g-in-g-prove-gc-gsg-1-c-gg
|
Question: <blockquote>
<p>If we know <span class="math-container">$\mathbb{Z}_{n}/ \mathbb{Z}_{m} \cong \mathbb{Z}_{k}$</span>, can we conclude that <span class="math-container">$\mathbb{Z}_{n} \cong \mathbb{Z}_{m} \times \mathbb{Z}_{k} $</span>?</p>
</blockquote>
<p>I think not, but I can not find right counterexample. Any hint helps!</p>
Answer: <p><strong>Counterexample:</strong> <span class="math-container">$n =4, k=2 $</span>.</p>
<p>Notably, <span class="math-container">$\Bbb Z_m \times \Bbb Z_k$</span> is cyclic if and only if <span class="math-container">$m, k$</span> are relatively prime. </p>
|
https://math.stackexchange.com/questions/3333937/if-we-know-mathbbz-n-mathbbz-m-cong-mathbbz-k-can-we-conclud
|
Question: <p>Let <span class="math-container">$S_5$</span> denote the group of bijections of the {1, 2, 3, 4, 5} under composition. Find all the elements <span class="math-container">$f ∈ S_5$</span> of order two such that <span class="math-container">$f(1) = 2$</span></p>
<p>I know for order of 2 we need that <span class="math-container">$f^2$</span> sends all elements back to themselves so <span class="math-container">$f(f(1))=1, f(f(2))=2....$</span></p>
<p>and since we know <span class="math-container">$f(1)=2$</span> then <span class="math-container">$f(f(1))$</span> becomes <span class="math-container">$f(2)$</span> which must equal 1</p>
<p>so we have <span class="math-container">$f(1)=2$</span> and <span class="math-container">$f(2)=1$</span> and I could do the same for <span class="math-container">$f(3)$</span> and <span class="math-container">$f(4)$</span> but since there is an odd number of elements I am left with an odd one out. What am I doing wrong?</p>
Answer: <p>Elements of <span class="math-container">$S_5$</span> of order <span class="math-container">$2$</span> such that <span class="math-container">$f(1)=2$</span> are the following maps </p>
<p>(elements not indicated are fixed):</p>
<p><span class="math-container">$1\mapsto2, 2\mapsto1\tag1$</span></p>
<p><span class="math-container">$1\mapsto2, 2\mapsto1, 3\mapsto4, 4\mapsto3\tag2$</span></p>
<p><span class="math-container">$1\mapsto2, 2\mapsto1, 3\mapsto5, 5\mapsto4\tag3$</span></p>
<p><span class="math-container">$1\mapsto2, 2\mapsto1, 4\mapsto5, 5\mapsto4\tag4$</span></p>
|
https://math.stackexchange.com/questions/3358628/find-all-elements-f-%e2%88%88-s-5-of-order-2-such-that-f1-2
|
Question: <p>Given groups <span class="math-container">$H$</span> and <span class="math-container">$K$</span>, define a group <span class="math-container">$G$</span> based on <span class="math-container">$H$</span> and <span class="math-container">$K$</span> as <span class="math-container">$G:=\{(h,k):h\in H, k\in K\}$</span> such that </p>
<p><span class="math-container">$$(h_1,k_1)(h_2,k_2)=(h_1h_2,\phi(h_1,k_1,h_2,k_2))$$</span></p>
<p>Given <span class="math-container">$G$</span> is a well-defined group , what properties we can derive that <span class="math-container">$\phi$</span> must have?</p>
Answer:
|
https://math.stackexchange.com/questions/3369151/define-group-g-based-on-h-and-k
|
Question: <blockquote>
<p>Let <span class="math-container">$G$</span> be a group and <span class="math-container">$a,b\in G$</span>. Prove that <span class="math-container">$|a|=|a^{-1}|$</span> and that <span class="math-container">$|ab|=|ba|$</span>.</p>
</blockquote>
<p>I said that <span class="math-container">$|a|=n$</span> where <span class="math-container">$n$</span> is the smallest integer such that <span class="math-container">$a^n=e$</span> and <span class="math-container">$|a^{-1}|=m$</span> where <span class="math-container">$m$</span> is the smallest integer such that <span class="math-container">$(a^{-1})^{m}=a^{-m}=e$</span> but I don't know how to continue. Any suggest are appreciated.</p>
Answer: <p><strong>Hint.</strong> By definition, <span class="math-container">$a^{-n}a^n=e$</span>, where <span class="math-container">$e$</span> is the identity element. But <span class="math-container">$a^n=e$</span>, which gives you one direction. Can you show the other direction? For the second question, consider <span class="math-container">$(ab)^na=a(bab\dotsm aba)$</span>. What happens if <span class="math-container">$(ba)^n=e$</span>?</p>
|
https://math.stackexchange.com/questions/3383518/let-g-be-a-group-and-a-b-in-g-prove-that-a-a-1-and-that-ab-ba
|
Question: <p>I know the order say m is some integral integer that gives gives an element an identity, say 0 in the additive case. Can somebody give an example as to why it has infinite order? Is that because there is infinity amount of elements?</p>
Answer: <p>Name <span class="math-container">$G$</span> one of those groups and take <span class="math-container">$0 \neq g \in G$</span>. If <span class="math-container">$g$</span> has a finite order <span class="math-container">$n$</span> we get</p>
<p><span class="math-container">$$n \cdot g = (n\cdot 1) \cdot g=0$$</span></p>
<p>That can't be as those groups do not have zero divisors for the multiplication.</p>
<p>Now regarding your question <em>Is that because there is infinity amount of elements?</em>, the answer is negative. A group can have an infinite number of elements but have elements of finite orders. An example is <span class="math-container">$(\mathbb U, \cdot)$</span> where <span class="math-container">$\mathbb U$</span> is the set of the complex numbers of modulus equal to one endowed with the multiplication.</p>
|
https://math.stackexchange.com/questions/3388664/in-the-additive-groups-mathbbz-mathbbq-mathbbr-or-mathbbc
|
Question: <p>I am working my way through Charles Pinter's book: A Book of Abstract Algebra. From recommendations on this site, I found a page/web address on Wisconsin University's Math Department that provides solutions to many (perhaps all) of the abundant exercises that are present in Pinter's book. </p>
<p>One of the proposed solutions to Pinter's exercises is the following generalization:</p>
<p><strong>If the product of n elements of a group is the identity element, it remains so no matter in what order the terms are multiplied.</strong></p>
<p>I take issue with this claim and think it is only valid if the group is abelian. Using a simple 3 element example, consider:</p>
<p><span class="math-container">$a\circ b \circ c = e$</span></p>
<p>Using the definition of inverses (and knowing that inverses commute...by defintion) and the associative law, I generated the following cases that must be true:</p>
<p><span class="math-container">$a \circ (b \circ c) =e$</span> and therefore <span class="math-container">$(b \circ c) \circ a =e$</span></p>
<p><span class="math-container">$(a \circ b) \circ c=e$</span> and therefore <span class="math-container">$c \circ (a \circ b)=e$</span></p>
<p>However, there are still several permutations of this list of elements that require consideration...for example:</p>
<p><span class="math-container">$a \circ (c \circ b) =e$</span></p>
<p>It seems to me this can only be true if the group is abelian. Therefore, should the solution manual be amended to say: </p>
<p>If the product of n elements of <strong>an abelian group</strong> is the identity element, it remains so no matter in what order the terms are multiplied.</p>
<p>?</p>
Answer: <p>Since this is a true or false question, it is not that the question is phrased incorrectly, but rather that the answer is that it is false. </p>
<p>Your claim that it can only hold if the group is abelian is not true for all such <span class="math-container">$a, b, c$</span>, which we can see in any group by <span class="math-container">$a=b=c=e$</span> and other less trivial examples. What you need to do to show it is false is to pick a specific nonabelian group and find three specific elements where it doesn't work.</p>
<p>Edit: I just noticed that this was not an exercise, but part of the solution manual. In <span class="math-container">$S_3$</span>, the elements <span class="math-container">$(12),(23),(321)$</span> show the original claim is invalid. In fact, if <span class="math-container">$abc=e$</span> and <span class="math-container">$bc\neq cb$</span>, then it is never true that <span class="math-container">$acb=e$</span> because <span class="math-container">$bc$</span> is the unique inverse of <span class="math-container">$a$</span>. By similar reasoning, if the claim holds for <span class="math-container">$a, b, c$</span>, then all three elements commute pairwise. You can show that the group is abelian if this holds true for all triples such that <span class="math-container">$abc=e$</span>. </p>
|
https://math.stackexchange.com/questions/3397903/true-or-false-if-the-product-of-n-elements-of-a-group-is-the-identity-element
|
Question: <p>Explicitly construct the regular representation of <span class="math-container">$\mathbb Z_3$</span> and diagonalize it. Since we are now fully diagonal every entry must furnish a one dimensional irreducible representation. Do you think you recovered all the irreps we have found in class? </p>
Answer: <p>The regular representation of <span class="math-container">$\mathbb Z_3$</span> is
<span class="math-container">$$\begin{align}0 &\mapsto \left(\begin{matrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{matrix}\right),\;\; \\
1 &\mapsto \left(\begin{matrix}
0 & 0 & 1\\
1 & 0 & 0 \\
0 & 1 & 0
\end{matrix}\right), \\
2 &\mapsto \left(\begin{matrix}
0 & 1 & 0\\
0 & 0 & 1 \\
1 & 0 & 0
\end{matrix}\right).
\end{align}$$</span>
Its diagonal form is
<span class="math-container">$$n \mapsto \left( \begin{matrix}
1 & 0 & 0 \\
0 & e^{2\pi n / 3} & 0\\
0 & 0 &e^{-2\pi n / 3}
\end{matrix} \right).$$</span>
Do you see why?</p>
|
https://math.stackexchange.com/questions/3412322/construct-the-regular-representation-of-z3-and-diagonalize-it
|
Question: <p>Does converse of Lagrange's theorem hold in <span class="math-container">$A_{4} \times \Bbb Z_{2}?$</span></p>
<p>The order of this group is <span class="math-container">$24$</span> and I'm unable to find a subgroup of order 4.Does there exist any group of order <span class="math-container">$4$</span> in this group<span class="math-container">$?$</span></p>
<p>Moreover, Is there any quick way to know the orders of subgroups of a particular group<span class="math-container">$?$</span></p>
Answer: <p>Assuming <span class="math-container">$A_4$</span> is the alternating group of <span class="math-container">$4$</span> elements, it contains an subgroup of order <span class="math-container">$4$</span>. In fact, this subgroup is normal, and the only (proper) normal subgroup of <span class="math-container">$A_4$</span>. (Hint: think <span class="math-container">$(2,2)$</span>-cycles.) This subgroup is isomorphic to the Klein-4 group.</p>
<p>By taking the product of this group with the trivial subgroup of <span class="math-container">$\mathbb{Z}/2$</span>, it follows that the product still has order <span class="math-container">$4$</span> and is a subgroup of <span class="math-container">$A_4 \times \mathbb{Z}/2$</span>.</p>
|
https://math.stackexchange.com/questions/3436892/does-converse-of-lagranges-theorem-hold-in-a-4-times-bbb-z-2
|
Question: <p>Suppose I am trying to construct a semidirect product and I have two homomorphisms <span class="math-container">$\varphi_1:K\rightarrow\text{Aut}(H)$</span> and <span class="math-container">$\varphi_2:K\rightarrow\text{Aut}(H)$</span>. Furthermore, suppose that <span class="math-container">$\ker\varphi_1\cong\ker\varphi_2$</span>. Are the semidirect products constructed by these homomorphisms equivalent?</p>
<p>For an explicit example, suppose I wanted to construct <span class="math-container">$Z_7\rtimes (Z_2\times Z_2)$</span>, Aut(<span class="math-container">$Z_7)\cong Z_6$</span>. If we have <span class="math-container">$Z_2\times Z_2=\langle a\rangle\times\langle b\rangle$</span>, the following homomorphisms both work (defined on the generators):
<span class="math-container">$$\varphi_1(a)=1\quad \varphi_1(b)=x$$</span>
<span class="math-container">$$\varphi_1(a)=x\quad \varphi_1(b)=x$$</span>
where <span class="math-container">$x$</span> is the automorphism of order <span class="math-container">$2$</span> in Aut<span class="math-container">$(H)$</span>. In both of these cases, the kernel is isomorphic to <span class="math-container">$Z_4$</span>. Is it true that <span class="math-container">$Z_7\rtimes_{\varphi_1} (Z_2\times Z_2)\cong Z_7\rtimes_{\varphi_2} (Z_2\times Z_2)$</span>?</p>
Answer: <p>The answer to your question is no: actually you can have <span class="math-container">$\varphi_1$</span> and <span class="math-container">$\varphi_2$</span> both injective (so with the same kernel) and still have non-isomorphic semi-direct products.</p>
<p>For instance, take <span class="math-container">$K=\mathbb{Z}/2\mathbb{Z}$</span> and <span class="math-container">$H=\mathbb{Z}/3\mathbb{Z}\times \mathbb{Z}/4\mathbb{Z}$</span>, and define <span class="math-container">$\varphi_1$</span> such that the non-trivial element in <span class="math-container">$K$</span> acts non-trivially on the <span class="math-container">$\mathbb{Z}/3\mathbb{Z}$</span> factor (and trivially on the other one), and <span class="math-container">$\varphi_2$</span> such that it acts non-trivially on <span class="math-container">$\mathbb{Z}/4\mathbb{Z}$</span> (and trivially on the other one).</p>
<p>Then <span class="math-container">$\varphi_1$</span> and <span class="math-container">$\varphi_2$</span> are injective, but the semi-direct products are respectively <span class="math-container">$D_6\times \mathbb{Z}/4\mathbb{Z}$</span> and <span class="math-container">$D_8\times \mathbb{Z}/3\mathbb{Z}$</span> (where <span class="math-container">$D_{2n}$</span> is the dihedral group with <span class="math-container">$2n$</span> elements) which are not isomorphic (for instance we can look at the center).</p>
|
https://math.stackexchange.com/questions/3440119/homomorphisms-with-the-same-kernel-has-the-same-semidirect-product
|
Question: <blockquote>
<p>Find all subgroups of <span class="math-container">$(\Bbb{Z}_2\times\Bbb{Z}_4,\overline{+})$</span>.</p>
</blockquote>
<p>I could find the following subgroups:</p>
<p><span class="math-container">$$\begin{array}{ll}
H_1=\langle(0,0)\rangle=\{(0,0)\}&\text{(Trivial subgroup)}\\
H_2=\langle(0,1)\rangle=\{(0,1),(0,2),(0,3),(0,0)\}=\langle(0,3)\rangle=H_4\\
H_3=\langle(0,2)\rangle=\{(0,2),(0,0)\}\\
H_5=\langle(1,0)\rangle=\{(1,0),(0,0)\}\\
H_6=\langle(1,1)\rangle=\{(1,1),(0,2),(1,3),(0,0)\}=\langle(1,3)\rangle=H_8\\
H_7=\langle(1,2)\rangle=\{(1,2),(0,0)\}\\
H_9=\Bbb{Z}_2\times\Bbb{Z}_4&\text{(Improper subgroup)}
\end{array}$$</span></p>
<p>That is, <span class="math-container">$7$</span> subgroups in total.</p>
<p>I found all the CYCLIC subgroups, but since the group IS NOT CYCLIC, then I need to find the NOT CYCLIC subgroups.</p>
<p>The final answer should be <span class="math-container">$8$</span> subgroups; the only subgroup that is NOT CYCLIC is: <span class="math-container">$$\text{Subgroup that is not cyclic}=\{(0,0),(0,2),(1,0),(1,2)\}.$$</span> So there are <span class="math-container">$8$</span> subgroups in total.</p>
<p>My question is:</p>
<blockquote>
<p>How can we find <span class="math-container">$\{(0,0),(0,2),(1,0),(1,2)\}$</span>? I mean, what should we put inside <span class="math-container">$\color{red}{\langle(\ldots)\rangle}$</span>?</p>
</blockquote>
Answer: <p>We write <span class="math-container">$\langle A \rangle$</span> for a subset (not necessarily subgroup) <span class="math-container">$A \subseteq G$</span> to mean the smallest subgroup which contains <span class="math-container">$A$</span>. So it would be correct in your case to write
<span class="math-container">$$ \{(0,0),(0,2),(1,0),(1,2)\} = \langle \{(0,0),(0,2),(1,0),(1,2)\} \rangle $$</span></p>
<p>However, this is somewhat unsatisfying, since in the cyclic case we only needed <em>one</em> generator (that is, we only needed one element inside the <span class="math-container">$\langle - \rangle$</span>).
If you are looking for a <em>small</em> set of elements to put inside, you can get away with</p>
<p><span class="math-container">$$ \{(0,0),(0,2),(1,0),(1,2)\} = \langle (0,2), (1,0) \rangle.$$</span></p>
<p>Any subgroup contains <span class="math-container">$(0,0)$</span>, and any subgroup containing <span class="math-container">$(1,0)$</span> and <span class="math-container">$(0,2)$</span> must <em>also</em> contain their sum <span class="math-container">$(1,2)$</span>. Thus {(0,0),(0,2),(1,0),(1,2)} is the smallest subgroup containing <span class="math-container">$(0,2)$</span> and <span class="math-container">$(1,0)$</span>.</p>
<p><strong>Edit:</strong>
In a comment you ask if we should operate on the elements inside the <span class="math-container">$\langle - \rangle$</span>. The answer is "no". As you have noticed, if we reduce the elements inside, we can only ever get cyclic subgroups. We allow ourselves multiple elements inside so that we can express more complicated subgroups, such as the example above.</p>
<p><span class="math-container">$\langle x + y \rangle$</span> is always a subgroup of <span class="math-container">$\langle x, y \rangle$</span>. (Since <span class="math-container">$x+y \in \langle x, y \rangle$</span>). It is very rarely the case that they are equal, however.</p>
<hr>
<p>I hope this helps ^_^</p>
|
https://math.stackexchange.com/questions/3441019/find-all-subgroups-of-bbbz-2-times-bbbz-4
|
Question: <p>Question is that we have an element <span class="math-container">$a$</span>, where <span class="math-container">$a = p-1$</span> and <span class="math-container">$p$</span> is a prime number, and I need to prove that order of <span class="math-container">$a$</span> is <span class="math-container">$2$</span>. </p>
<p>I know how do we calculate order of an element where we know exact number of elements in a group but I do not understand how do we prove it if we don't know the exact value of <span class="math-container">$p$</span>.</p>
Answer: <p>I am going to assume that you are in the group <span class="math-container">$\mathbb{Z}/p\mathbb{Z}$</span>, and that <span class="math-container">$p$</span> is odd.</p>
<p>Now <span class="math-container">$(p-1)^2 = p^2 - 2p + 1 = 1 \pmod p$</span>, so the order of <span class="math-container">$p-1$</span> is indeed <span class="math-container">$2$</span>.</p>
|
https://math.stackexchange.com/questions/3445321/how-to-show-that-order-of-an-element-is-2-where-the-element-is-p-1-for-prim
|
Question: <p>Let <span class="math-container">$G$</span> be a finite group of order <span class="math-container">$n$</span> given by generator-relator representation. </p>
<p><strong>Question :</strong> Let <span class="math-container">$a$</span> and <span class="math-container">$b$</span> are two elements of <span class="math-container">$G$</span> and we want to know the result of <span class="math-container">$ab$</span>.</p>
<p>Is there any way to know the result, without writing down the group as a table?</p>
<p>What other information I need to store so that I can get the result of the <span class="math-container">$ab$</span>(<span class="math-container">$a$</span> group operation <span class="math-container">$b$</span>)?</p>
Answer: <p>This is called "<a href="https://en.wikipedia.org/wiki/Word_problem_for_groups" rel="nofollow noreferrer">the word problem</a>", and it is hard. In some finitely presented groups it is actually undecidable. Of course, for finite groups you can always check by hand or write algorithms that terminate, giving you an answer. But it is not much different from "writing down the group as a table" (and in fact it is harder than that).</p>
|
https://math.stackexchange.com/questions/3460030/group-operation-in-generator-relator-representation
|
Question: <p>This is part of an assignment, so please no full answers just hints (c:</p>
<p>Let <span class="math-container">$a=(1234)$</span>, <span class="math-container">$b=(13)(5678)$</span>, <span class="math-container">$G=\langle a,b\rangle$</span>.
Show that the quotient group <span class="math-container">$G/\langle a\rangle$</span> is cyclic</p>
<p>I found 4 left cosets, so <span class="math-container">$|G/\langle a\rangle|=4$</span>. These are:</p>
<p><span class="math-container">$\langle a\rangle = \{(1234), (13)(24), (1432), e\}$</span></p>
<p><span class="math-container">$(57)(68)\langle a\rangle = \{(1234)(57)(68),(13)(24)(57)(68),(1432)(57)(68),(57)(68)\}$</span></p>
<p><span class="math-container">$(13)(5678)\langle a\rangle=\{(12)(34)(5678), (24)(5678), (14)(23)(5678), (13)(5678)\}$</span></p>
<p><span class="math-container">$(13)(5876)\langle a\rangle=\{(12)(34)(5876), (24)(5876), (14)(23)(5876), (13)(5876)\}$</span></p>
<p>I know <span class="math-container">$\langle a\rangle$</span> is a normal subgroup of G... but I still don't know how to actually answer the question (ie. show that this quotient group is cyclic). I guess the problem is that I don't quite understand this type of group... I thought I could show that it's cyclic by finding an element <span class="math-container">$g \in G/\langle a\rangle$</span> so that <span class="math-container">$g^4 = e$</span>... but the identity is <span class="math-container">$\langle a\rangle$</span>... I'm a bit lost...</p>
<p>Thanks!</p>
Answer: <p><strong>Hint</strong>:</p>
<p>The last two elements <span class="math-container">$(1\,3)(5\,6\,7\,8)\langle a\rangle$</span> and <span class="math-container">$(1\,3)(5\,8\,7\,6)\langle a\rangle$</span> have order <span class="math-container">$4$</span>.</p>
|
https://math.stackexchange.com/questions/3485924/show-that-a-quotient-group-is-cyclic
|
Question: <p>Let T be the set of "twists" that a curve can express. A twist of a curve is essentially a normal vector to the curve parameterized by the curve. The normal vector may include angle in 3D.</p>
<p>In a mobius strip there is a twist vector that simply is the normal vector to the strip. In 2D it would be simply a magnitude and sign, in 3D a magnitude and angle. The angle is protected in to a sign bit. Higher dimensions apply but is relatively irrelevant to this question.</p>
<p>The idea is that composing twists is a twist, there is the identity twist of not twisting, which can be considered a twist, and every twist has an inverse twist. </p>
<p>Hence twisting forms an abelian group under composition.</p>
<p>Has this been studied?</p>
<p>Specially what I'm interesting in is how 1D curves can be embedded in to higher dimensions and can be given different "twists" when they are done. E.g., The mobius strip and the normal strip both have the same 1D curve but when embedded in to 3D space they end up with different properties. These properties are due precisely because the 1D curve can be "twisted" idempotent(acting on the 1D curve it does nothing, but acting on the 3D embedded curve they give rise to different surfaces).</p>
<p>I'm seeing the Mobius strip vs the normal strip precisely as two different representations of the 1D curve where the representations are due to the twist group and these interact with embedded in particular ways.</p>
<p>(The curve itself embedded is irrelevant as when the mobius strip's width is shrunk to 0 it is the same curve as when the normal strip is shrunk... at least up to homotopy. Hence when we "flatten" out the curve it can give rise to different classes of surfaces in the embedded space. This seems to be precisely due to the fact that the 1D curves pass through the twisting group unaltered).</p>
Answer:
|
https://math.stackexchange.com/questions/3514356/the-twist-group
|
Question: <p>We defined a Folner sequence by the condition that <span class="math-container">$$ \lim_{n \to \infty} \frac{ | g A_n \Delta A_n |}{ | A_n|} = 0 $$</span> for all <span class="math-container">$g \in G$</span>, where <span class="math-container">$\left\{A_n \right\}$</span> is a sequence of finite subsets of <span class="math-container">$G$</span>. </p>
<p><strong>Problem:</strong> Let <span class="math-container">$G = \mathbb{Z}$</span>. Show that the sequence <span class="math-container">$(A_n)$</span> given by <span class="math-container">$$ A_n = \left\{0, \ldots, n-1 \right\} $$</span> is a Folner sequence. </p>
<p>Using this Folner sequence, explain how we can define a Folner sequence for the group <span class="math-container">$ \mathbb{Z}^k$</span> with <span class="math-container">$k \geq 2$</span>.</p>
<p><strong>Attempt:</strong> I think I have <span class="math-container">$$g A_n = \left\{ g, g+1, \ldots, g + n-1 \right\}. $$</span> Then I wish to know what <span class="math-container">$| g A_n \Delta A_n|$</span> equals for large <span class="math-container">$n$</span>. Also, I have <span class="math-container">$| A_n| = n$</span>. </p>
Answer:
|
https://math.stackexchange.com/questions/3526927/how-to-show-this-is-a-folner-sequence
|
Question: <p>Let U be the submodule of the group algebra <span class="math-container">$ℂ[S_3]$</span> generated by</p>
<p><span class="math-container">$u_1 := 1−(2 1) + (3 2)−(3 1 2)$</span></p>
<p>For all <span class="math-container">$τ ∈ S_3$</span>, let <span class="math-container">$U_τ$</span>
:= {<span class="math-container">$τuτ^{-1} ∈ ℂ[S_n] | u ∈ U$</span>}</p>
<p>How would I prove that <span class="math-container">$U ⊕U_{(2 1)} = U ⊕U_{(3 1)}$</span> but</p>
<p><span class="math-container">$U_{(2 1)} ≠ U_{(3 1)}$</span></p>
<p>I have no idea how to go about doing this and I don't really know what the direct sum means in this example. Any help would be great!</p>
Answer:
|
https://math.stackexchange.com/questions/3528930/prove-that-u-%e2%8a%95u-2-1-u-%e2%8a%95u-3-1
|
Question: <p>It's easy to construct a countable series of distinct groups - the cyclic groups, for instance - and it's also easy to create a family of groups parametrized by the reals, but most such constructions will have isomorphisms betwteen most if not all of the groups.</p>
<p>As the category Group is very large, one would expect that somewhere inside it lies a continuum-sized collection of groups which can be parametrized in some natural way and are all distinct, that is, nonisomorphic. However, I've pondered this for a bit and asked some friends and haven't come up with any clear constructions. So I'm curious if anyone can provide a collection of groups which is "natural" in some sense (ideally continuous, and not a composition of a bijection from a set of groups to <span class="math-container">$\{0,1\}^{\mathbb{N}}$</span> and from there to <span class="math-container">$\mathbb{R}$</span>), continuum-sized, and whose elements are not isomorphic. </p>
<p>(Answers which provide a family parametrized by the nonnegative reals, or the positive reals, or the interval <span class="math-container">$[0,1]$</span>, would also suffice.)</p>
Answer: <p>As proved in <a href="https://math.stackexchange.com/questions/3831720/are-these-permutation-groups-defined-by-asymptotic-properties-isomorphic?noredirect=1&lq=1">this</a> question, the groups <span class="math-container">$G_{x^r}$</span> for <span class="math-container">$r\in (0,1]$</span> (defined as permutations of <span class="math-container">$\mathbb{N}$</span> that fix all but <span class="math-container">$o(n^r)$</span> elements) are pairwise non-isomorphic.</p>
|
https://math.stackexchange.com/questions/3546089/is-there-a-natural-family-of-nonisomorphic-groups-parametrized-by-mathbbr
|
Question: <p>On page 40 of the first edition of Artin, he writes: </p>
<blockquote>
<p>Going back to a general law of composition, suppose we want to define a product of a string of <span class="math-container">$n$</span> elements of a set:
<span class="math-container">$$a_1 a_2 \ldots a_n = ?$$</span>
There are various ways to do this using the given law, which tells us how to multiply two elements. For instance, we could first use the law to find the product <span class="math-container">$a_1 a_2$</span>, then multiply this element by <span class="math-container">$a_3$</span>, and so on:
<span class="math-container">$$((a_1 a_2)a_3)a_4 \ldots.$$</span>
<strong>When <span class="math-container">$n = 4$</span>, there are four other ways to combine the same elements; <span class="math-container">$(a_1 a_2)(a_3 a_4)$</span> is one of them.</strong></p>
</blockquote>
<p>My confusion is on the bold statement. I do not see where Artin gets that there are five ways to combine the numbers <span class="math-container">$a_1, \ldots a_4$</span>. I get, for example:
<span class="math-container">\begin{align*}
& ((a_1 a_2)a_3)a_4 \\
& ((a_2 a_1)a_3)a_4 \\
& ((a_3 a_1))a_2)a_4 \\
& ((a_3 a_1))a_4)a_2 \\
& \vdots \\
& (a_3 a_4)(a_1 a_2) \\
& (a_1 a_4)(a_2 a_3) \\
& \vdots
\end{align*}</span>
and so on. There are too many to write out on this page. </p>
<p>Where does Artin get that there are five ways? Am I misinterpreting him? </p>
Answer: <p>Artin is not commuting the elements, he is only changing the location of the parentheses.</p>
|
https://math.stackexchange.com/questions/3550592/artin-on-the-associative-law-of-n-elements
|
Question: <p>I am studying about subgroup. My definition of subgroup is that:</p>
<p>Let a set <span class="math-container">$G$</span>, with a binary operation<span class="math-container">$
×:G×G→G,(a,b)↦×(a,b)=:a×b$</span> be a group. Then <span class="math-container">$H⊂G$</span> is a subgroup iff <span class="math-container">$H$</span> with a restriction of <span class="math-container">$×$</span> to <span class="math-container">$H×H$</span>, that is,<span class="math-container">$×|_{H×H}$</span> is also a group.</p>
<p>And my book states that if <span class="math-container">$H$</span> is a subgroup of a group <span class="math-container">$G$</span>, then <span class="math-container">$1_H=1_G$</span>. Why is this true?</p>
Answer: <p>Here are two ways to see this:</p>
<ol>
<li><p>If <span class="math-container">$h\in H$</span>, then <span class="math-container">$h=h1_H = h1_G$</span>. From <span class="math-container">$h1_H=h1_G$</span>, multiply by <span class="math-container">$h^{-1}$</span> on the left to get <span class="math-container">$1_H=1_G$</span>.</p>
<p>Note that the final product happens in <span class="math-container">$G$</span>, not in <span class="math-container">$H$</span>. As you note in comments, if we do not know yet that <span class="math-container">$1_H=1_G$</span>, then we don't know that the inverse of <span class="math-container">$h$</span> in <span class="math-container">$H$</span> is necessarily the same as the inverse of <span class="math-container">$h$</span> in <span class="math-container">$G$</span>; but we still have <span class="math-container">$h1_H=h1_G$</span> <em>in <span class="math-container">$G$</span></em>, from which we get <span class="math-container">$1_H=1_G$</span> as elements of <span class="math-container">$G$</span>.</p>
</li>
<li><p>In <span class="math-container">$G$</span>, the only element that satisfies <span class="math-container">$xx=x$</span> is <span class="math-container">$x=1_G$</span>. Since <span class="math-container">$1_H1_H=1_H$</span>, and this holds inside <span class="math-container">$G$</span> as well, that means <span class="math-container">$1_H=1_G$</span>.</p>
</li>
</ol>
|
https://math.stackexchange.com/questions/3591734/why-is-it-true-that-if-h-is-a-subgroup-of-a-group-g-then-1-h-1-g
|
Question: <p>Øystein Ore in 'Some remarks on commutators' proof that:</p>
<p>''any element of the alternating group <span class="math-container">$A_n$</span> with n ≥ 5 is a commutator of two elements''.</p>
<p>I quest the case <span class="math-container">$n=4$</span>. I reasoned that:</p>
<p>1) <span class="math-container">$A_4$</span> is costituited from identity <span class="math-container">$1$</span>, 3 permutation of type <span class="math-container">$(i,j)(h,k)$</span> [i,j,h,k distinct] and 8 of type <span class="math-container">$(i,j,h)$</span>.</p>
<p>2) <span class="math-container">$[(i,j),(i,k)]=(i,k,j)$</span>.</p>
<p>3) <span class="math-container">$[(i,j,k),(i,k,l)]=(i,l)(j,k)$</span>.</p>
<p>Therefore ''any element of the alternating group <span class="math-container">$A_4$</span> is a commutator of two elements''.
Is correct?</p>
<p>Futhermore the cases <span class="math-container">$n=2,3$</span> appear obvious.
Therefore:</p>
<p>''any element of the alternating group <span class="math-container">$A_n$</span> with n ≥ 2 is a commutator of two elements''.</p>
<p>I ask a confirm, please.</p>
Answer:
|
https://math.stackexchange.com/questions/3591908/commutator-element-in-permutation-group
|
Question: <p>As order of <span class="math-container">$Aut(G)$</span> is prime no. Then this implies <span class="math-container">$Aut(G)$</span> is cyclic this means <span class="math-container">$Aut(G)$</span> is abelian this implies inner automorphism group is also cyclic, as cyclic subgroup of cyclic group is cyclic hence as <span class="math-container">$Inn(G)$</span> is isomorphic to <span class="math-container">$G/Z(G)$</span>.
And as <span class="math-container">$G/Z(G)$</span> is cyclic therefore <span class="math-container">$G$</span> is abelian.
Then how can we say group order cannot exceed <span class="math-container">$3$</span>. </p>
Answer: <p>Some hints: since <span class="math-container">$G$</span> must be abelian, the map <span class="math-container">$g \mapsto g^{-1}$</span> gives rise to an automorphism of order <span class="math-container">$2$</span>. Hence either this map is the trivial one, that is <span class="math-container">$g^2=1$</span> for all <span class="math-container">$g \in G$</span>, and <span class="math-container">$G$</span> must be a direct product of copies of <span class="math-container">$C_2$</span>. Or <span class="math-container">$p=2$</span>. Can you finish?</p>
|
https://math.stackexchange.com/questions/3618620/let-g-be-finite-group-and-order-of-autg-is-prime-number-p-then-prove-tha
|
Question: <p>In my book, "Elements of Modern Algebra, 7th ed.- /Gilberts"</p>
<p>Characterization of a Subring is given as following (that is, conditions to be a subring of a ring <span class="math-container">$R$</span>)</p>
<blockquote>
<p>A subset <span class="math-container">$S$</span> of a ring <span class="math-container">$R$</span> is a subring of <span class="math-container">$R$</span> if and only if these conditions are satisfied:</p>
<ul>
<li><span class="math-container">$S$</span> is not empty</li>
<li><span class="math-container">$x,y\in S\Rightarrow x-y,xy\in S$</span></li>
</ul>
</blockquote>
<p>in the second condition if we do not check if <span class="math-container">$yx$</span> is in <span class="math-container">$S$</span>, <em><strong>will there be a problem?</strong></em> Because I guess there might be a counter-example considered with matrix groups or something.</p>
Answer: <p>We actually check both $xy$ and $yx$ with this axiom, but it isn't that apparent.</p>
<p>Look closely on what's said:</p>
<p>$$\forall x,y \in S \quad x-y\in S, xy \in S$$</p>
<p>See that "for all" symbol? It means that the same should be true if we swap $x$ and $y$! So, if $a,b \in S$, we have</p>
<ul>
<li>$ab \in S$, if we take $x=a$ and $y=b$, and</li>
<li>$ba \in S$, if we take $x=b$ and $y=a$.</li>
</ul>
|
https://math.stackexchange.com/questions/2768948/characterization-of-a-subring-shouldnot-we-concern-ab-and-ba-simultaneously
|
Question: <p>Let $G$ be a group, $N\triangleleft G$ an index-two (normal) subgroup, and $H_1,H_2<G$ two subgroups.
Is it true that
$$H_1\cap N = H_2\cap N \Rightarrow H_1 = H_2\ ?$$
If no, is it true with the extra hypothesis that $H_2=gH_1g^{-1}$ for some $g\in G$?</p>
<p>Proofs or couterexamples would be appreciated!</p>
Answer: <p>It's not true, even with the additional hypothesis.</p>
<p>For a counterexample, take $G = S_3$.</p>
<p>Let $N$ be the subgroup of order $3$, which is normal because its index is $2$.</p>
<p>Let $H_1$ and $H_2$ be two of the three subgroups of order $2$. Then $H_1$ and $H_2$ are conjugate because they are Sylow (or use the fact that any elements with the same cycle structure are conjugate).</p>
<p>We have $H_1 \cap N = H_2 \cap N = 1$ by Lagrange, but $H_1 \neq H_2$.</p>
|
https://math.stackexchange.com/questions/2735423/index-two-subgroup-implies-full-group
|
Question: <p>What is the order of element 2 in group ($\Bbb{Z^*_{47}}$,x) ?</p>
<p>Got no clue from where to start any help would be great</p>
Answer: <p>Let us use that computer help. This and similar problems can be best covered by using computer algebra systems, like <a href="http://www.mathsage.org" rel="nofollow noreferrer">sage</a>. So here is the sage dialog, then there will come come comments:</p>
<pre><code>sage: 47.is_prime()
True
sage: F = GF(47)
sage: G = F.unit_group()
sage: G.order()
46
sage: G.is_cyclic()
True
sage: F(2).multiplicative_order()
23
</code></pre>
<p>Comments: First of all, we note that $47$ is a prime number. Then $\Bbb Z/47$ (integers taken modulo $47$) is a field, "the" field $F=\Bbb F_{47}$ with $47$ elements. There is a known structure theorem, its unit group $F^\times$ (i.e. $F$ "without zero", with the multiplication inherited from $F$) is a cyclic group. This group has $47-1=46$ elements. Due to a theorem of Lagrange, the order of each element in this (or another finite) group divides the order of the group.</p>
<p>So the element $2$ (modulo $47$) has a multiplicative order among the divisors of $46$. These divisors are $1,2,23,46$. One has to exclude $1,2,46$. The $1$, $2$ are immediately excluded, $2^1$ and $2^4$ are not $1$ modulo $47$. The argument in the comments also exclude the $46$. In the following way. (If one did not see this before, then it is didactically better to give the answer.)</p>
<pre><code>sage: sqrt(F(2))
7
</code></pre>
<p>There is a "square root" of $2$ in the field, $7^2 =49=47+2=0+2=2$ in $F$. This $7$ also has a multiplicative order, which divides $46$. It is $23$ or $46$. (It is $23$, but we do not need this.) In both cases, the order of $2=7^2$ is $23$.</p>
|
https://math.stackexchange.com/questions/2763690/order-of-element-2-in-group-bbbz-47-under-multiplication
|
Question: <blockquote>
<p>Let $\langle G,\ast \rangle$ be a group and $X$ be a set. We define $F$ to be the set of all the functions from $X$ to $G$, meaning $F=\{ f\mid f:X \rightarrow G\}$.<br>
We define the operation $\bullet$ on the elements of $F$ by the following:
$$\forall f,g \in F, \forall x \in X: (f \bullet g)(x)=f(x)\ast g(x).$$
Prove that $\langle F,\bullet \rangle$ is a group.</p>
</blockquote>
<p>I am a bit confused on how to prove <em>closure</em>, since the outcome of $(f \bullet g)(x)$ would be in $G$.</p>
<p>I hope that I am not missing something basic here, full solution, partial solution or even a hint on solving this question would be appreciated.</p>
Answer:
|
https://math.stackexchange.com/questions/2807293/prove-that-langle-f-bullet-rangle-is-a-group-where-f-f-mid-fx%e2%86%92g-and
|
Question: <p>Question: Consider the following presentation $G=\langle a,b|aba^{-1}b^{-1} \rangle$. Does there exist a Van Kampen diagram over the presentation whose boundary label is the word $w=a^{2}ba^{-1}b^{-2}a^{-1}b$?</p>
<p>What I tried: First, the relators are $aba^{-1}b^{-1}$, $bab^{-1}a^{-1}$ and all their cyclic permutations. rewriting the word w in its full form gives $w=aaba^{-1}b^{-1}b^{-1}a^{-1}b$. Since the words $aba^{-1}b^{-1}$ and $ab^{-1}a^{-1}b$ are relators, they are equal to the identity 1 of G. It follows that $w=a(aba^{-1}b^{-1})b^{-1}a^{-1}b=1.1$ which is just the identity 1 in G and hence $w=1$ in G. By the Van Kampen diagram, yes there is a Van Kampen diagram over G whose boundary label is w. Can someone please help and see if the approach I used is the correct one? Subsidiary question: Since $w=1$ in G, can we say that the word $w=a^{2}ba^{-1}b^{-2}a^{-1}b$ is a relator? Also, anyone knows where (documentation, book, etc) I can solve more of these similar problems?</p>
<p>I appreciate your help.</p>
Answer:
|
https://math.stackexchange.com/questions/2808012/van-kampen-diagram-van-kampen-lemma
|
Question: <p>Apparently $\langle t_L,t_{L/2}r_h, r_v \rangle $ is equal to $ \langle t_L,t_{L/2}r_h,r_hr_v \rangle $ where each of these groups is described by a set of generators with $t_L$ representing a translation by a distance L, $t_{L/2}r_h$ is a glide reflection, $r_v$ is vertical reflection and $r_hr_v$ is a rotation, but I'm having a hard time seeing how these two are equal. </p>
<p>I tried some things... but apparently $\langle t_L,t_{L/2}r_h, r_v \rangle $ is also equal to $\langle t_L,t_{L/2}r_h,t_{L/2}r_v,r_hr_v \rangle$ which I then see how $\langle t_L,t_{L/2}r_h, r_v \rangle $ is equal to $ \langle t_L,t_{L/2}r_h,r_hr_v \rangle$ because the composition of $ t_{L/2}r_h$ and $r_hr_v$ is $t_{L/2}r_v$. But I don't understand how to get $\langle t_L,t_{L/2}r_h,t_{L/2}r_v,r_hr_v \rangle$.</p>
<p>Can someone please help me understand/demonstrate how $\langle t_L,t_{L/2}r_h, r_v \rangle $ is equal to $ \langle t_L,t_{L/2}r_h,r_hr_v \rangle $</p>
Answer:
|
https://math.stackexchange.com/questions/2764504/equal-generator-sets
|
Question: <p>I have some issues visualizing how to compute quotient of subgroups in general, safe making explicit the cosets which is not really satisfactory nor geometrically meaningful.</p>
<p><strong>Here is a specific instance.</strong> Let $k$ and $K$ two fields, $K$ extending $k$. This could also be rings, like $\mathbb{Z}$ and $n\mathbb{Z}$ Consider $GL(2)$ and its unipotent subgroup
$$U = \left( \begin{matrix} 1 & *\\ & 1 \end{matrix} \right) \subset GL(2)$$</p>
<p>I want to compute $U(K) / U(k)$. This is easy because of the identification of $U$ with the multiplicative subgroup, giving a quotient isomorphic to $K/k$. However, this example is too particular.</p>
<p><strong>Another instance.</strong> Consider the similar question in $GL(3)$ now. An take the unipotent subgroup
$$U_1 = \left( \begin{matrix} 1 & \alpha x & \beta y \\ & 1 & \gamma z \\ & & 1 \end{matrix} \right)_{\{x, y, z \in k\}} \subset GL(2, k)$$</p>
<p>for fixed $\alpha, \beta, \gamma \in k$, and also the whole unipotent subgroup
$$U = \left( \begin{matrix} 1 & * & * \\ & 1 & * \\ & & 1 \end{matrix} \right)_{\{x, y, z \in K\}} \subset GL(3, K)$$</p>
<blockquote>
<p>How do I compute the quotient $U / U_1$? (even in the case $k=K$)</p>
</blockquote>
<p>Indeed in this case, there is no obvious identification with product of simple multiplicative groups.</p>
Answer:
|
https://math.stackexchange.com/questions/2754340/on-quotients-of-unipotent-subgroups
|
Question: <blockquote>
<p>Let G = $<a>$ be a cyclic group of order n. Prove that the cyclic subgroup generated by $a^m$ is the same as the cyclic subgroup generated by $a^d$, where d = (m, n)</p>
</blockquote>
<p>The book said it suffices to show that $a^d$ is a power of $a^m$. <br>
I proved it using d = mu + nv. But I don't know why it suffices</p>
Answer: <p>If $a^d$ is a power of $a^m$, then $a^d\in \langle a^m\rangle$, and hence $\langle a^d \rangle \subseteq \langle a^m \rangle$.</p>
<p>On the other hand, $a^m$ is clearly a power of $a^d$, so by a similar argument $\langle a^m \rangle \subseteq \langle a^d \rangle$.</p>
|
https://math.stackexchange.com/questions/2809732/cyclic-subgroup-generated-by-am-is-the-same-as-the-cyclic-subgroup-generated
|
Question: <p>Let $K$ be an algebraically closed field of characteristic $p>0$ and let
$n$ be a power $p$. In this case, the abstract groups $SL_{n}(K)$ and $PGL_{n}(K)$ are isomorphic under the natural map $\alpha:SL_{n}(K)\to PGL_{n}(K)$.</p>
<p><strong>Question 1.</strong> Are they also isomorphic as algebraic groups?</p>
<p>If the answer is "no", then I have another question.
Since $\alpha$ is an isogeny (in particular, a morphism), its inverse $\alpha^{-1}$ must not be a morphism
of algebraic varieties.</p>
<p><strong>Question 2.</strong> Can it be seen directly why $\alpha^{-1}$ is not a morphism?</p>
Answer:
|
https://math.stackexchange.com/questions/2810059/isomorphy-of-abstract-vs-algebraic-groups
|
Question: <p>Let $G$ be a group and let $g\in G$. Prove that if $g\neq 1$, then there exists a subgroup of $G$ which is maximal with respect to the property of not containing $g$.</p>
<p>I built a chain of subgroups $M_\lambda$ for $\lambda\in \Lambda$ (where each subgroup not contain $g$) partially ordered by inclusion, i try to use the <strong>Zorn's lemma</strong>
but since $G$ can be an infinite group, i don't know how to go on with the proof. Any ideas or hints? Thank you</p>
Answer: <p>You shouldn't index by $\Bbb N$ since the chain may be uncountably infinite.</p>
<hr>
<p>To use Zorn's lemma, you show that there is a subgroup not containing $g$, and you show that every chain of subgroups not containing $g$ has its union also being a subgroup not containing $g$.</p>
|
https://math.stackexchange.com/questions/2810304/existence-of-maximal-subgroup-of-g-with-the-property-of-not-containing-g-in-g
|
Question: <p>In the proof that the multiplication defined on HK (H and K being groups) is associative (Dummit and Foote p. 176) I do not understand the following step (the dot operation is the action of H on K, a, b and c are elements of H and x, y and z are elements of K):
(a x.b x.(y.c), xyz) = (a x.(b y.c), xyz)</p>
<p>Is it true in general that when a group H acts on a group K, the action (dot operation) preserves the group operation, i.e. h.(kl) = h.k h.l? Why so?</p>
Answer: <p>One general defining principle of group actions is that for any action of a group $G$ on a mathematical object $X$, the action preserves all of the mathematical structure on $X$. (This is formalized using category theory, but I won't go there with this answer... well, I did just go there a tiny bit...)</p>
<p>So, for example, suppose that $X$ is itself a group, so it has a binary group operation and a unary inversion operation. Then an action of $G$ on $X$ preserves both of those operations. </p>
<p>There are many, many other examples of this principle. Another is the concept of a linear action, also known as a "linear representation", meaning an action of a group $G$ on a vector space $V$. The vector space has a binary addition operation, and it has a scalar multiplication operation where the scalars are chosen from some field such as the real numbers, and the action of $G$ on $V$ is required to preserve both of those operations.</p>
|
https://math.stackexchange.com/questions/2791868/group-action-on-another-group-preserves-the-latters-group-operation
|
Question: <p>Is finiteness a Quasi-Isometric invariant property?!
i.e.
Let $G$,$H$ be two groups which $G$ is finite and $G\sim_{QI} H$, is $H$ finite?!</p>
Answer: <p>A f.g. group $G$ is quasi-isometric to the trivial group iff every $g\in G$ has bounded length (with respect to some generating set), i.e., iff $G$ is finite.</p>
|
https://math.stackexchange.com/questions/2825097/quasi-isometry-and-finiteness
|
Question: <p>Let $n=2^7 \cdot 3^5 \cdot 11^3 \cdot 35$. In how many ways can the cyclic group $C_n$ can be written as a direct product of two or more nontrivial groups? List all these direct products.</p>
<p>Can someone guide me how to do this question please. I am not looking for a straight answer obviously. </p>
<p>Also, I know what cyclic groups are when its like for $\mathbb{Z}_4$ for example but what does it mean by $C_n$ in this case?</p>
Answer: <p>You can think of $C_{n}$ to be $\mathbb{Z}_{n}$; they are both cyclic groups of the same order.</p>
<p>I think this question wants you to use the fact that $\mathbb{Z}_{km} \cong \mathbb{Z}_{k} \times \mathbb{Z}_{m}$ iff $\gcd(k,m) = 1$ for $k,m\in\mathbb{Z}_{\geq 2}$. Using this you should be able to decompose $\mathbb{Z}_{2^7\cdot 3^5 \cdot 11^3 \cdot 35}$ into several direct products of smaller cyclic groups.</p>
<p>For example, $\mathbb{Z}_{2^7\cdot 3^5} \cong \mathbb{Z}_{2^7} \times \mathbb{Z}_{3^5}$, and $\mathbb{Z}_{35} = \mathbb{Z}_{5\cdot 7} \cong \mathbb{Z}_{5}\times \mathbb{Z}_{7}$.</p>
|
https://math.stackexchange.com/questions/1502639/direct-product-of-nontrivial-groups
|
Question: <p>Continuing for my study on practicing group theory...
I am now stuck on this problem about composition factors,</p>
<blockquote>
<p>$G$ is a group such that $|G|=p^2q$ where $p \neq q$ and $p,q$ are prime. Prove that the composition factors of $G$ are $C_p,C_p,C_q$ in some order.</p>
</blockquote>
<p>Well, Burnside's theorem tells me that such a $G$ is soluble as $2,1$ are not negative. But I'm stuck with proceeding to the "main" bit, namely to show that the factors are specifically $C_p,C_p,C_q$ in some order.</p>
<p>Does anyone know how to go about showing this? And in what order it is?</p>
<p>Help is very much appreciated, thank you in advance.</p>
Answer:
|
https://math.stackexchange.com/questions/1555306/application-of-burnsides-theorem-composition-factors-of-g-p2q-are
|
Question: <p>I'm trying to prove that if $G$ is an Abelian group under $\cdot$, $\forall a,b \in G. \forall z \in \mathbb{Z}. (a \cdot b)^n = a^n \cdot b^n.$ I was originally considering doing this problem using an AFSOC, but I realized that originally assuming that $(a \cdot b)^n \neq a^n \cdot b^n$ would be rather difficult to contradict with the given properties of an Abelian group. Thus, I considered inducting in two ways on $n$, first going through all the positives and then going through all the negatives in the integers. However, I'm worried about what I do with the case of $n = 0.$ I understand that in algebraic terms $a^0$ is an abbrevation for ``$a \cdot a \cdot ... \cdot a$ with 0 many $a$'s'', but I am confused as to what this represents in the Abelian group $G.$ Perhaps I need to prove some properties of $a^0$ for all $a \in G$?</p>
Answer: <p>$G$ is an abelian group, so let $a,b\in G$ be given, and fix $n$. Then
$$ (ab)^n=(ab)(ab)(ab)\cdots(ab).$$
Because $G$ is abelian
$$ (ab)(ab)(ab)\cdots(ab)=(aa\cdots a)(bb\cdots b)=a^nb^n.$$
This follows naturally from the definition of commutativity.</p>
|
https://math.stackexchange.com/questions/1555799/properties-of-abelian-groups
|
Question: <p>I've been starting to play around with some properties of groups, and I wanted to prove this claim to make things simpler for me in the future. If $G$ is closed under and operation $\cdot$ that is also associative, I want to show that $\forall a_1, a_2, ..., a_n \in G$, no matter how I choose to bracket $a_1 \cdot a_2 \cdot ... \cdot a_n$ while retaining the order, I will end up with the same element in $G.$ My friend told me that induction on the number of terms would probably be my best bet, and I see how the $n = 1 \wedge n = 2$ cases are trivial with the $n = 3$ case implied by associativity. However I am having some problems Figuring out how to work out the induction step. It seems to me that my induction hypothesis is not helping me when proving for brackets that include the $n+1$th element together with the $n$th element.</p>
Answer: <p>It's pretty messy and most books just say "it's obvious". If you do want do do it, probably the easiest way is to be specific, with something like this.</p>
<p><strong>Theorem</strong>. A product of terms $a_1,\ldots,a_n$, in that order, with any order of bracketing, is equal to
$$(\cdots((a_1*a_2)*a_3)*\cdots*a_{n-1})*a_n\ .$$</p>
<p><strong>Sketch of proof</strong>. In a given product, identify the "last" $*$ to be performed. Then the product looks like
$$\eqalign{
(\cdots)*(\cdots*a_n)
&=(\cdots)*((\cdots)*a_n)\qquad\hbox{[using inductive assumption for the}\cr
&\qquad\qquad\qquad\qquad\qquad\hbox{number of terms in the second bracket]}\cr
&=((\cdots)*(\cdots))*a_n\qquad\hbox{[associative law for three terms]}\cr
&=(\cdots(a_1*a_2)*\cdots)*a_n\qquad\hbox{[inductive assumption for $n-1$ terms]}\cr}$$</p>
|
https://math.stackexchange.com/questions/1555868/induction-with-an-associative-operator
|
Question: <p>I have a problem on the notion of FCelement.</p>
<p>let $G$ be a group and $a\in G$ of finite order, so $\langle a \rangle$ is an FC group.</p>
<p>My qeustion is : Why $a$ could be a non FC element in $G$? </p>
Answer: <p>Consider the group of all permutations of natural numbers, $G=S_\mathbb{N}$. Let $a$ be the permutation swapping $1$ and $2$ and fixing all other numbers. Then:</p>
<ul>
<li><p>$a$ has order $2$; but</p></li>
<li><p>$a$ has infinitely many conjugates (in particular, $a$ is conjugate to every <em>other</em> permutation which swaps two elements and preserves all others).</p></li>
</ul>
|
https://math.stackexchange.com/questions/1590843/about-the-concept-of-an-fc-element
|
Question: <p>Good day all,</p>
<p>Wikipedia states:
There are 2328 groups of order 128 up to ismorphism. How is this calculated? Also, what does "up to isomorphism" mean? </p>
<p>(I know what an isomorphism is...i'm just unsure of the phrasing "up to...")
Thanks!</p>
Answer: <p>The groups $(\Bbb Z/5\Bbb Z)^\times$ and $\Bbb Z/4\Bbb Z$ are different, but they are isomorphic. If you literally counted all the groups of a given order, that would be like counting sets of a given order - given any cardinal numbers $\kappa$ and $\lambda$ one can exhibit $\kappa$-many sets of size $\lambda$ that are all different, so this is a pointless question to ask because there is no limit to how many sets of some size there can be. Instead, when we count groups we count isomorphic groups as the same.</p>
<p>Anyway, we use the fundamental theorem of finite abelian groups. If $n$ is a whole number with prime factorization $\prod p^e$, and $f(n)$ denotes the number of abelian groups of size $n$, then we can say that $f(n)=\prod f(p^e)$, and moreover that $f(p^e)$ equals the number of integer partitions of the integer $e$. In this case, $128=2^7$ and there are $15$ integer partitions of $7$, so there are $15$ abelian groups of order $2^7$.</p>
<p>Note that $2328$ counts the number of finite groups of order $2^7$, not the number of finite abelian groups of that order. Counting those is very nontrivial.</p>
|
https://math.stackexchange.com/questions/1534592/counting-the-number-of-abelian-groups-of-a-certain-order
|
Question: <p>How do I prove that any finite subgroup of $SO(2)$ must be cyclic? </p>
<p>Also, what are all the finite subgroups of $O(2)$?</p>
Answer: <p>I am not quite sure what you mean by "prove" in this case, but $SO(2)$ is the group of rotations in the plane. A finite subgroup must be the group of rotations by an integral fraction of $2\pi$, and will always be a cyclic group because you can see rotations as cycles.</p>
<p>Take, for example, any regular $n$-gon centered at the origin in the plane and number its vertices consecutively from $1$ to $n$. Then a rotation by $2\pi/n$ will take $1$ into $2$, $2$ into $3$, etc, in a cyclic fashion.</p>
<p>The group $O(2)$ includes rotations and reflections. Its finite subgroups will be dihedral groups.</p>
|
https://math.stackexchange.com/questions/1571081/proof-that-any-finite-subgroup-of-so2-must-be-cyclic
|
Question: <p>This may be a stupid question, but </p>
<p>let's consider the cyclic group $G=(\mathbb{Z}/10\mathbb{Z},+)=\{0,1,2,3,4,5,6,7,8,9\}$.</p>
<p>By Lagrange's Theorem this group can only have subgroups of order $2$ or $5$, since its order is $10$.</p>
<p>So for example the group $H=(\mathbb{Z}/3\mathbb{Z},+)=\{0,1,2\}$ is not a subgroup of $G$, since its order is $3$ and $3$ is not a divisor of $10$.</p>
<p>But it is closed under the group operation, has the neutral element $0$ of $G$ and every element has its inverse, so the subgroup properties are there.</p>
<p>Clearly, I am making some kind of mistake here, but I can't see it.</p>
Answer: <p>For $H$ to be a subgroup of $G$ it must not only be the case that every element of $H$ is in $G$ but also that <em>the group operation is the same</em>.</p>
<p>This is not the case here -- for example, we have $2+2=4$ in $G$, but $2+2=1$ in $H$.</p>
<p>So the group operation on $H$ is not just the restriction of $G$'s group operation to the elements of $H$, and therefore $H$ is not a subgroup.</p>
|
https://math.stackexchange.com/questions/1535672/why-is-h-not-a-subgroup-of-g
|
Question: <p>An amenable group is at most contable group $G$ for which exist a sequence $\{F_n\}_{n\in \mathbb{N}}$ of finite sets such that
$$\displaystyle\lim_{n\rightarrow \infty}\frac{|(g\cdot F_n)\triangle F_n|}{|F_n|}=0 $$
for every $g\in G$.</p>
<p>There is something that I am not understanding about this definition. The "almost trvial" example is $\mathbb{Z}$. Consider the sequence of sets of the form $[-N,N]=\{-N,-N+1,...,N-1, N\}$.</p>
<p>I tried to prove that indeed this sequence shows that $\mathbb{Z}$ is amenable. </p>
<p>But if we do the calculation $[-N,N]\triangle [-N+a,N+a]|=|[-N,N]\cup [-N+a,N+a]|=4N$ for every $a$ big enough and $|[-N,N]|=2N$ but this implies that</p>
<p>$$\displaystyle\lim_{n\rightarrow \infty}\frac{|[-N,N]\cup [-N+a,N+a]|}{|[-N,N]|}=2 $$</p>
<p>What am I doing wrong?</p>
<p>Thanks!</p>
Answer: <p>If you set $F_N := [-N, N]$, then for big enough $N$ you will have that $a F_N \Delta F_N$ has $2 |a|$ elements ($a$ is fixed!!), while $F_N$ has <em>always</em> $2N + 1$ elements (don't forget to count $0$). So for $N \to \infty$ this goes to $0$ as desired.</p>
|
https://math.stackexchange.com/questions/1593043/about-the-definition-of-amenable-group
|
Question: <blockquote>
<p>Show that $D_{2n}$ has two conjugacy classes if $n$ is even, but only one if $n$ is odd.</p>
</blockquote>
<p>My questions:</p>
<ol>
<li><p>What is meant by 'two conjugacy classes of reflections'</p></li>
<li><p>How is this question related to whether $n$ is even or odd</p></li>
</ol>
<p>I am thinking about orbit-stabiliser theorem, and find that, if $D_{2n}=<r,s|r^n=e=s^2,sr=r^{-1}s>$, $R_n=<r>$ and $K \leq R_n$, then$$|\text{ccl}(s)|=n$$ when $n$ is odd, and $$|\text{ccl}(s)|={n \over 2}$$ when $n$ is even.</p>
<p>However I cannot see how is this result related to the question, can someone please give me a hint?</p>
Answer: <p>You've done almost all the work you need to do, it's just a matter of relating it to the goal.</p>
<p>You know that, no matter whether $n$ is even or odd, $D_{2n}$ has exactly $n$ reflections. You should also be able to show that the only elements conjugate to a reflection are themselves reflections. So if we let $s$ be a fixed but arbitrary reflection and $S$ the set of reflections, make sure you mention/show that $\operatorname{ccl}(s) \subseteq S$.</p>
<p>When $n$ is odd, you've shown correctly that $\lvert \operatorname{ccl}(s) \rvert = n$ while $\operatorname{ccl}(s)$ must consist entirely of reflections, of which there are only $n$, and we must have $\operatorname{ccl}(s) = S$.</p>
<p>A very similar argument happens when $n$ is even. Still $\operatorname{ccl(s)} \subseteq S$ for any reflection $s \in S$, and since each reflection is conjugate to exactly $\frac{n}{2}$ others, $S$ must be the disjoint union $S = \operatorname{ccl}(s) \sqcup \operatorname{ccl}(t)$ for two rotations $s, t \in S$, purely for counting/elementary set theoretical reasons.</p>
<p>And it helps to keep the geometry in mind.
<a href="https://i.sstatic.net/XHvUM.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/XHvUM.png" alt="enter image description here"></a></p>
<p>It's pretty clear that, geometrically, there are "two kinds" of reflections for a square, and the cool thing is that this is reflected in the group structure of $D_{8}$. But if you think about, say, a pentagon, every reflection axis goes from a vertex to the midpoint of the opposite edge, which causes all reflections to be conjugate in $D_{10}$.</p>
|
https://math.stackexchange.com/questions/1559139/show-that-d-2n-has-two-conjugacy-classes-of-reflections-if-n-is-even-but
|
Question: <p>Calculate how groups up isomorphisms exist of order 88 such that has least one element of order 8.</p>
<p>I have two groups: $\mathbb{Z}_{88}$ (abelian) and $\mathbb{Z_{11}}\rtimes_{\phi}\mathbb{Z}_8$, where $\phi:\mathbb{Z}_{8}\rightarrow Aut(\mathbb{Z}_{11})$, defined by $y\mapsto \phi(y)(a)=a^{10}$, $a\in \mathbb{Z}_{11}$.</p>
<p>There are other groups?</p>
<p>Thank you by some hints.</p>
Answer: <p>The number $n_{11}$ of 11-Sylow subgroup should satisfy $n_{11} \equiv _{11} 1$ and $n_{11} \mid 88/11=8$ so you must have that $n_11=1$, or equivalently there is only 1 normal 11-Sylow subgroup. Similarly, the number of 2-Sylow group is odd and divides $88/8=11$ so it is either 1 or 11. </p>
<p>If there is one 2-Sylow subgroup, then it is normal and you get that $G=P_{11}\times P_8$ (since $P_{11},P_8$ are both normal, their intersection is $\{e\}$ and generate $G$. Since 11 is prime you must have that $P_{11}$ is cyclic, and there are exactly 5 groups of order 8 (3 abelian groups, the quaternions and dihedral). So you already have 5 non isomorphic groups.
EDIT: just noticed the condition about element of order 8, so this group must be the cyclic group of order 8</p>
<p>Assume now that there is more then one 2-Sylow subgroup. Since $P_{11}$ is normal in $G$, you still get a semidirect product $P_{11}\rtimes P_8$, so you need to find all the action of a group of order 8 on a cyclic group of order 11. Any such action is defined by a homomorphism $\psi:P_8\to Aut(\mathbb{Z}_{11})$. You can now use the fact that $Aut(\mathbb{Z}_{11})\cong \mathbb{Z}_{10} \cong \mathbb{Z}_5 \times \mathbb{Z}_2$. Since 8,5 are coprime, the image of $P_8$ must be contained in the $\mathbb{Z}_2$ part. You can now go over all the group of order 8 and find all the homomorphisms to $\mathbb{Z}_2$ (namely, find the subgroup of index 2), and try to understand what is the group.
EDIT: Since $P_8$ must be cyclic, it has exactly 1 (nontrivial) such homomorphism, so there is exactly one non abelian group with element of order 8.</p>
|
https://math.stackexchange.com/questions/1572728/how-many-groups-up-isomorphisms-exist-such-that-has-least-one-element-of-order-8
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.