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Proposition 6.2. A nonempty subset \( H \) of a group \( G \) is a subgroup if and only if\n\n\[ \left( {\forall a, b \in H}\right) : \;a{b}^{-1} \in H. \] | Proof. It is clear that if \( H \) is a subgroup, then the stated condition holds: indeed, if \( b \in H \), then the inverse of \( b \) must also be in \( H \) and \( H \) is closed under the operation of \( G \) .\n\nConversely, assume the stated condition holds; we have to check that \( H \) is closed under the oper... | Yes |
Lemma 6.3. If \( {\left\{ {H}_{\alpha }\right\} }_{\alpha \in A} \) is any family of subgroups of a group \( G \), then\n\n\[ H = \mathop{\bigcap }\limits_{{\alpha \in A}}{H}_{\alpha } \]\n\nis a subgroup of \( G \) . | Proof. This follows right away from Proposition 6.2. \( H \) is nonempty, because \( e \in {H}_{\alpha } \) for all \( \alpha \), so \( e \in H \) ; and\n\n\[ a, b \in H \Rightarrow \left( {\forall \alpha \in A}\right) : a, b \in {H}_{\alpha } \Rightarrow \left( {\forall \alpha \in A}\right) : a{b}^{-1} \in {H}_{\alpha... | Yes |
Lemma 6.4. Let \( \varphi : G \rightarrow {G}^{\prime } \) be a group homomorphism, and let \( {H}^{\prime } \) be a subgroup of \( {G}^{\prime } \) . Then \( {\varphi }^{-1}\left( {H}^{\prime }\right) \) is a subgroup of \( G \) . | Proof. Recall (end of [12.5] that \( {\varphi }^{-1}\left( {H}^{\prime }\right) \) consists of all \( g \in G \) such that \( \varphi \left( g\right) \in \) \( {H}^{\prime } \) . Since \( \varphi \left( {e}_{G}\right) = {e}_{{G}^{\prime }} \in {H}^{\prime } \), this set is nonempty. If \( a, b \in {\varphi }^{-1}\left(... | Yes |
Proposition 6.6. Let \( \varphi : G \rightarrow {G}^{\prime } \) be a homomorphism. Then the inclusion \( i \) : \( \ker \varphi \hookrightarrow G \) is final in the category 26 of group homomorphisms \( \alpha : K \rightarrow G \) such that \( \varphi \circ \alpha \) is the trivial map. | Proof. If \( \alpha : K \rightarrow G \) is such that \( \varphi \circ \alpha \) is the trivial map, then \( \forall k \in K \)\n\n\[ \varphi \circ \alpha \left( k\right) = \varphi \left( {\alpha \left( k\right) }\right) = {e}_{{G}^{\prime }}, \]\n\nthat is, \( \alpha \left( k\right) \in \ker \varphi \) . We can (and m... | Yes |
Proposition 6.9. Let \( G \subseteq \mathbb{Z} \) be a subgroup. Then \( G = d\mathbb{Z} \) for some \( d \geq 0 \) . | Proof of Proposition 6.9. If \( G = \{ 0\} \), then \( G = 0\mathbb{Z} \) . If not, note that \( G \) must contain positive integers: indeed, if \( a \in G \) and \( a \leq 0 \), then \( - a \in G \) and \( - a > 0 \) . We can then let \( d \) be the smallest positive integer 29 in \( G \), and I claim \( G = d\mathbb{... | Yes |
Proposition 6.11. Let \( n > 0 \) be an integer and let \( G \subseteq \mathbb{Z}/n\mathbb{Z} \) be a subgroup. Then \( G \) is the cyclic subgroup of \( \mathbb{Z}/n\mathbb{Z} \) generated by \( {\left\lbrack d\right\rbrack }_{n} \), for some divisor \( d \) of \( n \) . | Proof. Let \( {\pi }_{n} : \mathbb{Z} \rightarrow \mathbb{Z}/n\mathbb{Z} \) be the quotient map, and consider \( {G}^{\prime } \mathrel{\text{:=}} {\pi }_{n}^{-1}\left( G\right) \) . By Lemma 6.4. \( {G}^{\prime } \) is a subgroup of \( \mathbb{Z} \) ; by Proposition 6.9, \( {G}^{\prime } \) is a cyclic subgroup of \( ... | Yes |
Proposition 6.12. The following are equivalent:\n\n(a) \( \\varphi \) is a monomorphism;\n\n(b) \( \\ker \\varphi = \\left\\{ {e}_{G}\\right\\} \) ;\n\n(c) \( \\varphi : G \\rightarrow {G}^{\\prime } \) is injective (as a set-function). | Proof. (a) \( \\Rightarrow \) (b): Assume (a) holds, and consider the two parallel compositions\n\n\[ \n\\ker \\varphi \\xrightarrow[e]{i}G\\overset{\\varphi }{ \\rightarrow }{G}^{\\prime }\n\]\nwhere \( i \) is the inclusion and \( e \) is the trivial map. Both \( \\varphi \\circ i \) and \( \\varphi \\circ e \) are t... | Yes |
Lemma 7.2. If \( \varphi : G \rightarrow {G}^{\prime } \) is any group homomorphism, then \( \ker \varphi \) is a normal subgroup of \( G \) . | Proof. We already know that \( \ker \varphi \) is a subgroup of \( G \) ; to verify it is normal note that \( \forall g \in G,\forall n \in \ker \varphi \)\n\n\[ \varphi \left( {{gn}{g}^{-1}}\right) = \varphi \left( g\right) \varphi \left( n\right) \varphi \left( {g}^{-1}\right) = \varphi \left( g\right) {e}_{{G}^{\pri... | Yes |
Proposition 7.4. Let \( \sim \) be an equivalence relation on a group \( G \), satisfying \( \left( \dagger \right) \) . Then\n\n- the equivalence class of \( {e}_{G} \) is a subgroup \( H \) of \( G \) ; and\n\n- \( a \sim b \Leftrightarrow {a}^{-1}b \in H \Leftrightarrow {aH} = {bH} \) . | Proof. Let \( H \subseteq G \) be the equivalence class of the identity; \( H \neq \varnothing \) as \( {e}_{G} \in H \) . For \( a, b \in H \), we have \( {e}_{G} \sim b \) and hence \( {b}^{-1} \sim {e}_{G} \) (applying \( \left( \dagger \right) \), multiplying on the left by \( {b}^{-1} \) ); hence \( a{b}^{-1} \sim... | Yes |
Proposition 7.6. If \( H \) is any subgroup of a group \( G \), the relation \( { \sim }_{L} \) defined by\n\n\[ \n\\left( {\\forall a, b \\in G}\\right) : \\;a{ \\sim }_{L}b \\Leftrightarrow {a}^{-1}b \\in H \n\]\n\nis an equivalence relation satisfying \( \\left( \\dagger \\right) \) . | Proof. This is straightforward and is mostly left to the reader (Exercise 7.8). To see that the relation satisfies \( \\left( \\dagger \\right) \), note that\n\n\[ \na{ \\sim }_{L}b \\Rightarrow {a}^{-1}b \\in H \\Rightarrow {a}^{-1}\\left( {{g}^{-1}g}\\right) b \\in H \\Rightarrow {\\left( ga\\right) }^{-1}\\left( {gb... | No |
Proposition 7.8. There is a one-to-one correspondence between subgroups of \( G \) and equivalence relations on \( G \) satisfying \( \left( {\dagger \dagger }\right) \) ; for the relation \( { \sim }_{R} \) corresponding to a subgroup \( H, G/{ \sim }_{R} \) may be described as the set of right-cosets \( {Ha} \) of \(... | The relation corresponding to \( H \) in this second way is defined by\n\n\[ \n a{ \sim }_{R}b \Leftrightarrow a{b}^{-1} \in H \Leftrightarrow {Ha} = {Hb}. \n\] | No |
Let \( G = {S}_{3} \), and let \( H \) be the subgroup consisting of the identity and the \( 1 \leftrightarrow 2 \) switch:\n\n\[ H = \left\{ {\left( \begin{array}{lll} 1 & 2 & 3 \\ 1 & 2 & 3 \end{array}\right) ,\left( \begin{array}{lll} 1 & 2 & 3 \\ 2 & 1 & 3 \end{array}\right) }\right\} \]\n\nThen\n\n\[ \left( \begin... | This state of affairs simply reflects the fact that the two conditions \( \left( \dagger \right) \) and \( \left( {\dagger \dagger }\right) \) are different: there is no reason to expect that if one holds, the other one should also hold (unless \( G \) is commutative, of course). Once more, keep in mind that both have ... | No |
Proposition 7.10. The relations \( { \sim }_{L},{ \sim }_{R} \) corresponding to a subgroup \( H \) coincide if and only if \( H \) is normal. | Proof. Two relations coincide if the corresponding partitions agree. Therefore\n\n\( { \sim }_{L} = { \sim }_{R} \Leftrightarrow \) left- and right-cosets of \( H \) coincide \( \Leftrightarrow \left( {\forall g \in G}\right) : {gH} = {Hg}. \)\n\nBut this is one of the equivalent conditions defining the notion of norma... | Yes |
Theorem 7.12. Let \( H \) be a normal subgroup of a group \( G \) . Then for every group homomorphism \( \varphi : G \rightarrow {G}^{\prime } \) such that \( H \subseteq \ker \varphi \) there exists a unique group homomorphism \( \widetilde{\varphi } : G/H \rightarrow {G}^{\prime } \) so that the diagram\n\n![cc115a52... | Proof. We only need to match the stated universal property with the one we proved in Proposition 7.3, and indeed,\n\n\[ H \subseteq \ker \varphi \Leftrightarrow \left( {\forall h \in H}\right) : \varphi \left( h\right) = {e}_{{G}^{\prime }} \]\n\nis equivalent to\n\n\[ \left( {\forall a, b \in G}\right) : a{b}^{-1} \in... | Yes |
Theorem 8.1. Every group homomorphism \( \varphi : G \rightarrow {G}^{\prime } \) may be decomposed as follows: \n\nwhere the isomorphism \( \widetilde{\varphi } \) in the middle is the homomorphism induced by \( \va... | It is important that the reader agree that we have already proved anything that deserves to be proven here. We know that the projection on the left and the inclusion on the right are homomorphisms and \( \widetilde{\varphi } \) comes from Theorem 7.12. The decomposition is the same one obtained at the level of set-func... | No |
Corollary 8.2. Suppose \( \varphi : G \rightarrow {G}^{\prime } \) is a surjective group homomorphism. Then\n\n\[ \n{G}^{\prime } \cong \frac{G}{\ker \varphi }.\n\] | Proof. \( {im\varphi } = {G}^{\prime } \) in Theorem 8.1 | No |
Claim 8.4. If \( {H}_{1} \subseteq {G}_{1} \) and \( {H}_{2} \subseteq {G}_{2} \) are normal subgroups, then \( {H}_{1} \times {H}_{2} \) is a normal subgroup of the group \( {G}_{1} \times {G}_{2} \) and\n\n\[ \frac{{G}_{1} \times {G}_{2}}{{H}_{1} \times {H}_{2}} \cong \frac{{G}_{1}}{{H}_{1}} \times \frac{{G}_{2}}{{H}... | Indeed, composing the projections\n\n\[ {\pi }_{1} : {G}_{1} \times {G}_{2} \rightarrow {G}_{1},\;{\pi }_{2} : {G}_{1} \times {G}_{2} \rightarrow {G}_{2} \]\n\nwith the morphisms to the quotients gives surjective homomorphisms\n\n\[ {\pi }_{1} : {G}_{1} \times {G}_{2} \rightarrow \frac{{G}_{1}}{{H}_{1}},\;{\pi }_{2} : ... | Yes |
As a particular case of Claim 8.4, take \( {H}_{1} = \left\{ {e}_{{G}_{1}}\right\} \subseteq {G}_{1} \) and \( {H}_{2} = {G}_{2} \subseteq {G}_{2} \) | \[ \frac{{G}_{1} \times {G}_{2}}{{G}_{2}} \cong \frac{{G}_{1}}{\left\{ {e}_{{G}_{1}}\right\} } \times \frac{{G}_{2}}{{G}_{2}} \cong {G}_{1} \] where on the left we identify \( {G}_{2} \) with the subgroup \( \left\{ {e}_{{G}_{1}}\right\} \times {G}_{2} \) | Yes |
The cyclic group \( {C}_{3} \) may be viewed as a subgroup of the dihedral group \( {D}_{6} \) : the rotations of a triangle give a copy of \( {C}_{3} \) inside \( {D}_{6} \) . Then \( {C}_{3} \) is normal in \( {D}_{6} \), and \[ \frac{{D}_{6}}{{C}_{3}} \cong {C}_{2} \] | This can of course be checked 'by hand'. But note that there is an evident surjective homomorphism \( {D}_{6} \rightarrow {C}_{2} \), whose kernel is \( {C}_{3} \) : map an element \( \sigma \) of \( {D}_{6} \) to the identity in \( {C}_{2} \) if it does not flip the triangle (that is, precisely when \( \sigma \in {C}_... | Yes |
One can give a circle (denoted \( {S}^{1} \) ) a group structure by identifying its points with rotations of a plane about a point and adding them accordingly. The function\n\n\[ \rho : {\mathbb{R}}^{1} \rightarrow {S}^{1} \]\n\nmapping a number \( r \) to the result of a rotation by \( {2\pi r} \) radians is then a su... | By Corollary 8.2, therefore,\n\n\[ \frac{\mathbb{R}}{\mathbb{Z}} \cong {S}^{1} \]\n\n(Cf. Exercise 11.6.) Geometrically, this amounts to 'wrapping' \( \mathbb{R} \) infinitely many times around the circle, realizing \( \mathbb{R} \) as the ’universal cover’ of \( {S}^{1} \) ; here, \( \mathbb{Z} \) plays the role of ’f... | No |
Example 8.8. Here is the effect of this operation on the lattice of subgroups of \( {C}_{12} \cong \mathbb{Z}/{12}\mathbb{Z} \) (labeled by generators; cf. 6.4), after quotienting by \( H = \langle \left\lbrack 6\right\rbrack \rangle \cong {C}_{2} \) : | The result matches the lattice of subgroups of \( {C}_{6} \cong {C}_{12}/{C}_{2} \) . | Yes |
Proposition 8.9. Let \( H \) be a normal subgroup of a group \( G \) . Then for every subgroup \( K \) of \( G \) containing \( H \) , \( K/H \) may be identified with a subgroup of \( G/H \) . The function \[ u : \{ \text{subgroups}K\text{of}G\text{containing}H\} \rightarrow \{ \text{subgroups of}G/H\} \] defined by \... | Proof. The group \( K/H \) consists of the cosets \( {aH} \in G/H \) with \( a \in K \), and in this sense it is a subset (and clearly a subgroup) of \( G/H \) . It is also clear that if \( H \subseteq K \subseteq L \), then \( u\left( K\right) = K/H \subseteq L/H = u\left( L\right) \) ; that is, \( u \) preserves incl... | Yes |
Proposition 8.10. Let \( H \) be a normal subgroup of a group \( G \), and let \( N \) be a subgroup of \( G \) containing \( H \) . Then \( N/H \) is normal in \( G/H \) if and only if \( N \) is normal in \( G \), and in this case \[ \frac{G/H}{N/H} \cong \frac{G}{N} \] | Proof. If \( N \) is normal, then consider the projection \[ G \rightarrow \frac{G}{N} : \] the subgroup \( H \) is contained in \( N \), which is the kernel of this homomorphism, so we get (by the universal property of quotients, Theorem 7.12) an induced homomorphism \[ \frac{G}{H} \rightarrow \frac{G}{N} \] The subgr... | Yes |
Proposition 8.11. Let \( H, K \) be subgroups of a group \( G \), and assume that \( H \) is normal in \( G \) . Then\n\n- \( {HK} \) is a subgroup of \( G \), and \( H \) is normal in \( {HK} \) ;\n\n- \( H \cap K \) is normal in \( K \), and\n\n\[ \frac{HK}{H} \cong \frac{K}{H \cap K} \] | Proof. To verify that \( {HK} \) is a subgroup of \( G \) when \( H \) is normal, note that \( {HK} \) is the union of all cosets \( {Hk} \), with \( k \in K \) ; that is,\n\n\[ {HK} = {\pi }^{-1}\left( {\pi \left( K\right) }\right) \]\n\nwhere \( \pi : G \rightarrow G/H \) is the canonical projection. Since \( \pi \le... | Yes |
Lemma 8.13. Let \( H \) be a subgroup of a group \( G \) . Then \( \forall g \in G \) the functions\n\n\[ H \rightarrow {gH},\;h \mapsto {gh},\]\n\n\[ H \rightarrow {Hg},\;h \mapsto {hg}\]\n\nare bijections. | Proof. Both functions are surjective by definition of coset. Cancellation implies that they are injective. | No |
Corollary 8.14 (Lagrange’s theorem). If \( G \) is a finite group and \( H \subseteq G \) is a subgroup, then \( \left| G\right| = \left\lbrack {G : H}\right\rbrack \cdot \left| H\right| \) . In particular, \( \left| H\right| \) is a divisor of \( \left| G\right| \) . | Proof. Indeed, \( G \) is the disjoint union of \( \left| {G/H}\right| \) distinct cosets \( {gH} \), and \( \left| {gH}\right| = \left| H\right| \) by Lemma 8.13 | Yes |
The order \( \left| g\right| \) of any element \( g \) of a finite group \( G \) is a divisor of \( \left| G\right| \) | indeed, \( \left| g\right| \) equals the order of the subgroup \( \langle g\rangle \) generated by \( g \) | No |
If \( \left| G\right| \) is a prime integer \( p \), then necessarily \( G \cong \mathbb{Z}/p\mathbb{Z} \) . | Indeed, let \( g \in G \) be any element other than the identity; then \( \langle g\rangle \) is a subgroup of \( G \), of order \( > 1 \) . By Lagrange’s theorem, \( \left| {\langle g\rangle }\right| = p = \left| G\right| \) ; that is, \( G \cong \langle g\rangle \) is cyclic of order \( p \), as claimed. | Yes |
Example 8.17 (Fermat’s little theorem). Let \( p \) be a prime integer, and let \( a \) be any integer. Then \( {a}^{p} \equiv a{\;\operatorname{mod}\;p} \) . | Indeed, this is immediate if \( a \) is a multiple of \( p \) ; if \( a \) is not a multiple of \( p \), then the class \( {\left\lbrack a\right\rbrack }_{p} \) modulo \( p \) is nonzero, so it is an element of the group \( {\left( \mathbb{Z}/p\mathbb{Z}\right) }^{ * } \), which has order \( p - 1 \) . Thus\n\n\[{\left... | Yes |
Proposition 8.18. Let \( \varphi : G \rightarrow {G}^{\prime } \) be a homomorphism of abelian groups. The following are equivalent:\n\n(a) \( \varphi \) is an epimorphism;\n\n(b) \( \operatorname{coker}\varphi \) is trivial;\n\n(c) \( \varphi : G \rightarrow {G}^{\prime } \) is surjective (as a set-function). | Proof. (a) \( \Rightarrow \) (b): Assume (a) holds, and consider the two parallel compositions\n\n\[ G\overset{\varphi }{ \rightarrow }{G}^{\prime }\xrightarrow[e]{\pi }\operatorname{coker}\varphi \]\n\nwhere \( \pi \) is the canonical projection and \( e \) is the trivial map. Both \( \pi \circ \varphi \) and \( e \ci... | Yes |
Every group \( G \) acts in a natural way on the underlying set \( G \) . The function \( \rho : G \times G \rightarrow G \) is simply the operation in the group: | \[ \left( {\forall g, a \in G}\right) : \;\rho \left( {g, a}\right) = {ga}. \] | Yes |
Theorem 9.5 (Cayley's theorem). Every group acts faithfully on some set. That is, every group may be realized as a subgroup of a permutation group. | Proof. Indeed, simply observe that the left-multiplication action of \( G \) on itself is manifestly faithful. | Yes |
Proposition 9.9. Every transitive left-action of \( G \) on a nonempty set \( A \) is isomorphic to the left-multiplication of \( G \) on \( G/H \), for \( H = \) the stabilizer of any \( a \in A \) . | Proof. Let \( G \) act transitively on a set \( A \), let \( a \in A \) be any element, and let \( H = {\operatorname{Stab}}_{G}\left( a\right) \) . I claim that there is an equivariant bijection\n\n\[ \varphi : G/H \rightarrow A \]\n\ndefined by\n\n\[ \varphi \left( {gH}\right) \mathrel{\text{:=}} {ga} \]\n\nfor all \... | Yes |
Corollary 9.10. If \( O \) is an orbit of the action of a finite group \( G \) on a set \( A \), then \( O \) is a finite set and\n\n\[ \left| O\right| \text{divides}\left| G\right| \text{.} \] | Proof. By Proposition 9.9 there is a bijection between \( O \) and \( G/{\operatorname{Stab}}_{G}\left( a\right) \) for any element \( a \in O \) ; thus\n\n\[ \left| O\right| \cdot \left| {{\operatorname{Stab}}_{G}\left( a\right) }\right| = \left| G\right| \]\n\nby Corollary 8.14 | Yes |
Example 9.11. There are no transitive actions of \( {S}_{3} \) on a set with 5 elements. | Indeed, 5 does not divide 6. | Yes |
Proposition 9.12. Suppose a group \( G \) acts on a set \( A \), and let \( a \in A, g \in G \) , \( b = {ga} \) . Then\n\n\[ \n{\operatorname{Stab}}_{G}\left( b\right) = g{\operatorname{Stab}}_{G}\left( a\right) {g}^{-1}.\n\] | Proof. Indeed, assume \( h \in {\operatorname{Stab}}_{G}\left( a\right) \) ; then\n\n\[ \n\left( {{gh}{g}^{-1}}\right) \left( b\right) = {gh}\left( {{g}^{-1}g}\right) a = {gha} = {ga} = b :\n\]\n\nthus \( {gh}{g}^{-1} \in {\operatorname{Stab}}_{G}\left( b\right) \) . This proves the \( \supseteq \) inclusion; \( \subse... | Yes |
Lemma 1.2. In a ring \( R \) ,\n\n\[ 0 \cdot r = 0 = r \cdot 0 \]\n\nfor all \( r \in R \) . | Proof. Indeed, \( 0 = 0 + 0 \) ; hence, applying distributivity,\n\n\[ r \cdot 0 = r \cdot \left( {0 + 0}\right) = r \cdot 0 + r \cdot 0, \]\n\nfrom which \( r \cdot 0 = 0 \) by cancellation (in the group \( \left( {R, + }\right) \) ). The equality \( 0 \cdot r = 0 \) is proven similarly. | Yes |
Example 1.4. More interesting examples are the number-based groups such as \( \mathbb{Z} \) or \( \mathbb{R} \), with the usual operations. These are very well known to our reader, who will realize immediately that they satisfy the requirements given in Definition 1.1 but they are very special. Why? | To begin with, note that multiplication is commutative in these examples; this is not among the requirements we have posed on rings in the official definition given above. | Yes |
Proposition 1.9. In a ring \( R, a \in R \) is not a left- (resp., right-) zero-divisor if and only if left (resp., right) multiplication by a is an injective function \( R \rightarrow R \) . | Proof. Let's verify the 'left' statement (the 'right' statement is of course entirely analogous). Assume \( a \) is not a left-zero-divisor and \( {ab} = {ac} \) for \( b, c \in R \) . Then, by distributivity,\n\n\[ a\left( {b - c}\right) = {ab} - {ac} = 0, \]\n\nand this implies \( b - c = 0 \) since \( a \) is not a ... | Yes |
In a ring \( R \): - \( u \) is a left- (resp., right-) unit if and only if left- (resp., right-) multiplication by \( u \) is a surjective functions \( R \rightarrow R \) ; - if \( u \) is a left- (resp., right-) unit, then right- (resp., left-) multiplication by \( u \) is injective; that is, \( u \) is not a right- ... | These assertions are all straightforward. For example, denote by \( {\rho }_{u} : R \rightarrow \) \( R \) right-multiplication by \( u \), so that \( {\rho }_{u}\left( r\right) = {ru} \) . If \( u \) is a right-unit, let \( v \in R \) be such that \( {vu} = 1 \) ; then \( \forall r \in R \)\n\n\[ \n{\rho }_{u} \circ {... | No |
Proposition 1.15. Assume \( R \) is a finite commutative ring; then \( R \) is an integral domain if and only if it is a field. | Proof. One implication holds for all rings, as pointed out above; thus we only have to verify that if \( R \) is a finite integral domain, then it is a field. This amounts to verifying that if \( a \) is a non-zero-divisor in a finite (commutative) ring \( R \), then it is a unit in \( R \) .\n\nNow, if \( a \) is a no... | Yes |
The group of units in the ring \( \mathbb{Z}/n\mathbb{Z} \) is precisely the group \( {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ * } \) | indeed, a class \( {\left\lbrack m\right\rbrack }_{n} \) is a unit if and only if (right-) multiplication by \( {\left\lbrack m\right\rbrack }_{n} \) is surjective (by Proposition 1.12), if and only if the map \( a \mapsto a{\left\lbrack m\right\rbrack }_{n} \) is surjective, if and only if \( {\left\lbrack m\right\rbr... | Yes |
Proposition 2.1. \( \\left( {i,\\mathbb{Z}\\left\\lbrack {{x}_{1},\\cdots ,{x}_{n}}\\right\\rbrack }\\right) \) is initial in \( {\\mathcal{R}}_{A} \) . | Proof. Let \( \\left( {j, R}\\right) \) be an arbitrary object of \( {\\mathcal{R}}_{A} \) ; we have to show that there is a unique morphism \( \\left( {i,\\mathbb{Z}\\left\\lbrack {{x}_{1},\\cdots ,{x}_{n}}\\right\\rbrack }\\right) \\rightarrow \\left( {j, R}\\right) \), that is, there exists exactly one ring homomorp... | Yes |
Proposition 2.4. For a ring homomorphism \( \varphi : R \rightarrow S \), the following are equivalent:\n\n(a) \( \varphi \) is a monomorphism;\n\n(b) \( \ker \varphi = \{ 0\} \) ;\n\n(c) \( \varphi \) is injective (as a set-function). | Proof. We prove (a) \( \Rightarrow \) (b), leaving the rest to the reader. Assume \( \varphi : R \rightarrow S \) is a monomorphism and \( r \in \ker \varphi \) . Applying the extension property of Example 2.2, we obtain unique ring homomorphisms \( {\operatorname{ev}}_{r} : \mathbb{Z}\left\lbrack x\right\rbrack \right... | No |
Proposition 2.7. Let \( R \) be a ring. Then the function \( r \mapsto {\lambda }_{r} \) is an injective ring homomorphism | Proof. For any \( r \in R \) and for all \( a, b \in R \), distributivity gives\n\n\[{\lambda }_{r}\left( {a + b}\right) = r\left( {a + b}\right) = {ra} + {rb} = {\lambda }_{r}\left( a\right) + {\lambda }_{r}\left( b\right) :\]\n\nthis shows that \( {\lambda }_{r} \) is indeed an endomorphism of the group \( \left( {R,... | Yes |
Example 3.3. Let \( \varphi : R \rightarrow S \) be any ring homomorphism. Then \( \ker \varphi \) is an ideal of \( R \) . | Indeed, we know already that \( \ker \varphi \) is a subgroup; we have to verify the absorption properties. These are an immediate consequence of Lemma 1.2 for all \( r \in R \) , all \( a \in \ker \varphi \), we have\n\n\[ \varphi \left( {ra}\right) = \varphi \left( r\right) \varphi \left( a\right) = \varphi \left( r\... | Yes |
Example 3.4. It need not be, if \( I \) is an arbitrary subgroup of \( R \) . For example, take \( \mathbb{Z} \) as a subgroup of \( \mathbb{Q} \) ; then\n\n\[ 0 + \mathbb{Z} = 1 + \mathbb{Z} \]\n\n\( \left( { = \mathbb{Z}}\right) \) as elements of the group \( \mathbb{Q}/\mathbb{Z} \), and \( \frac{1}{2} + \mathbb{Z} ... | Answer: \( I \) is the kernel of \( R \rightarrow R/I \), so necessarily \( I \) must be an ideal, as seen in Example 3.3 | Yes |
We know that all subgroups of \( \left( {\mathbb{Z}, + }\right) \) are of the form \( n\mathbb{Z} \) for a nonnegative integer \( n \) (Proposition 116.9). It is immediately verified that all subgroups of \( \mathbb{Z} \) are in fact ideals of the ring \( \left( {\mathbb{Z},+, \cdot }\right) \) . | The fact that \( \mathbb{Z} \) is initial in Ring now prompts a natural definition. For a ring \( R \) , let \( f : \mathbb{Z} \rightarrow R \) be the unique ring homomorphism, defined by \( a \mapsto a \cdot {1}_{R} \) . Then \( \ker f = n\mathbb{Z} \) for a well-defined nonnegative integer \( n \) determined by \( R ... | No |
Theorem 3.8. Let \( I \) be a two-sided ideal of a ring \( R \) . Then for every ring homomorphism \( \varphi : R \rightarrow S \) such that \( I \subseteq \ker \varphi \) there exists a unique ring homomorphism \( \widetilde{\varphi } : R/I \rightarrow S \) so that the diagram\n\n![cc115a52-9d62-431d-bb8a-29ed52f048ec... | As a reminder to the lazy reader, \( \widetilde{\varphi } \) is defined by\n\n\[ \widetilde{\varphi }\left( {r + I}\right) \mathrel{\text{:=}} \varphi \left( r\right) \]\n\n(part of) the content of the theorem is that this function is well-defined (if \( I \subseteq \) \( \ker \varphi \) ), and it is a ring homomorphis... | Yes |
Theorem 3.9. Every ring homomorphism \( \varphi : R \rightarrow S \) may be decomposed as follows:  where the isomorphism \( \widetilde{\varphi } \) in the middle is the homomorphism induced by \( \varphi \) (as in T... | The reader will realize that this statement requires no proof at this point: the decomposition holds at the level of groups (by Theorem III8.1) and the maps are all ring homomorphisms as observed earlier in this section. | No |
Proposition 3.11. Let \( I \) be an ideal of a ring \( R \), and let \( J \) be an ideal of \( R \) containing \( I \). Then \( J/I \) is an ideal of \( R/I \), and\n\n\[ \frac{R/I}{J/I} \cong \frac{R}{J} \] | Proof. Since \( I \subseteq J = \ker \left( {R \rightarrow R/J}\right) \), we have an induced ring homomorphism\n\n\[ \varphi : R/I \rightarrow R/J \]\n\nby Theorem 3.8 Explicitly, \( \varphi \left( {r + I}\right) = r + J;\varphi \) is manifestly surjective. Since\n\n\( \ker \varphi = \{ r + I \mid \varphi \left( {r + ... | Yes |
For example, let \( R \) be a commutative ring, and let \( a, b \in R \) ; denote by \( \bar{b} \) the class of \( b \) in \( R/\left( a\right) \) . Then\n\n\[ \left( {R/\left( a\right) }\right) /\left( \bar{b}\right) \cong R/\left( {a, b}\right) \] | Indeed, this is a particular case of Proposition 3.11 since\n\n\[ \left( \bar{b}\right) = \frac{\left( a, b\right) }{\left( a\right) } \]\n\nas ideals of \( R/\left( a\right) \) . | Yes |
Proposition 4.4. \( \mathbb{Z} \) is a PID. | Proof. Let \( I \subseteq \mathbb{Z} \) be an ideal. Since \( I \) is a subgroup, \( I = n\mathbb{Z} \) for some \( n \in \mathbb{Z} \), by Proposition 116.9 Since \( n\mathbb{Z} = \left( n\right) \), this shows that \( I \) is principal. | Yes |
Lemma 4.5. Let \( f\left( x\right) \) be a monic polynomial, and assume\n\n\[ f\left( x\right) {q}_{1}\left( x\right) + {r}_{1}\left( x\right) = f\left( x\right) {q}_{2}\left( x\right) + {r}_{2}\left( x\right) \]\n\nwith both \( {r}_{1}\left( x\right) \) and \( {r}_{2}\left( x\right) \) polynomials of degree \( < \deg ... | Proof. Indeed, we have\n\n\[ f\left( x\right) \left( {{q}_{1}\left( x\right) - {q}_{2}\left( x\right) }\right) = {r}_{2}\left( x\right) - {r}_{1}\left( x\right) \]\n\nif \( {r}_{2}\left( x\right) \neq {r}_{1}\left( x\right) \), then \( {r}_{2}\left( x\right) - {r}_{1}\left( x\right) \) has degree \( < \deg f\left( x\ri... | Yes |
Proposition 4.6. Let \( R \) be a commutative ring, and let \( f\\left( x\\right) \\in R\\left\\lbrack x\\right\\rbrack \) be a monic polynomial of degree \( d \) . Then the function\n\n\[ \n\\varphi : R\\left\\lbrack x\\right\\rbrack \\rightarrow {R}^{\\oplus d} \n\]\n\ndefined by sending \( g\\left( x\\right) \\in R\... | Proof. The given function \( \\varphi \) is well-defined by Lemma 4.5, and it is surjective since it has a right inverse (that is, the function \( \\psi : {R}^{\\oplus d} \\rightarrow R\\left\\lbrack x\\right\\rbrack \) defined above).\n\nI claim that \( \\varphi \) is a homomorphism of abelian groups. Indeed, if\n\n\[... | Yes |
Assume \( f\left( x\right) \) is monic of degree 1: \( f\left( x\right) = x - a \) for some \( a \in R \) . Then the remainder of \( g\left( x\right) \) after division by \( f\left( x\right) \) is simply the ’evaluation’ \( g\left( a\right) \) (cf. Example 2.3). | Indeed, \n\n\[ \n g\left( x\right) = \left( {x - a}\right) q\left( x\right) + r \n\] \n\nfor some \( r \in R \) (the remainder must have degree \( < 1 \) ; hence it is a constant); evaluating at \( a \) gives \n\n\[ \n g\left( a\right) = \left( {a - a}\right) q\left( a\right) + r = 0 \cdot q\left( a\right) + r = r \n\]... | Yes |
For a concrete example, apply this procedure with \( f\left( x\right) = {x}^{2} + 1 \) : Proposition 4.6 gives an isomorphism of groups\n\n\[ R \oplus R \cong \frac{R\left\lbrack x\right\rbrack }{\left( {x}^{2} + 1\right) } \]\n\nwhat multiplication does this isomorphism induce on \( R \oplus R \) ? | Take two elements \( \left( {{a}_{0},{a}_{1}}\right) ,\left( {{b}_{0},{b}_{1}}\right) \) of \( R \oplus R \) . With the notation used in Proposition 4.6, we have\n\n\[ \left( {{a}_{0},{a}_{1}}\right) = \varphi \left( {{a}_{0} + {a}_{1}x}\right) ,\;\left( {{b}_{0},{b}_{1}}\right) = \varphi \left( {{b}_{0} + {b}_{1}x}\ri... | Yes |
For all \( a \in R \), the ideal \( \left( {x - a}\right) \) is prime in \( R\left\lbrack x\right\rbrack \) if and only if \( R \) is an integral domain; it is maximal if and only if \( R \) is a field. | Indeed, \( R\left\lbrack x\right\rbrack /\left( {x - a}\right) \cong R \), as we have seen in Example 4.7 | No |
Proposition 4.11. Let \( I \neq \left( 1\right) \) be an ideal of a commutative ring \( R \) . Then\n\n- I is prime if and only if for all \( a, b \in R \)\n\n\[ \n{ab} \in I \Rightarrow \left( {a \in I\text{ or }b \in I}\right) \n\]\n\n- I is maximal if and only if for all ideals \( J \) of \( R \)\n\n\[ \nI \subseteq... | Proof. The ring \( R/I \) is an integral domain if and only if \( \forall \bar{a},\bar{b} \in R/I \)\n\n\[ \n\bar{a} \cdot \bar{b} = 0 \Rightarrow \left( {\bar{a} = 0\text{ or }\bar{b} = 0}\right) .\n\]\n\nThis condition translates immediately to the given condition in \( R \), with \( \bar{a} = a + I \) , \( \bar{b} =... | Yes |
Proposition 4.12. Let \( I \) be an ideal of a commutative ring \( R \) . If \( R/I \) is finite, then \( I \) is prime if and only if it is maximal. | Proof. This follows immediately from Proposition 1.15 | No |
Proposition 4.13. Let \( R \) be a PID, and let \( I \) be a nonzero ideal in \( R \). Then \( I \) is prime if and only if it is maximal. | Proof. Maximal ideals are prime in every ring, so we only need to verify that nonzero prime ideals are maximal in a PID; we will use the characterization of prime and maximal ideals obtained in Proposition 4.11 Let \( I = \left( a\right) \) be a prime ideal in \( R \), with \( a \neq 0 \), and assume \( I \subseteq J \... | Yes |
Proposition 5.3. Every abelian group is a \( \mathbb{Z} \) -module, in exactly one way. | Proof. Let \( G \) be an abelian group. A \( \mathbb{Z} \) -module structure on \( G \) is a ring homomorphism\n\n\[ \mathbb{Z} \rightarrow {\operatorname{End}}_{\mathrm{{Ab}}}\left( G\right) \]\n\nSince \( \mathbb{Z} \) is initial in Ring (§2.1), there exists exactly one such homomorphism, proving the statement.\n\nTh... | Yes |
Any homomorphism of rings \( \alpha : R \rightarrow S \) may be used to define an interesting \( R \) -module: define \( \rho : R \times S \rightarrow S \) by\n\n\[ \rho \left( {r, s}\right) \mathrel{\text{:=}} \alpha \left( r\right) s \] \n\nfor all \( r \in R \) and \( s \in S \). | The operation on the right is simply multiplication in \( S \) , and the axioms of Definition 5.2 are immediate consequence of the ring axioms and of the fact that \( \alpha \) is a homomorphism. For instance, taking \( S = R \) and \( \alpha = {\operatorname{id}}_{R} \) makes \( R \) a (left-) module over itself. | Yes |
Theorem 5.14. Let \( N \) be a submodule of an \( R \) -module \( M \) . Then for every homomorphism of \( R \) -modules \( \varphi : M \rightarrow P \) such that \( N \subseteq \ker \varphi \) there exists a unique homomorphism of \( R \) -modules \( \widetilde{\varphi } : M/N \rightarrow P \) so that the diagram\n\n!... | As in previous appearances of such statements, this is an immediate consequence of the set-theoretic version ( 115.3) and of easy notation matching and compatibility checks. For an even faster proof, one can just apply Theorem 117.12 and verify that \( \widetilde{\varphi } \) is an \( R \) -module homomorphism. | No |
Proposition 6.1. The direct sum \( M \oplus N \) satisfies the universal properties of both the product and the coproduct of \( M \) and \( N \) . | Proof. Product: Let \( P \) be an \( R \) -module, and let \( {\varphi }_{M} : P \rightarrow M,{\varphi }_{N} : P \rightarrow N \) be two \( R \) -module homomorphisms. The definition of an \( R \) -module homomorphism\n\n\[ \n{\varphi }_{M} \times {\varphi }_{N} : P \rightarrow M \oplus N \n\]\n\nis forced by the need... | No |
Proposition 6.2. The following hold in \( R \) -Mod:\n\n- kernels and cokernels exist;\n\n- \( \varphi \) is a monomorphism \( \Leftrightarrow \ker \varphi \) is trivial \( \Leftrightarrow \varphi \) is injective as a set-function;\n\n- \( \varphi \) is an epimorphism \( \Leftrightarrow \operatorname{coker}\varphi \) i... | Kernels exist: indeed, the 'standard' definition of kernel satisfies the universal properties spelled out above (same argument as in Proposition 1116.6). Cokernels exist: indeed, let\n\n\[ \operatorname{coker}\varphi = \frac{N}{\operatorname{im}\varphi } \]\n\nif \( \beta : N \rightarrow P \) is such that \( \beta \cir... | No |
Proposition 6.4. \( R\left\lbrack A\right\rbrack \) is a free commutative \( R \) -algebra on the set \( A \) . | Proof. The statement translates into the following: for every commutative \( R \) - algebra \( S \) and every set-function \( f : A \rightarrow S \), there exists a unique \( R \) -algebra homomorphism \( \varphi : R\left\lbrack A\right\rbrack \rightarrow S \) such that the diagram\n\n![cc115a52-9d62-431d-bb8a-29ed52f0... | No |
Proposition 6.7. Let \( M \) be an \( R \) -module, and let \( N \) be a submodule of \( M \) . Then \( M \) is Noetherian if and only if both \( N \) and \( M/N \) are Noetherian. | Proof. If \( M \) is Noetherian, then so is \( M/N \) (same proof as for Exercise 4.2), and so is \( N \) (because every submodule of \( N \) is a submodule of \( M \), so it is finitely generated because \( M \) is Noetherian). This proves the ’only if’ part of the statement.\n\nFor the converse, assume \( N \) and \(... | No |
Corollary 6.8. Let \( R \) be a Noetherian ring, and let \( M \) be a finitely generated R-module. Then \( M \) is Noetherian (as an R-module). | Proof. Indeed, by hypothesis there is an onto homomorphism \( {R}^{\oplus n} \rightarrow M \) of \( R \) - modules; hence (by the first isomorphism theorem, Corollary 5.16) \( M \) is isomorphic to a quotient of \( {R}^{\oplus n} \) . By Proposition 6.7, it suffices to prove that \( {R}^{\oplus n} \) is Noetherian.\n\n... | No |
A complex\n\n\[ \cdots \rightarrow 0 \rightarrow L\xrightarrow[]{\alpha }M \rightarrow \cdots \]\n\nis exact at \( L \) if and only if \( \alpha \) is a monomorphism. | Indeed, exactness at \( L \) is equivalent to \( \ker \alpha = \) image of the trivial homomorphism \( 0 \rightarrow L \), that is, to\n\n\[ \ker \alpha = 0. \]\n\nThis is equivalent to the injectivity of \( \alpha \) (Proposition 6.2). | Yes |
A complex \[ \cdots \rightarrow M\overset{\beta }{ \rightarrow }N \rightarrow 0 \rightarrow \cdots \] is exact at \( N \) if and only if \( \beta \) is an epimorphism. | Indeed, the complex is exact at \( N \) if and only if \( \operatorname{im}\beta = \) kernel of the trivial homomorphism \( N \rightarrow 0 \), that is, \( \operatorname{im}\beta = N \). | Yes |
Proposition 7.5. Let \( \varphi : M \rightarrow N \) be an \( R \) -module homomorphism. Then\n\n- \( \varphi \) has a left-inverse if and only if the sequence\n\n\[ 0 \rightarrow M\overset{\varphi }{ \rightarrow }N \rightarrow \operatorname{coker}\varphi \rightarrow 0 \]\n\nsplits.\n\n- \( \varphi \) has a right-inver... | Proof. I will prove the first part and leave the other as an exercise to the reader (Exercise 7.6).\n\nIf the sequence splits, then \( \varphi \) may be identified with the embedding of \( M \) into a direct sum \( M \oplus {M}^{\prime } \), and the projection \( M \oplus {M}^{\prime } \rightarrow M \) gives a left-inv... | No |
In fact, homology should be thought of as a (vast) generalization of the notions of kernel and cokernel. Indeed, consider the (very) particular case in which \( {M}_{ \bullet } \) is the complex\n\n\[ 0 \rightarrow {M}_{1}\overset{\varphi }{ \rightarrow }{M}_{0} \rightarrow 0. \]\n\nThen\n\n\[ {H}_{1}\left( {M}_{ \bull... | I will end this very brief excursion into more abstract territories by indicating how a commutative diagram involving two short exact sequences generates a 'long exact sequence' in homology. This is actually a particular case of a more general construction-according to which a suitable commutative diagram involving thr... | No |
Corollary 7.12. In the same situation presented in the snake lemma (notation as in [7.3]), assume that \( \mu \) is surjective and \( \nu \) is injective. Then \( \lambda \) is surjective and \( \nu \) is an isomorphism. | Proof. Indeed, \( \mu \) surjective \( \Rightarrow \operatorname{coker}\mu = 0;\nu \) injective \( \Rightarrow \ker \nu = 0 \) (Proposition 6.2). Feeding this information into the sequence of the snake lemma gives an exact sequence\n\n\[ 0 \rightarrow \ker \lambda \rightarrow \ker \mu \rightarrow 0 \rightarrow \operato... | No |
Proposition 1.1. Let \( S \) be a finite set, and let \( G \) be a group acting on \( S \). With notation as above,\n\n\[ \left| S\right| = \left| Z\right| + \mathop{\sum }\limits_{{a \in A}}\left\lbrack {G : {G}_{a}}\right\rbrack \]\n\nwhere \( A \subseteq S \) has exactly one element for each nontrivial orbit of the ... | Proof. The orbits form a partition of \( S \), and \( Z \) collects the trivial orbits; hence\n\n\[ \left| S\right| = \left| Z\right| + \mathop{\sum }\limits_{{a \in A}}\left| {O}_{a}\right| \]\n\nwhere \( {O}_{a} \) denotes the orbit of \( a \). By Proposition 1119.9, the order \( \left| {O}_{a}\right| \) equals the i... | Yes |
Corollary 1.3. Let \( G \) be a p-group acting on a finite set \( S \), and let \( Z \) be the fixed point set of the action. Then\n\n\[ \left| Z\right| \equiv \left| S\right| \;{\;\operatorname{mod}\;p}. \] | Proof. Indeed, each summand \( \left\lbrack {G : {G}_{a}}\right\rbrack \) in Proposition 1.1 is a number larger than 1, and a power of \( p \) ; hence it is 0 mod \( p \) . | Yes |
Lemma 1.5. Let \( G \) be a finite group, and assume \( G/Z\left( G\right) \) is cyclic. Then \( G \) is commutative (and hence \( G/Z\left( G\right) \) is in fact trivial). | Proof. (Cf. Exercise 1.5) As \( G/Z\left( G\right) \) is cyclic, there exists an element \( g \in G \) such that the class \( {gZ}\left( G\right) \) generates \( G/Z\left( G\right) \) . Then \( \forall a \in G \)\n\n\[ \n{aZ}\left( G\right) = {\left( gZ\left( G\right) \right) }^{r} \n\]\n\nfor some \( r \in \mathbb{Z} ... | No |
Proposition 1.8 (Class formula). Let \( G \) be a finite group. Then\n\n\[ \left| G\right| = \left| {Z\left( G\right) }\right| + \mathop{\sum }\limits_{{a \in A}}\left\lbrack {G : Z\left( a\right) }\right\rbrack \]\n\nwhere \( A \subseteq G \) is a set containing one representative for each nontrivial conjugacy class i... | Proof. The set of fixed points is \( Z\left( G\right) \), and the stabilizer of \( a \) is the centralizer \( Z\left( a\right) \) ; apply Proposition 1.1 | No |
Corollary 1.9. Let \( G \) be a nontrivial p-group. Then \( G \) has a nontrivial center. | Proof. Since \( \left| {Z\left( G\right) }\right| \equiv \left| G\right| {\;\operatorname{mod}\;p} \) and \( \left| G\right| > 1 \) is a power of \( p \), necessarily \( \left| {Z\left( G\right) }\right| \) is a multiple of \( p \) . As \( Z\left( G\right) \neq \varnothing \) (since \( {e}_{G} \in Z\left( G\right) \) )... | Yes |
Consider a group \( G \) of order 6 ; what are the possibilities for its class formula? | If \( G \) is commutative, then the class formula will tell us very little:\n\n\[ 6 = 6\text{.}\]\n\nIf \( G \) is not commutative, then its center must be trivial (as a consequence of Lagrange’s theorem and Lemma 1.5); so the class formula is \( 6 = 1 + \cdots \), where \( \cdots \) collects the sizes of the nontrivia... | No |
Lemma 1.13. Let \( H \subseteq G \) be a subgroup. Then (if finite) the number of subgroups conjugate to \( H \) equals the index \( \left\lbrack {G : {N}_{G}\left( H\right) }\right\rbrack \) of the normalizer of \( H \) in \( G \) . | Proof. This is again an immediate consequence of Proposition 119.9 | No |
Corollary 1.14. If \( \left\lbrack {G : H}\right\rbrack \) is finite, then the number of subgroups conjugate to \( H \) is finite and divides \( \left\lbrack {G : H}\right\rbrack \) . | Proof.\n\n\[ \left\lbrack {G : H}\right\rbrack = \left\lbrack {G : {N}_{G}\left( H\right) }\right\rbrack \cdot \left\lbrack {{N}_{G}\left( H\right) : H}\right\rbrack \]\n\n(cf. \( §\overline{118.5} \) ). | Yes |
Theorem 2.1 (Cauchy’s theorem). Let \( G \) be a finite group, and let \( p \) be a prime divisor of \( \left| G\right| \) . Then \( G \) contains an element of order \( p \) . | Proof of Theorem 2.1. Consider the set \( S \) of ordered \( p \) -tuples of elements of \( G \) :\n\n\[ \n\left( {{a}_{1},\ldots ,{a}_{p}}\right) \n\]\n\nsuch that \( {a}_{1}\cdots {a}_{p} = e \) . I claim that \( \left| S\right| = {\left| G\right| }^{p - 1} \) : indeed, once \( {a}_{1},\ldots ,{a}_{p - 1} \) are chos... | Yes |
Example 2.4. Let \( p \) be a positive prime integer. If \( \left| G\right| = {mp} \), with \( 1 < m < p \) , then \( G \) is not simple. | Indeed, consider the subgroups of \( G \) with \( p \) elements. By Claim 2.2, the number of such subgroups is \( \equiv 1{\;\operatorname{mod}\;p} \) . Thus, if there is more than one such subgroup, then there must be at least \( p + 1 \) . Any two distinct subgroups of prime order can only meet at the identity (why?)... | No |
Proposition 2.6. If \( {p}^{k} \) divides the order of \( G \), then \( G \) has a subgroup of order \( {p}^{k} \) . | Proof of Proposition 2.6. If \( k = 0 \), there is nothing to prove, so we may assume \( k \geq 1 \) and in particular that \( \left| G\right| \) is a multiple of \( p \) .\n\nArgue by induction on \( \left| G\right| \) : if \( \left| G\right| = p \), again there is nothing to prove; if \( \left| G\right| > p \) and \(... | Yes |
Theorem 2.8 (Second Sylow theorem). Let \( G \) be a finite group, let \( P \) be a p-Sylow subgroup, and let \( H \subseteq G \) be a p-group. Then \( H \) is contained in a conjugate of \( P \) : there exists \( g \in G \) such that \( H \subseteq {gP}{g}^{-1} \) . | Proof. Act with \( H \) on the set of left-cosets of \( P \), by left-multiplication. Since there are \( \left\lbrack {G : P}\right\rbrack \) cosets and \( p \) does not divide \( \left\lbrack {G : P}\right\rbrack \), we know this action must have fixed points (Exercise 1.1): let \( {gP} \) be one of them. This means t... | No |
Lemma 2.9. Let \( H \) be a p-group contained in a finite group \( G \) . Then\n\n\[ \left\lbrack {{N}_{G}\left( H\right) : H}\right\rbrack \equiv \left\lbrack {G : H}\right\rbrack \;{\;\operatorname{mod}\;p}. \] | Proof. If \( H \) is trivial, then \( {N}_{G}\left( H\right) = G \) and the two numbers are equal.\n\nAssume then that \( H \) is nontrivial, and act with \( H \) on the set of left-cosets of \( H \) in \( G \), by left-multiplication. The fixed points of this action are the cosets \( {gH} \) such that \( \forall h \in... | Yes |
Proposition 2.10. Let \( H \) be a p-subgroup of a finite group \( G \), and assume that \( H \) is not a p-Sylow subgroup. Then there exists a p-subgroup \( {H}^{\prime } \) of \( G \) containing \( H \) , such that \( \left\lbrack {{H}^{\prime } : H}\right\rbrack = p \) and \( H \) is normal in \( {H}^{\prime } \) . | Proof. Since \( H \) is not a \( p \) -Sylow subgroup of \( G, p \) divides \( \left\lbrack {{N}_{G}\left( H\right) : H}\right\rbrack \), by Lemma 2.9. Since \( H \) is normal in \( {N}_{G}\left( H\right) \), we may consider the quotient group \( {N}_{G}\left( H\right) /H \), and \( p \) divides the order of this group... | Yes |
Theorem 2.11 (Third Sylow theorem). Let \( p \) be a prime integer, and let \( G \) be a finite group of order \( \left| G\right| = {p}^{r}m \) . Assume that \( p \) does not divide \( m \) . Then the number of p-Sylow subgroups of \( G \) divides \( m \) and is congruent to 1 modulo \( p \) . | Proof. Let \( {N}_{p} \) denote the number of \( p \) -Sylow subgroups of \( G \) .\n\nBy Theorem 2.8, the \( p \) -Sylow subgroups of \( G \) are the conjugates of any given \( p \) -Sylow subgroup \( P \) . By Lemma 1.13, \( {N}_{p} \) is the index of the normalizer \( {N}_{G}\left( P\right) \) of \( P \) ; thus (Cor... | Yes |
Example 2.13. There are no simple groups of order 2002. | Indeed9,\n\n\[ \n{2002} = 2 \cdot 7 \cdot {11} \cdot {13} \n\]\n\nthe divisors of \( 2 \cdot 7 \cdot {13} \) are\n\n\[ \n1,2,7,{13},{14},{26},{91},{182} :\n\]\n\nof these, only 1 is congruent to \( 1{\;\operatorname{mod}\;{11}} \) . Thus there is a normal subgroup of order 11 in every group of order 2002. | Yes |
There are no simple groups of order 12. | Note that \( 3 \equiv 1{\;\operatorname{mod}\;2} \) and \( 4 \equiv 1{\;\operatorname{mod}\;3} \) : thus the argument used above does not guarantee the existence of either a normal 2-Sylow subgroup or a normal 3-Sylow subgroup.\n\nHowever, suppose that there is more than one 3-Sylow subgroup. Then there must be 4 , by ... | Yes |
Example 2.15. There are no simple groups of order 24. | Indeed, let \( G \) be a group of order 24, and consider its 2-Sylow subgroups; by the third Sylow theorem, there are either 1 or 3 such subgroups. If there is 1 , the 2-Sylow subgroup is normal and \( G \) is not simple. Otherwise, \( G \) acts (nontrivially) by conjugation on this set of three 2-Sylow subgroups; this... | Yes |
Theorem 3.2 (Jordan-Hölder). Let \( G \) be a group, and let\n\n\[ G = {G}_{0} \supsetneq {G}_{1} \supsetneq {G}_{2} \supsetneq \cdots \supsetneq {G}_{n} = \{ e\} ,\]\n\n\[ G = {G}_{0}^{\prime } \supsetneq {G}_{1}^{\prime } \supsetneq {G}_{2}^{\prime } \supsetneq \cdots \supsetneq {G}_{m}^{\prime } = \{ e\} \]\n\nbe tw... | Proof. Let\n\n\( \left( *\right) \)\n\n\[ G = {G}_{0} \supsetneq {G}_{1} \supsetneq {G}_{2} \supsetneq \cdots \supsetneq {G}_{n} = \{ e\} \]\n\nbe a composition series. Argue by induction on \( n \) : if \( n = 0 \), then \( G \) is trivial, and there is nothing to prove. Assume \( n > 0 \), and let\n\n\( \left( {* * }... | Yes |
Example 3.3. Let \( G = \mathbb{Z}/6\mathbb{Z} = \{ \left\lbrack 0\right\rbrack ,\left\lbrack 1\right\rbrack ,\left\lbrack 2\right\rbrack ,\left\lbrack 3\right\rbrack ,\left\lbrack 4\right\rbrack ,\left\lbrack 5\right\rbrack \} \) . Then \[ \{ \left\lbrack 0\right\rbrack ,\left\lbrack 1\right\rbrack ,\left\lbrack 2\rig... | The (normal) subgroup \( N = \{ \left\lbrack 0\right\rbrack ,\left\lbrack 2\right\rbrack ,\left\lbrack 4\right\rbrack \} \) ’turns off’ the second factor: indeed, intersecting the series with \( N \) gives \[ \{ \left\lbrack 0\right\rbrack ,\left\lbrack 2\right\rbrack ,\left\lbrack 4\right\rbrack \} \supsetneq \{ \left... | Yes |
Let \( G \) be a group, and let \( N \) be a normal subgroup of \( G \) . Then \( G \) has a composition series if and only if both \( N \) and \( G/N \) have composition series. Further, if this is the case, then\n\n\[ \ell \left( G\right) = \ell \left( N\right) + \ell \left( {G/N}\right) \]\n\nand the composition fac... | Proof. If \( G/N \) has a composition series, the subgroups appearing in it correspond to subgroups of \( G \) containing \( N \), with isomorphic quotients, by Proposition 118.10 (the \ | No |
Proposition 3.5. Any two normal series of a finite group ending with \( \{ e\} \) admit equivalent refinements. | Proof. Refine the series to a composition series; then apply the Jordan-Hölder theorem. | No |
Proposition 3.8. Let \( {G}^{\prime } \) be the commutator subgroup of \( G \) . Then\n\n- \( {G}^{\prime } \) is normal in \( G \) ;\n\n- the quotient \( G/{G}^{\prime } \) is commutative;\n\n- if \( \alpha : G \rightarrow A \) is a homomorphism of \( G \) to a commutative group, then \( {G}^{\prime } \subseteq \) \( ... | Proof. These are all easy consequences of Lemma 3.7\n\n-By Lemma 3.7, the commutator subgroup is characteristic, hence normal (cf. Exercise 2.2).\n\n-By Lemma 3.7, the commutator of any two cosets \( g{G}^{\prime }, h{G}^{\prime } \) is the coset of the commutator \( \left\lbrack {g, h}\right\rbrack \) ; hence it is th... | Yes |
Proposition 3.11. For a finite group \( G \), the following are equivalent:\n\n(i) All composition factors of \( G \) are cyclic.\n\n(ii) \( G \) admits a cyclic series ending in \( \{ e\} \) .\n\n(iii) \( G \) admits an abelian series ending in \( \{ e\} \) .\n\n(iv) \( G \) is solvable. | Proof. (i) \( \Rightarrow \) (ii) \( \Rightarrow \) (iii) are trivial. (iii) \( \Rightarrow \) (i) is obtained by refining an abelian series to a composition series (keeping in mind that the simple abelian groups are cyclic \( p \) -groups).\n\n(iv) \( \Rightarrow \) (iii) is also trivial, since the derived series is a... | Yes |
Corollary 3.13. Let \( N \) be a normal subgroup of a group \( G \) . Then \( G \) is solvable if and only if both \( N \) and \( G/N \) are solvable. | Proof. This follows immediately from Proposition 3.4 and the formulation of solvability in terms of composition factors given in Proposition 3.11 | Yes |
Lemma 4.3. Every \( \sigma \in {S}_{n},\sigma \neq e \), can be written as a product of disjoint nontrivial cycles, in a unique way up to permutations of the factors. | Proof. As we have seen, every \( \sigma \in {S}_{n} \) determines a partition of \( \{ \mathbf{1},\ldots ,\mathbf{n}\} \) into orbits under the action of \( \langle \sigma \rangle \) . If \( \sigma \neq e \), then \( \langle \sigma \rangle \) has nontrivial orbits. As \( \sigma \) acts as a cycle on each orbit, it foll... | No |
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