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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 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 \( H = {\operatorname{Stab}}_{G}\left( a\right) \) . We 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\( \left| O\right| \) divides \( \left| G\right| \) . | 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. | No |
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 \( R \rightarrow R \) . In other words, \( a \) is not a left- (resp., right-) zero-divisor if and only if multiplicative left- (resp., 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, \[ a\left( {b - c}\right) = {ab} - {ac} = 0, \] and this implies \( b - c = 0 \) since \( a \) is not a left-z... | Yes |
Proposition 1.12. In a ring \( R \) :\n\n- \( u \) is a left- (resp., right-) unit if and only if left- (resp., right-) multiplication by \( u \) is a surjective functions \( R \rightarrow R \) ;\n\n- if \( u \) is a left- (resp., right-) unit, then right- (resp., left-) multiplication by \( u \) is injective; that is,... | Proof. 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\[{\rho }_{u} \ci... | 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 homomorpshisms \( {\mathrm{{ev}}}_{r} : \mathbb{Z}\left\lbrack x\right\rbrack \rightarr... | No |
Proposition 2.6. \( {\operatorname{End}}_{\mathrm{{Ab}}}\left( \mathbb{Z}\right) \cong \mathbb{Z} \) as rings. | Proof. Consider the function\n\n\[ \varphi : {\operatorname{End}}_{\mathrm{{Ab}}}\left( \mathbb{Z}\right) \rightarrow \mathbb{Z} \]\n\ndefined by\n\n\[ \varphi \left( \alpha \right) = \alpha \left( 1\right) \]\n\nfor all group homomorphisms \( \alpha : \mathbb{Z} \rightarrow \mathbb{Z} \) . Then \( \varphi \) is a grou... | Yes |
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) = 0 \cdot a = 0, \... | 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 |
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![ed3b132a-22da-440c-b51e-c335d3c56ae7... | 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:\n\n\n\nwhere the isomorphism \( \widetilde{\varphi } \) in the middle is the homomorphism induced by \( \varphi \) (a... | The reader will realize that this statement requires no proof at this point: the decomposition holds at the level of groups (by Theorem II.8.1) 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\[\n\frac{R/I}{J/I} \cong \frac{R}{J}\n\] | 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\;|\;\varphi \left( {r + ... | Yes |
It is a good idea to get used to a bit of 'calculus' of ideals and quotients in terms of generators; judicious use of the isomorphism theorems yields convenient statements. For example, let \( R \) be a commutative ring, and let \( a, b \in R \) ; denote by \( \bar{b} \) the class of \( b \) in \( R/\left( a\right) \) ... | 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 II.6.9. 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 |
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\[ \varphi : R\left\lbrack x\right\rbrack \rightarrow {R}^{\oplus d} \]\n\ndefined by sending \( g\left( x\right) \in R\left\lbrack x\right\rbrack \) to the r... | 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\nWe claim that \( \varphi \) is a homomorphism of abelian groups. Indeed, if\n\n\[ {g}_{1}\left( ... | 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’ \( g\left( a\right) \) | indeed,\n\n\[ g\left( x\right) = \left( {x - a}\right) q\left( x\right) + r \]\n\nfor some \( r \in R \) (the remainder must have degree \( < 1 \) ; hence it is a constant); evaluating at \( a \) gives\n\n\[ g\left( a\right) = \left( {a - a}\right) q\left( a\right) + r = 0 \cdot q\left( a\right) + r = r \]\n\nas claime... | 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\[{ab} \in I \Rightarrow \left( {a \in I\text{ or }b \in I}\right)\]\n\n- I is maximal if and only if for all ideals \( J \) of \( R \)\n\n\[I \subseteq J \Right... | Proof. The ring \( R/I \) is an integral domain if and only if \( \forall \bar{a},\bar{b} \in R/I \)\n\n\[ \bar{a} \cdot \bar{b} = 0 \Rightarrow \left( {\bar{a} = 0\text{ or }\bar{b} = 0}\right) .\]\n\nThis condition translates immediately to the given condition in \( R \), with \( \bar{a} = a + I \) , \( \bar{b} = b +... | No |
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 (§I.5.3) and of easy notation matching and compatibility checks. For an even faster proof, one can just apply Theorem II.7.12 and verify that \( \widetilde{\varphi } \) is an \( R \) -module homomorphism. | No |
Proposition 5.17. Let \( N \) be a submodule of an \( R \) -module \( M \), and let \( P \) be a submodule of \( R \) containing \( N \) . Then \( P/N \) is a submodule of \( M/N \), and | \[ \frac{M/N}{P/N} \cong \frac{M}{P} \] | Yes |
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... | This proposition of course simply generalizes to \( R \) -Mod facts we know already from our study of \( \mathrm{{Ab}} \), and a quick review should suffice for the careful reader. Kernels exist: indeed, the 'standard' definition of kernel satisfies the universal properties spelled out above (same argument as in Propos... | 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![ed3b132a-22da-440c-b51e-c335d3c5... | 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 \[ \cdots \rightarrow 0 \rightarrow L\xrightarrow[]{\alpha }M \rightarrow \cdots \] is 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 \[ \ker \alpha = 0\text{.} \] This 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. We 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-in... | No |
Lemma 7.8 (The snake lemma). With notation as above, there is an exact sequence\n\n\[ 0 \rightarrow \ker \lambda \rightarrow \ker \mu \rightarrow \ker \nu \overset{\delta }{ \rightarrow }\operatorname{coker}\lambda \rightarrow \operatorname{coker}\mu \rightarrow \operatorname{coker}\nu \rightarrow 0. \] | Proving the snake lemma is something that should not be done in public, and it is notoriously useless to write down the details of the verification for others to read: the details are all essentially obvious, but they lead quickly to a notational quagmire. Such proofs are collectively known as the sport of diagram chas... | 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 \[ 0 \rightarrow \ker \lambda \rightarrow \ker \mu \rightarrow 0 \rightarrow \operatorna... | 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 II.9.9, 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 power of \( p \) larger than 1 ; 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 |
Example 1.10. 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 II.9.9. | 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. §II.8.5). | 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 \( p \) -tuples of elements of \( G \) :\n\n\[ \n\left\{ {{a}_{1},\ldots ,{a}_{p}}\right\} \n\]\n\nsuch that \( {a}_{1}\cdots {a}_{p} = e \) . We claim that \( \left| S\right| = {\left| G\right| }^{p - 1} \) : indeed, once \( {a}_{1},\ldots ,{a}_{p - 1} \) are chosen (a... | 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?)... | Yes |
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. 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. | Indeed, 2002 = 2 \cdot 7 \cdot 11 \cdot 13; the divisors of 2 \cdot 7 \cdot 13 are 1,2,7,13,14,26,91,182: of these, only 1 is congruent to 1 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 \( 3 \equiv 1\; \) mod \( 2 \) and \( 4 \equiv 1\; \) 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 the third Sylow theorem... | Yes |
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 two... | Proof. Let\n\n(*) \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\( \left( {* * }\right) \)\n\n\[ G... | No |
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 II.8.10 (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 t... | 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 |
All \( p \) -groups are solvable. | Indeed, the composition factors of a \( p \) -group are simple \( p \) -groups (what else could they be?), hence cyclic. | No |
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. | No |
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 |
Lemma 4.5. Let \( \tau \in {S}_{n} \), and let \( \left( {{a}_{1}\ldots {a}_{r}}\right) \) be a cycle. Then\n\n\[ \tau \left( {{a}_{1}\ldots {a}_{r}}\right) {\tau }^{-1} = \left( {{a}_{1}{\tau }^{-1}\ldots {a}_{r}{\tau }^{-1}}\right) . \] | Proof. This is verified by checking that both sides act in the same way on \( \{ \mathbf{1},\ldots ,\mathbf{n}\} \) . For example, for \( 1 \leq i < r \)\n\n\[ \left( {{a}_{i}{\tau }^{-1}}\right) \left( {\tau \left( {{a}_{1}\ldots {a}_{r}}\right) {\tau }^{-1}}\right) = {a}_{i}\left( {{a}_{1}\ldots {a}_{r}}\right) {\tau... | No |
Proposition 4.6. Two elements of \( {S}_{n} \) are conjugate in \( {S}_{n} \) if and only if they have the same type. | Proof. The 'only if' part of this statement follows immediately from the preceding considerations: conjugating a permutation yields a permutation of the same type. As for the 'if' part, suppose\n\n\[ \n{\sigma }_{1} = \left( {{a}_{1}\ldots {a}_{r}}\right) \left( {{b}_{1}\ldots {b}_{s}}\right) \cdots \left( {{c}_{1}\ldo... | Yes |
In \( {S}_{8} \), \(\left( {18632}\right) \left( {47}\right) \text{ and }\left( {12345}\right) \left( {67}\right)\) must be conjugate, since they have the same type. | The proof of Proposition 4.6 tells us that \(\tau \left( {18632}\right) \left( {47}\right) {\tau }^{-1} = \left( {12345}\right) \left( {67}\right)\) for \(\tau = \left( \begin{array}{llllllll} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ 1 & 8 & 6 & 3 & 2 & 4 & 7 & 5 \end{array}\right)\) and of course this may be checked by hand i... | Yes |
Corollary 4.8. The number of conjugacy classes in \( {S}_{n} \) equals the number of partitions of \( n \) . | For example, there are 7 conjugacy classes in \( {S}_{5} \), indexed by the Tetris look-alikes drawn above. It is also reasonably straightforward to compute the number of elements in each conjugacy class, in terms of the type. For example, in order to count the number of permutations of type \( \left\lbrack {2,2,1}\rig... | No |
There are no normal subgroups of size 30 in \( {S}_{5} \) . | Indeed, normal subgroups are unions of conjugacy classes (§1.3); since the identity is in every subgroup and \( {30} - 1 = {29} \) cannot be written as a sum of the numbers appearing in the class formula for \( {S}_{5} \), there is no such subgroup. | Yes |
Lemma 4.11. Transpositions generate \( {S}_{n} \) . | Proof. Indeed, by Lemma 4.3 it suffices to show that every cycle is a product of transpositions, and indeed\n\n\[ \left( {{a}_{1}\ldots {a}_{r}}\right) = \left( {{a}_{1}{a}_{2}}\right) \left( {{a}_{1}{a}_{3}}\right) \cdots \left( {{a}_{1}{a}_{r}}\right) \]\n\nas may be checked by applying \( {}^{18} \) both sides to ev... | No |
Lemma 4.12. Let \( \sigma = {\tau }_{1}\cdots {\tau }_{r} \) be a product of transpositions. Then \( \sigma \) is even, resp., odd, according to whether \( r \) is even, resp., odd. | Proof. This follows immediately from the facts that \( \epsilon \) is a homomorphism and the sign of a transposition is -1 : indeed, \( \left( {ij}\right) \) acts on \( {\Delta }_{n} \) by permuting its factors and changing the sign of exactly one factor (namely, the factor \( \pm \left( {{x}_{i} - {x}_{j}}\right) \) )... | Yes |
Lemma 4.14. Let \( n \geq 2 \), and let \( \sigma \in {A}_{n} \) . Then \( {\left\lbrack \sigma \right\rbrack }_{{A}_{n}} = {\left\lbrack \sigma \right\rbrack }_{{S}_{n}} \) or the size of \( {\left\lbrack \sigma \right\rbrack }_{{A}_{n}} \) is half the size of \( {\left\lbrack \sigma \right\rbrack }_{{S}_{n}} \), acco... | Proof. (Cf. Exercise 1.16.) Note that\n\n\[ \n{Z}_{{A}_{n}}\left( \sigma \right) = {A}_{n} \cap {Z}_{{S}_{n}}\left( \sigma \right) \]\n\nthis follows immediately from the definition of centralizer (Definition 1.6). Now recall that the centralizer of \( \sigma \) is its stabilizer under conjugation, and therefore the si... | Yes |
Looking again at \( {A}_{5} \), we have noted in \( §{4.3} \) that the types of the even permutations in \( {S}_{5} \) are \( \left\lbrack {1,1,1,1,1}\right\rbrack ,\left\lbrack {2,2,1}\right\rbrack ,\left\lbrack {3,1,1}\right\rbrack \), and \( \left\lbrack 5\right\rbrack \) . By Proposition 4.15 the conjugacy classes ... | Therefore there are exactly 5 conjugacy classes in \( {A}_{5} \), and the class formula for \( {A}_{5} \) is\n\n\[ \n{60} = 1 + {15} + {20} + {12} + {12}.\n\] | Yes |
Corollary 4.17. The alternating group \( {A}_{5} \) is a simple noncommutative group of order 60. | Proof. A normal subgroup of \( {A}_{5} \) is necessarily the union of conjugacy classes, contains the identity, and has order equal to a divisor of 60 (by Lagrange’s theorem). The divisors of 60 other than 1 and 60 are\n\n\[2,3,4,5,6,{10},{12},{15},{20},{30}\]\n\ncounting the elements other than the identity would give... | Yes |
Lemma 4.18. The alternating group \( {A}_{n} \) is generated by 3-cycles. | Proof. Since every even permutation is a product of an even number of 2-cycles, it suffices to show that every product of two 2-cycles may be written as product of 3-cycles. Therefore, consider a product\n\n\[ \left( {ab}\right) \left( {cd}\right) \]\n\nwith \( a \neq b, c \neq d \) . If \( \left( {ab}\right) = \left( ... | Yes |
Theorem 4.20. The alternating group \( {A}_{n} \) is simple for \( n \geq 5 \) . | Proof. We have already checked this for \( n = 5 \), and the reader has checked it for \( n = 6 \) . For \( n > 6 \), let \( N \) be a nontrivial normal subgroup of \( {A}_{n} \) ; we will show that necessarily \( N = {A}_{n} \), by proving that \( N \) contains 3-cycles.\n\nLet \( \tau \in N,\tau \neq \left( 1\right) ... | No |
Corollary 4.21. For \( n \geq 5 \), the group \( {S}_{n} \) is not solvable. | Proof. Since \( {A}_{n} \) is simple, the sequence \[ {S}_{n} \supsetneq {A}_{n} \supsetneq \{ \left( 1\right) \} \] is a composition series for \( {S}_{n} \) . It follows that the composition factors of \( {S}_{n} \) are \( \mathbb{Z}/2\mathbb{Z} \) and \( {A}_{n} \) . By Proposition 3.11, \( {S}_{n} \) is not solvabl... | Yes |
Lemma 5.1. Let \( N \) , \( H \) be normal subgroups of a group \( G \) . Then\n\n\[ \left\lbrack {N, H}\right\rbrack \subseteq N \cap H \] | Proof. It suffices to verify this on generators; that is, it suffices to check that\n\n\[ \left\lbrack {n, h}\right\rbrack = n\left( {h{n}^{-1}{h}^{-1}}\right) = \left( {{nh}{n}^{-1}}\right) {h}^{-1} \in N \cap H \]\n\nfor all \( n \in N, h \in H \) . But the first expression and the normality of \( N \) show that \( \... | Yes |
Corollary 5.2. Let \( N, H \) be normal subgroups of a group \( G \) . Assume \( N \cap H = \{ e\} \) . Then \( N, H \) commute with each other:\n\n\[ \left( {\forall n \in N}\right) \left( {\forall h \in H}\right) \;{nh} = {hn}. \] | Proof. By Lemma 5.1, \( \left\lbrack {N, H}\right\rbrack = \{ e\} \) if \( N \cap H = \{ e\} \) ; the result follows immediately. | Yes |
Proposition 5.3. Let \( N, H \) be normal subgroups of a group \( G \), such that \( N \cap H = \) \( \{ e\} \) . Then \( {NH} \cong N \times H \) . | Proof. Consider the function\n\n\[ \varphi : N \times H \rightarrow {NH} \]\n\ndefined by \( \varphi \left( {n, h}\right) = {nh} \) . Under the stated hypothesis, \( \varphi \) is a group homomorphism: indeed\n\n\[ \varphi \left( {\left( {{n}_{1},{h}_{1}}\right) \cdot \left( {{n}_{2},{h}_{2}}\right) }\right) = \varphi ... | Yes |
Lemma 5.7. Let \( N \) be a normal subgroup of a group \( G \), and let \( H \) be a subgroup of \( G \) such that \( G = {NH} \) and \( N \cap H = \{ e\} \) . Then \( G \) is a split extension of \( H \) by \( N \) . | Proof. We have to construct an exact sequence\n\n\[ 1 \rightarrow N \rightarrow G \rightarrow H \rightarrow 1\text{; } \]\n\nwe let \( N \rightarrow G \) be the inclusion map, and we prove that \( G/N \cong H \) . For this, consider the composition\n\n\[ \alpha : H \hookrightarrow G \rightarrow G/N. \]\n\nThen \( \alph... | Yes |
Lemma 5.8. The resulting structure \( \left( {N \times H,{ \bullet }_{\theta }}\right) \) is a group, with identity element \( \left( {{e}_{N},{e}_{H}}\right) \) . | Proof. The reader should carefully verify this. For example, inverses exist because\n\n\[ \left( {{n}_{1},{h}_{1}}\right) { \bullet }_{\theta }\left( {{\theta }_{{h}_{1}^{-1}}\left( {n}_{1}^{-1}\right) ,{h}_{1}^{-1}}\right) = \left( {{n}_{1}{\theta }_{{h}_{1}}\left( {{\theta }_{{h}_{1}^{-1}}\left( {n}_{1}^{-1}\right) }... | No |
Proposition 5.10. Let \( N, H \) be groups, and let \( \theta : H \rightarrow {\operatorname{Aut}}_{\mathrm{{Grp}}}\left( N\right) \) be a homomorphism; let \( G = N{ \rtimes }_{\theta }H \) be the corresponding semidirect product. Then\n\n- \( G \) contains isomorphic copies of \( N \) and \( H \) ;\n\n- the natural p... | Proof. The functions \( N \rightarrow G, H \rightarrow G \) defined for \( n \in N, h \in H \) by\n\n\[ n \mapsto \left( {n,{e}_{H}}\right) ,\;h \mapsto \left( {{e}_{N}, h}\right) \]\n\nare manifestly injective homomorphisms, allowing us to identify \( N, H \) with the corresponding subgroups of \( G \) . It is clear t... | Yes |
Proposition 5.11. Let \( N \) , \( H \) be subgroups of a group \( G \) , with \( N \) normal in \( G \) . Assume that \( N \cap H = \{ e\} \), and \( G = {NH} \) . Let \( \gamma : H \rightarrow {\operatorname{Aut}}_{\mathrm{{Grp}}}\left( N\right) \) be defined by conjugation: for \( h \in H, n \in N \) ,\n\n\[ {\gamma... | Proof. Define a function\n\n\[ \varphi : N{ \rtimes }_{\gamma }H \rightarrow G \]\n\nby \( \varphi \left( {n, h}\right) = {nh} \) ; this is clearly a bijection. We need to verify that \( \varphi \) is a homomorphism, and indeed \( \left( {\forall n \in N}\right) ,\left( {\forall h \in H}\right) \) :\n\n\[ \varphi \left... | Yes |
The automorphism group of \( {C}_{3} \) is isomorphic to the cyclic group \( {C}_{2} \) : if \( {C}_{3} = \left\{ {e, y,{y}^{2}}\right\} \), then the two automorphisms of \( {C}_{3} \) are | \[ \text{ id : }\left\{ {\begin{aligned} e & \mapsto e, \\ y & \mapsto y, \\ {y}^{2} & \mapsto {y}^{2}, \end{aligned}\;\sigma : \begin{cases} e & \mapsto e, \\ y & \mapsto {y}^{2}, \\ {y}^{2} & \mapsto y. \end{cases}}\right. \] | Yes |
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