category stringclasses 107
values | title stringlengths 15 179 | question_link stringlengths 59 147 | question_body stringlengths 53 33.8k | answer_html stringlengths 0 28.8k | __index_level_0__ int64 0 1.58k |
|---|---|---|---|---|---|
group-theory | Frattini Subgroup is trivial hen the group is elementary abelian | https://math.stackexchange.com/questions/2197158/frattini-subgroup-is-trivial-hen-the-group-is-elementary-abelian | <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>
| <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>
| 100 |
group-theory | If $G$ and $G'$ are two finite group of same cardinal, then $G\cong G'$. | https://math.stackexchange.com/questions/1649636/if-g-and-g-are-two-finite-group-of-same-cardinal-then-g-cong-g | <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>
| <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 i... | 101 |
group-theory | If $G$ is perfect, show $G/N$ is also perfect (for a $N\trianglelefteq G$) | https://math.stackexchange.com/questions/2573644/if-g-is-perfect-show-g-n-is-also-perfect-for-a-n-trianglelefteq-g | <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-con... | <p>If $G$ is perfect, that is $G=G'$, then $(G/N)'=G'N/N=GN/N=G/N$.</p>
| 102 |
group-theory | Find the dimension of the system with equations and variables | https://math.stackexchange.com/questions/4168724/find-the-dimension-of-the-system-with-equations-and-variables | <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>... | <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 clas... | 103 |
group-theory | Looking for examples of Free groups, Ordered groups, and Unique product groups | https://math.stackexchange.com/questions/3544023/looking-for-examples-of-free-groups-ordered-groups-and-unique-product-groups | <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 o... | <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 ... | 104 |
group-theory | Homomorphism between $S_n$ to $\mathbb Z$ | https://math.stackexchange.com/questions/2660136/homomorphism-between-s-n-to-mathbb-z | <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>
| <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>
| 105 |
group-theory | Let $(\mathbb R^3 , *)$ be a group with operation $(x,y,t)*(x',y',t'):=( x+x',y+y',t+t'+\dfrac 12(xy'-x 'y ) ) $, what is the center of this group? | https://math.stackexchange.com/questions/1474870/let-mathbb-r3-be-a-group-with-operation-x-y-tx-y-t-xx-y | <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>
| <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... | 106 |
group-theory | Quotient a group by a proper subgroup. | https://math.stackexchange.com/questions/676442/quotient-a-group-by-a-proper-subgroup | <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>
| <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 an... | 107 |
group-theory | Example of a homomorphism from $GL_n(\mathbb{R}) \rightarrow \mathbb{R^+}$ with kernel $K = \{A\in GL_n(\mathbb{R})| \det(A) = \pm 1\}$ | https://math.stackexchange.com/questions/1539892/example-of-a-homomorphism-from-gl-n-mathbbr-rightarrow-mathbbr-with | <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>
| <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>
| 108 |
group-theory | Understanding Semidirect Products in Group-Theory through exercises. | https://math.stackexchange.com/questions/1482008/understanding-semidirect-products-in-group-theory-through-exercises | <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 $H$ normaliz... | <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 s... | 109 |
group-theory | Proving the composition of two group homomorphisms is a group homomorphism. | https://math.stackexchange.com/questions/2130210/proving-the-composition-of-two-group-homomorphisms-is-a-group-homomorphism | <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) &... | <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... | 110 |
group-theory | Show that the order of G = order f(G) times order ker(G). | https://math.stackexchange.com/questions/1327197/show-that-the-order-of-g-order-fg-times-order-kerg | <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... | <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... | 111 |
group-theory | Suppose that G is a finite, nonabelian group with odd order. Show s is surjective, and hence bijective | https://math.stackexchange.com/questions/57913/suppose-that-g-is-a-finite-nonabelian-group-with-odd-order-show-s-is-surjectiv | <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 ... | <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>
| 112 |
group-theory | What are the sylow subgroups of $Z_{24}$? | https://math.stackexchange.com/questions/1744801/what-are-the-sylow-subgroups-of-z-24 | <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... | <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 lar... | 113 |
group-theory | List of all Abelian groups of order $1188$? | https://math.stackexchange.com/questions/2293862/list-of-all-abelian-groups-of-order-1188 | <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... | <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>
| 114 |
group-theory | Groups that have a non-zero even number of elements of order two | https://math.stackexchange.com/questions/4184824/groups-that-have-a-non-zero-even-number-of-elements-of-order-two | <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 browse... | 115 | |
group-theory | compute $\limsup\limits_{k\rightarrow \infty}$ ${l_{G,S} (k)}$ / ${l_{G,T} (k)}$ $<$ $ \infty$ | https://math.stackexchange.com/questions/4189011/compute-limsup-limits-k-rightarrow-infty-l-g-s-k-l-g-t-k | <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 \v... | 116 | |
group-theory | Groups with operators $\Omega \ni \omega$ that commute with positive powers, does $(g^{-1})^\omega$ always equal $(g^\omega)^{-1}$ | https://math.stackexchange.com/questions/4193994/groups-with-operators-omega-ni-omega-that-commute-with-positive-powers-doe | <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="m... | 117 | |
group-theory | Suppose that $H$ is a subgroup of a group $G$ and $|H|=10$. If $a$ belongs to $G$ and $a^6$ belongs to $H$, then what are the possibilities for $|a|$? | https://math.stackexchange.com/questions/2903641/suppose-that-h-is-a-subgroup-of-a-group-g-and-h-10-if-a-belongs-to-g | <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>
| <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-conta... | 118 |
group-theory | Non empty set and group | https://math.stackexchange.com/questions/2909462/non-empty-set-and-group | <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>
| <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 $\subse... | 119 |
group-theory | Is this true that $N_G(H) \subseteq N_G(H \cap K)$ if not what is the counter-example? | https://math.stackexchange.com/questions/2929568/is-this-true-that-n-gh-subseteq-n-gh-cap-k-if-not-what-is-the-counter-ex | <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>
| <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>. ... | 120 |
group-theory | Difficulty in understanding the definition of this sequence | https://math.stackexchange.com/questions/2946994/difficulty-in-understanding-the-definition-of-this-sequence | <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... | <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... | 121 |
group-theory | Proving something isn't isomorphic, $\varphi : 5\mathbb{Z} \rightarrow \mathbb{Z}$ | https://math.stackexchange.com/questions/2959686/proving-something-isnt-isomorphic-varphi-5-mathbbz-rightarrow-mathbbz | <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 ... | <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 i... | 122 |
group-theory | Inequality involving the number of subgroups of a direct product of groups | https://math.stackexchange.com/questions/2968181/inequality-involving-the-number-of-subgroups-of-a-direct-product-of-groups | <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... | 123 | |
group-theory | Show that the set $H = \{2^n:𝑛 \in\mathbb{Z}\}$ is a subgroup of for $\mathbb{Q}\setminus\{0\}$ | https://math.stackexchange.com/questions/2993916/show-that-the-set-h-2n-in-mathbbz-is-a-subgroup-of-for-mathbbq | <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">$\mathbb{Q}\setminus\{0\}$... | <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 ... | 124 |
group-theory | Let $H$ a subgroup of $G$. $\exists g \in G$, $g \notin H$, such that $gH=Hg$, and $[G:H]$ is a prime p. Prove that $H$ is normal. | https://math.stackexchange.com/questions/2999699/let-h-a-subgroup-of-g-exists-g-in-g-g-notin-h-such-that-gh-hg-a | <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 tha... | <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 <... | 125 |
group-theory | Let $G$ be a finite group and $K$ a normal subgroup of $G$. Suppose $|K|^2 \nmid |G|$, show $K$ is characteristic subgroup | https://math.stackexchange.com/questions/3004107/let-g-be-a-finite-group-and-k-a-normal-subgroup-of-g-suppose-k2-nmid | <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-containe... | <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>
| 126 |
group-theory | Proof of group theory | https://math.stackexchange.com/questions/3005921/proof-of-group-theory | <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>
| <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 an... | 127 |
group-theory | A simple question on groups and quotient spaces. | https://math.stackexchange.com/questions/3008670/a-simple-question-on-groups-and-quotient-spaces | <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\longr... | 128 | |
group-theory | Is there such a group? | https://math.stackexchange.com/questions/3023788/is-there-such-a-group | <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>
| <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-cardi... | 129 |
group-theory | How to deduce information about center of group if only order of centraliser is given? | https://math.stackexchange.com/questions/3028409/how-to-deduce-information-about-center-of-group-if-only-order-of-centraliser-is | <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="mat... | 130 | |
group-theory | Prove that a map is defined properly | https://math.stackexchange.com/questions/3077227/prove-that-a-map-is-defined-properly | <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... | <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... | 131 |
group-theory | $G$ is a Group of order $pq$, if $G$ has exactly one subgroup of order $p$ and another with order $q$, then $G$ is cyclic ($p,q$ are prime) | https://math.stackexchange.com/questions/3141832/g-is-a-group-of-order-pq-if-g-has-exactly-one-subgroup-of-order-p-and-a | <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">... | <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$</span>, then we... | 132 |
group-theory | Modulo groups and non-prime numbers | https://math.stackexchange.com/questions/3145241/modulo-groups-and-non-prime-numbers | <p>Let <span class="math-container">$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 ... | <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 th... | 133 |
group-theory | How do I show that $\mathbb{Q}$ and $\mathbb{Q}\times\mathbb{Q}$ are not group isomorphic? | https://math.stackexchange.com/questions/3151523/how-do-i-show-that-mathbbq-and-mathbbq-times-mathbbq-are-not-group-i | <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>
| <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"... | 134 |
group-theory | If $cl(g)=\{g\}$ can we imply that $g\in Z(G)?$ | https://math.stackexchange.com/questions/3151972/if-clg-g-can-we-imply-that-g-in-zg | <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-containe... | <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">$$gh=hg.$$</s... | 135 |
group-theory | $\phi \in \operatorname{Aut}(\Bbb Z_{50})$ via $\phi(11) = 3$ Then $\phi(x) = $? For any $x \in\Bbb Z_{50}$ | https://math.stackexchange.com/questions/3153385/phi-in-operatornameaut-bbb-z-50-via-phi11-3-then-phix | <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="... | <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) = ... | 136 |
group-theory | Let $G$ be a group. $|G|=21$ and $|Z(G)| \neq 1$ $\Rightarrow$ $|Z(G)| = $? | https://math.stackexchange.com/questions/3153429/let-g-be-a-group-g-21-and-zg-neq-1-rightarrow-zg | <blockquote>
<p>Let <span class="math-container">$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)| = $</span> ?</p>
</blockquote>
<p>We know that <span clas... | <blockquote>
<p><strong>Lemma</strong>: Let <span class="math-container">$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 p... | 137 |
group-theory | Let $G$ be a group with order 12. Which of the following claims are false? | https://math.stackexchange.com/questions/3153581/let-g-be-a-group-with-order-12-which-of-the-following-claims-are-false | <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 a... | <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... | 138 |
group-theory | Are the two groups $G, G'$ isomorphic? | https://math.stackexchange.com/questions/3161957/are-the-two-groups-g-g-isomorphic | <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>
| <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>
| 139 |
group-theory | $H$ is a subgroup of $G$, prove $|gHg^{-1}| = |H|$, $\forall g \in G$ | https://math.stackexchange.com/questions/3194039/h-is-a-subgroup-of-g-prove-ghg-1-h-forall-g-in-g | <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">$H$</s... | <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 <spa... | 140 |
group-theory | $H$ is a subgroup of $G$, $N$ is a normal subgroup of $G$, $\gcd(|H|, |G/N|) = 1 \Rightarrow$ $H$ is a subgroup of $N$ | https://math.stackexchange.com/questions/3195123/h-is-a-subgroup-of-g-n-is-a-normal-subgroup-of-g-gcdh-g-n-1 | <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"... | <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-co... | 141 |
group-theory | Is $H_nH_m = G$ if $|G|=nm$ and $H_n, H_m$ are the only subgroups of $G$ of order $n, m$ respectively? | 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 | <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 <sp... | <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... | 142 |
group-theory | About centralizer of an infinite cyclic subgroup | https://math.stackexchange.com/questions/3222863/about-centralizer-of-an-infinite-cyclic-subgroup | <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>
| <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 ... | 143 |
group-theory | Are there equations which have solutions in all groups but which are not algebraicly solvable | https://math.stackexchange.com/questions/3240948/are-there-equations-which-have-solutions-in-all-groups-but-which-are-not-algebra | <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=... | <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 ther... | 144 |
group-theory | Proving that there is a homomorphism $f : \mathbb{Z}^2 \ast \mathbb{Z}^2 \to \mathbb{Z}^2$ with kernel free and not finitely generated | https://math.stackexchange.com/questions/3241462/proving-that-there-is-a-homomorphism-f-mathbbz2-ast-mathbbz2-to-ma | <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 c... | <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... | 145 |
group-theory | *-Groups or "Group with Involution"? | https://math.stackexchange.com/questions/3259366/groups-or-group-with-involution | <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 ... | 146 | |
group-theory | Disproving the converse of Lagrange's theorem | https://math.stackexchange.com/questions/3266823/disproving-the-converse-of-lagranges-theorem | <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>
| <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 d... | 147 |
group-theory | Prove that either $A=\Delta B$ or $A=\Delta C$ | https://math.stackexchange.com/questions/3274585/prove-that-either-a-delta-b-or-a-delta-c | <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>... | <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... | 148 |
group-theory | Prove or disprove that there is a element $y$ in $G$ such that , $y^2 = x$ | https://math.stackexchange.com/questions/3278714/prove-or-disprove-that-there-is-a-element-y-in-g-such-that-y2-x | <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$</span> such ... | <p>Suppose <span class="math-container">$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>
| 149 |
group-theory | What examples to use when learning group theory? | https://math.stackexchange.com/questions/3289459/what-examples-to-use-when-learning-group-theory | <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... | <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 uninterestin... | 150 |
group-theory | Group of $2 \times 2$ matrices with complex numbers | https://math.stackexchange.com/questions/3304910/group-of-2-times-2-matrices-with-complex-numbers | <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">$$\begin{bmatrix}0&1\\-1&0\end{bmatrix} \quad \text{ and } \quad \begin{bmatrix}i&... | <p><em>Hint:</em>
Those two matrices commute, because the second one is <span class="math-container">$iI$</span>.
Therefore, the group they generate is an abelian group.
What are the conjugacy classes of an abelian group?</p>
| 151 |
group-theory | Finite index of centralizer $C_H(g)$ in $C_G(g)$ | https://math.stackexchange.com/questions/3315817/finite-index-of-centralizer-c-hg-in-c-gg | <p>Let <span class="math-container">$H$</span> be a subgroup of finite index in group <span class="math-container">$G$</span> and let <span class="math-container">$g \in G$</span>. </p>
<p><strong>Question:</strong> Is it true, that <span class="math-container">$C_H(g)$</span> has finite index in <span class="math-con... | 152 | |
group-theory | Proof for $HK = \operatorname{grp}\{H \cup K\} \iff HK \text{ is a subgroup of } G$ | https://math.stackexchange.com/questions/3321901/proof-for-hk-operatornamegrp-h-cup-k-iff-hk-text-is-a-subgroup-of | <p>If <span class="math-container">$H$</span> and <span class="math-container">$K$</span> are subgroups of a group <span class="math-container">$G,*$</span>, how can one proof that:
<span class="math-container">$$HK = \operatorname{grp}\{H \cup K\} \iff HK \text{ is a subgroup of } G$$</span> </p>
<p>I've seen similar... | <p>If <span class="math-container">$HK$</span> is the subgroup generated by <span class="math-container">$H\cup K$</span>, then it is of course a subgroup.</p>
<p>On the other hand, if <span class="math-container">$HK$</span> is a subgroup of <span class="math-container">$G$</span>, then it is clearly the smallest sub... | 153 |
group-theory | Let G be a group. If order of the element in the group $G$ is prime | https://math.stackexchange.com/questions/3329586/let-g-be-a-group-if-order-of-the-element-in-the-group-g-is-prime | <p>Let G be a group. If order of the element in the group <span class="math-container">$G$</span> is prime (<span class="math-container">$G$</span> is not a <span class="math-container">$p$</span>-group, order of the elements are different prime). Is it true that every element of order <span class="math-container">$p$<... | <p>No. Not all elements of order 5 in the alternating group <span class="math-container">$A_5$</span> are conjugate.</p>
| 154 |
group-theory | Let $G$ be group, $S \subset G$, $g \in G $. Prove: $gC_{G}(S)(g)^{-1} = C_{G}(gS(g)^{-1}).$ | https://math.stackexchange.com/questions/3333011/let-g-be-group-s-subset-g-g-in-g-prove-gc-gsg-1-c-gg | <blockquote>
<p>Let <span class="math-container">$G$</span> be group, <span class="math-container">$S \subset G$</span>, <span class="math-container">$g \in G $</span>. Prove:
<span class="math-container">$$gC_{G}(S)(g)^{-1} = C_{G}(gS(g)^{-1}), \forall g\in G.$$</span></p>
</blockquote>
<p>I have managed to prove... | <p>Let <span class="math-container">$x \in C_G(gSg^{-1})$</span>, then <span class="math-container">$$\forall s \in S, g \in G: g^{-1}xgs = g^{-1}xgsg^{-1}g.$$</span> Since <span class="math-container">$x$</span> commutes with <span class="math-container">$gsg^{-1}$</span> this is <span class="math-container">$$\underb... | 155 |
group-theory | If we know $\mathbb{Z}_{n}/ \mathbb{Z}_{m} \cong \mathbb{Z}_{k}$, can we conclude that $\mathbb{Z}_{n} \cong \mathbb{Z}_{m} \times \mathbb{Z}_{k} $? | https://math.stackexchange.com/questions/3333937/if-we-know-mathbbz-n-mathbbz-m-cong-mathbbz-k-can-we-conclud | <blockquote>
<p>If we know <span class="math-container">$\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} $</span>?</p>
</blockquote>
<p>I think not, but I can not find right counterexample. Any h... | <p><strong>Counterexample:</strong> <span class="math-container">$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>
| 156 |
group-theory | find all elements $f ∈ S_5$ of order 2 such that $f(1)=2$ | https://math.stackexchange.com/questions/3358628/find-all-elements-f-%e2%88%88-s-5-of-order-2-such-that-f1-2 | <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... | <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... | 157 |
group-theory | Define group G based on H and K | https://math.stackexchange.com/questions/3369151/define-group-g-based-on-h-and-k | <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> su... | 158 | |
group-theory | Let $G$ be a group and $a,b\in G$. Prove that $|a|=|a^{-1}|$ and that $|ab|=|ba|$. | 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 | <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">$|a|=n$</spa... | <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 cla... | 159 |
group-theory | In the additive groups $\mathbb{Z}$, $\mathbb{Q}$, $\mathbb{R}$, or $\mathbb{C}$ every nonzero (i.e., nonidentity) element has infinite order. | https://math.stackexchange.com/questions/3388664/in-the-additive-groups-mathbbz-mathbbq-mathbbr-or-mathbbc | <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>
| <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>... | 160 |
group-theory | True or False: 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. | https://math.stackexchange.com/questions/3397903/true-or-false-if-the-product-of-n-elements-of-a-group-is-the-identity-element | <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 propose... | <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 ... | 161 |
group-theory | Construct the regular representation of Z3 and diagonalize it | https://math.stackexchange.com/questions/3412322/construct-the-regular-representation-of-z3-and-diagonalize-it | <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>
| <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... | 162 |
group-theory | Does converse of Lagrange's theorem hold in $A_{4} \times \Bbb Z_{2}?$ | https://math.stackexchange.com/questions/3436892/does-converse-of-lagranges-theorem-hold-in-a-4-times-bbb-z-2 | <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... | <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$</spa... | 163 |
group-theory | Homomorphisms with the same kernel has the same semidirect product? | https://math.stackexchange.com/questions/3440119/homomorphisms-with-the-same-kernel-has-the-same-semidirect-product | <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\va... | <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{... | 164 |
group-theory | Find all subgroups of $(\Bbb{Z}_2\times\Bbb{Z}_4,+)$ | https://math.stackexchange.com/questions/3441019/find-all-subgroups-of-bbbz-2-times-bbbz-4 | <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)\r... | <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"... | 165 |
group-theory | How to show that order of an element is $2$ where the element is $ p-1$ for prime $p$. | https://math.stackexchange.com/questions/3445321/how-to-show-that-order-of-an-element-is-2-where-the-element-is-p-1-for-prim | <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 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 inde... | 166 |
group-theory | Group operation in generator-relator representation | https://math.stackexchange.com/questions/3460030/group-operation-in-generator-relator-representation | <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-... | <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 an... | 167 |
group-theory | Show that a quotient group is cyclic | https://math.stackexchange.com/questions/3485924/show-that-a-quotient-group-is-cyclic | <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">$G/\langle a\ra... | <p><strong>Hint</strong>:</p>
<p>The last two elements <span class="math-container">$(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>
| 168 |
group-theory | The Twist Group | https://math.stackexchange.com/questions/3514356/the-twist-group | <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 magn... | 169 | |
group-theory | How to show this is a Folner sequence? | https://math.stackexchange.com/questions/3526927/how-to-show-this-is-a-folner-sequence | <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 cl... | 170 | |
group-theory | prove that $U ⊕U_{(2 1)} = U ⊕U_{(3 1)}$ | https://math.stackexchange.com/questions/3528930/prove-that-u-%e2%8a%95u-2-1-u-%e2%8a%95u-3-1 | <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-contain... | 171 | |
group-theory | Is there a natural family of nonisomorphic groups parametrized by $\mathbb{R}$? | https://math.stackexchange.com/questions/3546089/is-there-a-natural-family-of-nonisomorphic-groups-parametrized-by-mathbbr | <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... | <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 permutati... | 172 |
group-theory | Artin on the associative law of $n$ elements | https://math.stackexchange.com/questions/3550592/artin-on-the-associative-law-of-n-elements | <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 ... | <p>Artin is not commuting the elements, he is only changing the location of the parentheses.</p>
| 173 |
group-theory | Why is it true that if $H$ is a subgroup of a group $G$, then $1_H=1_G$? | 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 | <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$</span> is a subgroup iff <span class="math-container"... | <p>Here are two ways to see this:</p>
<ol>
<li><p>If <span class="math-container">$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">$1_H... | 174 |
group-theory | Commutator element in permutation group | https://math.stackexchange.com/questions/3591908/commutator-element-in-permutation-group | <p>Øystein Ore in 'Some remarks on commutators' proof that:</p>
<p>''any element of the alternating group <span class="math-container">$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">... | 175 | |
group-theory | Let $G$ be finite group and order of $Aut(G)$ is prime number $p$ then prove that order of $G$ is always less than equal to $3$? | https://math.stackexchange.com/questions/3618620/let-g-be-finite-group-and-order-of-autg-is-prime-number-p-then-prove-tha | <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... | <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 a... | 176 |
group-theory | Characterization of a Subring, shouldnot we concern $ab$ and $ba$ simultaneously? | https://math.stackexchange.com/questions/2768948/characterization-of-a-subring-shouldnot-we-concern-ab-and-ba-simultaneously | <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-co... | <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 ... | 177 |
group-theory | Index-two subgroup implies full group | https://math.stackexchange.com/questions/2735423/index-two-subgroup-implies-full-group | <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>
| <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 (... | 178 |
group-theory | Order of element 2 in group $\Bbb{Z^*_{47}}$ under multiplication | https://math.stackexchange.com/questions/2763690/order-of-element-2-in-group-bbbz-47-under-multiplication | <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>
| <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 ... | 179 |
group-theory | Prove that $\langle F,\bullet \rangle$ is a group where $F=\{f\mid f:X→G\}$ and $\langle G,\ast \rangle$ is a group | https://math.stackexchange.com/questions/2807293/prove-that-langle-f-bullet-rangle-is-a-group-where-f-f-mid-fx%e2%86%92g-and | <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... | 180 | |
group-theory | Van Kampen diagram/Van Kampen lemma | https://math.stackexchange.com/questions/2808012/van-kampen-diagram-van-kampen-lemma | <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 permu... | 181 | |
group-theory | Equal generator sets | https://math.stackexchange.com/questions/2764504/equal-generator-sets | <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, ... | 182 | |
group-theory | On quotients of unipotent subgroups | https://math.stackexchange.com/questions/2754340/on-quotients-of-unipotent-subgroups | <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... | 183 | |
group-theory | cyclic subgroup generated by $a^m$ is the same as the cyclic subgroup generated by $a^d$ | https://math.stackexchange.com/questions/2809732/cyclic-subgroup-generated-by-am-is-the-same-as-the-cyclic-subgroup-generated | <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 ... | <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>
| 184 |
group-theory | Isomorphy of abstract vs. algebraic groups | https://math.stackexchange.com/questions/2810059/isomorphy-of-abstract-vs-algebraic-groups | <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>... | 185 | |
group-theory | Existence of maximal subgroup of $G$ with the property of not containing $g\in G$. | https://math.stackexchange.com/questions/2810304/existence-of-maximal-subgroup-of-g-with-the-property-of-not-containing-g-in-g | <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 ... | <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>
| 186 |
group-theory | group action on another group preserves the latter's group operation | https://math.stackexchange.com/questions/2791868/group-action-on-another-group-preserves-the-latters-group-operation | <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... | <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, f... | 187 |
group-theory | Quasi-Isometry and finiteness | https://math.stackexchange.com/questions/2825097/quasi-isometry-and-finiteness | <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>
| <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>
| 188 |
group-theory | direct product of nontrivial groups | https://math.stackexchange.com/questions/1502639/direct-product-of-nontrivial-groups | <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 k... | <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 $\m... | 189 |
group-theory | Application of Burnside's theorem? Composition factors of $|G|=p^2q$ are... | https://math.stackexchange.com/questions/1555306/application-of-burnsides-theorem-composition-factors-of-g-p2q-are | <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, Burns... | 190 | |
group-theory | Properties of Abelian Groups | https://math.stackexchange.com/questions/1555799/properties-of-abelian-groups | <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... | <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.$$
This follows naturally from the definition of commutativity.</p>
| 191 |
group-theory | Induction with an associative operator | https://math.stackexchange.com/questions/1555868/induction-with-an-associative-operator | <p>I've been starting to play around with some properties of groups, and I wanted to prove this claim to make things simpler for me in the future. If $G$ is closed under and operation $\cdot$ that is also associative, I want to show that $\forall a_1, a_2, ..., a_n \in G$, no matter how I choose to bracket $a_1 \cdot a... | <p>It's pretty messy and most books just say "it's obvious". If you do want do do it, probably the easiest way is to be specific, with something like this.</p>
<p><strong>Theorem</strong>. A product of terms $a_1,\ldots,a_n$, in that order, with any order of bracketing, is equal to
$$(\cdots((a_1*a_2)*a_3)*\cdots*a_... | 192 |
group-theory | about the concept of an FC element | https://math.stackexchange.com/questions/1590843/about-the-concept-of-an-fc-element | <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>
| <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> permutat... | 193 |
group-theory | Counting the number of abelian groups of a certain order | https://math.stackexchange.com/questions/1534592/counting-the-number-of-abelian-groups-of-a-certain-order | <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>
| <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 a... | 194 |
group-theory | Proof that any finite subgroup of $SO(2)$ must be cyclic | https://math.stackexchange.com/questions/1571081/proof-that-any-finite-subgroup-of-so2-must-be-cyclic | <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>
| <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 ... | 195 |
group-theory | Why is $H$ not a subgroup of $G$? | https://math.stackexchange.com/questions/1535672/why-is-h-not-a-subgroup-of-g | <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 $H=(\mathbb{Z}/3\mathbb{Z},+)=\{0,1,2\}$... | <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 $... | 196 |
group-theory | About the definition of amenable group. | https://math.stackexchange.com/questions/1593043/about-the-definition-of-amenable-group | <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... | <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>
| 197 |
group-theory | Show that $D_{2n}$ has two conjugacy classes of reflections if $n$ is even, but only one if $n$ is odd | https://math.stackexchange.com/questions/1559139/show-that-d-2n-has-two-conjugacy-classes-of-reflections-if-n-is-even-but | <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 think... | <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 $... | 198 |
group-theory | How many groups up isomorphisms exist such that has least one element of order 8. | https://math.stackexchange.com/questions/1572728/how-many-groups-up-isomorphisms-exist-such-that-has-least-one-element-of-order-8 | <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 \mathb... | <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-S... | 199 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.