8.2: Solutions to the Exercises

These solutions fall somewhere between hints and complete solutions. Students were expected to write out complete solutions.

1-1 By inspection, the only element of order \(2\) is \(c=a^{2}=b^{2}\). Since \(gcg^{-1}\) also has order \(2\), it must equal \(c\), i.e., \(gcg^{-1}=c\) for all \(g\in Q\). Thus \(c\) commutes with all elements of \(Q\), and \(\{1,c\}\) is a normal subgroup of \(Q\). The remaining subgroups have orders \(1\), \(4\), or \(8\), and are automatically normal (see [bd24]a).

1-2 The product \(ab=% \begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix}\), and \(% \begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix} ^{n}=% \begin{pmatrix} 1 & n\\ 0 & 1 \end{pmatrix}\).

1-3 Consider the subsets \(\{g,g^{-1}\}\) of \(G\). Each set has exactly \(2\) elements unless \(g\) has order \(1\) or \(2\), in which case it has \(1\) element. Since \(G\) is a disjoint union of these sets, there must be a (nonzero) even number of sets with \(1\) element, and hence at least one element of order \(2\).

1-4 The symmetric group \(S_{n}\) contains a subgroup that is a direct product of subgroups \(S_{n_{1}}\), …, \(S_{n_{r}}\).

1-5 Because the group \(G/N\) has order \(n\), \((gN)^{n}=1\) for every \(g\in G\) (see [bd16]). But \((gN)^{n}=g^{n}N\), and so \(g^{n}\in N\). For the second statement, consider the subgroup \(\{1,s\}\) of \(D_{3}\). It has index \(3\) in \(D_{3}\), but the element \(t\) has order \(2\), and so \(t^{3}=t\notin\{1,s\}\).

The symmetric group \(S_{n}\) contains a subgroup that is a direct product of subgroups \(S_{n_{1}}\), …, \(S_{n_{r}}\).

1-5 Because the group \(G/N\) has order \(n\), \((gN)^{n}=1\) for every \(g\in G\) (see [bd16]). But \((gN)^{n}=g^{n}N\), and so \(g^{n}\in N\). For the second statement, consider the subgroup \(\{1,s\}\) of \(D_{3}\). It has index \(3\) in \(D_{3}\), but the element \(t\) has order \(2\), and so \(t^{3}=t\notin\{1,s\}\).

1-6 (a) Let \(a,b\in G\). We are given that \(a^{2}=b^{2}=(ab)^{2}=e\). In particular, \(abab=e\). On multiplying this on right by \(ba\), we find that \(ab=ba\). (b) Show by induction that

\[% \begin{pmatrix} 1 & a & b\\ 0 & 1 & c\\ 0 & 0 & 1 \end{pmatrix} ^{n}=% \begin{pmatrix} 1 & na & nb+\frac{n(n-1)}{2}ac\\ 0 & 1 & nc\\ 0 & 0 & 1 \end{pmatrix} . \nonumber \]

1-7 Commensurability is obviously reflexive and symmetric, and so it suffices to prove transitivity. We shall use that if a subgroup \(H\) of a group \(G\) has finite index in \(G\), then \(H\cap G^{\prime}\) has finite index in \(G^{\prime}\) for any subgroup \(G^{\prime}\) of \(G\) (because the natural map \(G^{\prime }/H\cap G^{\prime}\rightarrow G/H\) is injective). Using this, it follows that if \(H_{1}\) and \(H_{3}\) are both commensurable with \(H_{2}\), then \(H_{1}\cap H_{2}\cap H_{3}\) is of finite index in \(H_{1}\cap H_{2}\) and in \(H_{2}\cap H_{3}\) (and therefore also in \(H_{1}\) and \(H_{3}\)). As \(H_{1}\cap H_{3}\supset H_{1}\cap H_{2}\cap H_{3}\), it also has finite index in each of \(H_{1}\) and \(H_{3}\).

1-8 By assumption, the set \(G\) is nonempty, so let \(a\in G\). Because \(G\) satisfies the cancellation law, the map \(x\mapsto ax\colon G\rightarrow G\) is a permutuation of \(G\), and some power of this permutation is the identity permutation. Therefore, for some \(n\geq1\), \(a^{n}x=x\) for all \(x\in G\), and so \(a^{n}\) is a left neutral element. By counting, one sees that every element has a left inverse, and so we can apply ([bd3z]a).

1-9 Let \(b\) be such that the right multiplication \(x\mapsto xb\) is injective. Let \(a_{0}\in G\); there is a unique \(e\in G\) such that \(a_{0}e=a_{0}\). Then \(a_{0}eb=a_{0}b\), which implies that \(eb=b\). Then \(aeb=ab\) for all \(a\in A\), which implies that \(ae=a\). Therefore \(e\) is a right neutral element. For each \(a\in G\), there is a unique \(a^{\prime}\) such that \(aa^{\prime}=e\). Therefore \(G\) also has right inverses, and so it is a group ([bd3z]a).

Let \(G\) be a set, and consider the binary operation \(a,b\mapsto b\) on \(G\). This is associative, and all left multiplications are bijective (in fact, the identity map), but \(G\) is not a group if it has at least two elements.

2-1 The key point is that \(\langle a\rangle=\langle a^{2}\rangle\times\langle a^{n}\rangle\). Apply ([it05]) to see that \(D_{2n}\) breaks up as a product.

2-2 Note first that any group generated by a commuting set of elements must be commutative, and so the group \(G\) in the problem is commutative. According to ([fg09]), any map \(\{a_{1},\ldots,a_{n}\}\rightarrow A\) with \(A\) commutative extends uniquely to homomorphism \(G\rightarrow A\), and so \(G\) has the universal property that characterizes the free abelian group on the generators \(a_{i}\).

2-3 (a) If \(a\neq b\), then the word \(a\cdots ab^{-1}\cdots b^{-1}\) is reduced and \(\neq1\). Therefore, if \(a^{n}b^{-n}=1\), then \(a=b\). (b) is similar. (c) The reduced form of \(x^{n}\), \(x\neq1\), has length at least \(n\).

2-4 (a) Universality. (b) \(C_{\infty}\times C_{\infty}\) is commutative, and the only commutative free groups are \(1\) and \(C_{\infty}\). (c) Suppose \(a\) is a nonempty reduced word in \(x_{1},\ldots,x_{n}\), say \(a=x_{i}\cdots\) (or \(x_{i}^{-1}\cdots\)). For \(j\neq i\), the reduced form of \([x_{j},a]\overset {\text

ParseError: invalid DekiScript (click for details)

Callstack:
    at (Workbench/Group_Theory_4e_(Milne)/08:_Exercises/8.02:_Solutions_to_the_Exercises), /content/body/p[19]/span, line 1, column 1

}{=}x_{j}ax_{j}^{-1}a^{-1}\) can’t be empty, and so \(a\) and \(x_{j}\) don’t commute.

2-5 The unique element of order \(2\) is \(b^{2}\). Since \(gb^{2}g^{-1}\) also has order \(2\) for any \(g\in Q_{n}\), we see that \(gb^{2}g^{-1}=b^{2}\), and so \(b^{2}\) lies in the centre. [Check that it is the full centre.] The quotient group \(Q_{n}/\langle b^{2}\rangle\) has generators \(a\) and \(b\), and relations \(a^{2^{n-2}}=1\), \(b^{2}=1\), \(bab^{-1}=a^{-1}\), which is a presentation for \(D_{2^{n-2}}\) (see [fg10]).

2-6 (a) A comparison of the presentation \(D_{n}=\langle r,s\mid r^{n}% ,s^{2},srsr=1\rangle\) with that for \(G\) suggests putting \(r=ab\) and \(s=a\). Check (using [fg09]) that there are homomorphisms:

\[D_{n}\rightarrow G,\quad r\mapsto ab,\quad s\mapsto a,\qquad G\rightarrow D_{n},\quad a\mapsto s,\quad b\mapsto s^{-1}r\text{.}% \nonumber \]

The composites \(D_{n}\rightarrow G\rightarrow D_{n}\) and \(G\rightarrow D_{n}\rightarrow G\) are the both the identity map on generating elements, and therefore ([fg09] again) are identity maps. (b) Omit.

2-7 The hint gives \(ab^{3}a^{-1}=bc^{3}b^{-1}\). But \(b^{3}=1\). So \(c^{3}=1\). Since \(c^{4}=1\), this forces \(c=1\). From \(acac^{-1}=1\) this gives \(a^{2}=1\). But \(a^{3}=1\). So \(a=1\). The final relation then gives \(b=1\).

2-8 The elements \(x^{2}\), \(xy\), \(y^{2}\) lie in the kernel, and it is easy to see that \(\langle x,y|x^{2},xy,y^{2}\rangle\) has order (at most) \(2\), and so they must generate the kernel (at least as a normal group — the problem is unclear). One can prove directly that these elements are free, or else apply the Nielsen-Schreier theorem ([fg06]). Note that the formula on p.  (correctly) predicts that the kernel is free of rank \(2\cdot2-2+1=3\)

2-9 We have to show that if \(s\) and \(t\) are elements of a finite group satisfying \(t^{-1}s^{3}t=s^{5}\), then the given element \(g\) is equal to \(1\). Because the group is finite, \(s^{n}=1\) for some \(n\). If \(3|n\), the proof is easy, and so we suppose that \(\mathrm{gcd}(3,n)=1\). But then

\[3r+nr^{\prime}=1\text{, some }r,r^{\prime}\in\mathbb{Z}{}, \nonumber \]

and so \(s^{3r}=s\). Hence

\[t^{-1}st=t^{-1}s^{3r}t=(t^{-1}s^{3}t)^{r}=s^{5r}. \nonumber \]

Now,

\[g=s^{-1}(t^{-1}s^{-1}t)s(t^{-1}st)=s^{-1}s^{-5r}ss^{5r}=1, \nonumber \]

as required. [In such a question, look for a pattern. Note that \(g\) has two conjugates in it, as does the relation for \(G\), and so it is natural to try to relate them.]

3-1 Let \(N\) be the unique subgroup of order \(2\) in \(G\). Then \(G/N\) has order \(4\), but there is no subgroup \(Q\subset G\) of order \(4\) with \(Q\cap N=1\) (because every group of order \(4\) contains a group of order \(2\)), and so \(G\neq N\rtimes Q\) for any \(Q\). A similar argument applies to subgroups \(N\) of order \(4\).

3-2 For any \(g\in G\), \(gMg^{-1}\) is a subgroup of order \(m\), and therefore equals \(M\). Thus \(M\) (similarly \(N\)) is normal in \(G\), and \(MN\) is a subgroup of \(G\). The order of any element of \(M\cap N\) divides \(\gcd(m,n)=1\), and so equals \(1\). Now ([it06]) shows that \(M\times N\approx MN\), which therefore has order \(mn\), and so equals \(G\).

3-3 Show that \(\GL_{2}(\mathbb{F}_{2})\) permutes the \(3\) nonzero vectors in \(\mathbb{F}_{2}\times\mathbb{F}{}_{2}\) (\(2\)-dimensional vector space over \(\mathbb{F}{}_{2}\)).

3-4 I thank readers for their help with these solutions. We write the quaternion group as

\[Q=\{\pm1,\pm i,\pm j,\pm k\}. \nonumber \]

Note the \(Q\) has one element of order \(2\), namely, \(-1\), and six elements of order \(4\), namely, \(\pm i\), \(\pm j\), \(\pm k\). (A) The automorphism group of \(Q\) must permute the elements of order \(4\). Write the six elements of order \(4\) on the six faces of a cube with \(i\) opposite \(-i\), etc. Each rotation of the cube induces an automorphism of \(Q\), and \(\Aut(Q)\) is the symmetry group of the cube, \(S_{4}\). (B) Every automorphism of \(Q\) permutes the circularly ordered sets \(\{i,j,k\}\), \(\{-i,-k,-j\}\), \(\{-i,-j,k\}\), …. There are \(8\) of these which can be divided into two sets of \(4\), each of which is permuted transitively by \(\Aut(Q)\). (C). The action of \(\Aut(Q)\) on the set \(\{\langle i\rangle,\langle j\rangle,\langle k\rangle\}\) of subgroups of order \(3\) defines a homomorphism \(\varphi\colon\Aut{Q}\to S_3\). The automorphisms \(i\leftrightarrow j\) and \(j\leftrightarrow k\) of \(Q\) map onto generators of \(S_3\), and so they generate a subgroup of \(\Aut{Q}\) mapped isomorphically onto \(S_3\) by \(\varphi\). Automorphisms of \(Q\) in the kernel of \(\varphi\) map \(i\) to \(\pm i\) and \(j\) to \(\pm j\). This gives us four possibilities, and in fact there is a subgroup of \(\Aut{Q}\) isomorphic to \(V_4\) in the kernel of \(\varphi\). Now \(\Aut(Q)\approx V_4\rtimes S_3\approx S_4\) (see sx195932).

3-5 The pair

\[N=\left\{ \begin{pmatrix} 1 & 0 & b\\ 0 & 1 & c\\ 0 & 0 & 1 \end{pmatrix} \right\} \text{ and }Q=\left\{ \begin{pmatrix} a & 0 & 0\\ 0 & a & 0\\ 0 & 0 & d \end{pmatrix} \right\} \nonumber \]

satisfies the conditions (i), (ii), (iii) of ([it13]). For example, for (i) (Maple says that)

\[% \begin{pmatrix} a & 0 & b\\ 0 & a & c\\ 0 & 0 & d \end{pmatrix}% \begin{pmatrix} 1 & 0 & b\\ 0 & 1 & c\\ 0 & 0 & 1 \end{pmatrix}% \begin{pmatrix} a & 0 & b\\ 0 & a & c\\ 0 & 0 & d \end{pmatrix} ^{-1}\allowbreak=\allowbreak% \begin{pmatrix} 1 & 0 & -\frac{b}{d}+\frac{1}{d}\left( b+ab\right) \\ 0 & 1 & -\frac{c}{d}+\frac{1}{d}\left( c+ac\right) \\ 0 & 0 & 1 \end{pmatrix} \nonumber \]

It is not a direct product of the two groups because it is not commutative.

3-6 Let \(g\) generate \(C_{\infty}\). Then the only other generator is \(g^{-1}\), and the only nontrivial automorphism is \(g\mapsto g^{-1}\). Hence \(\Aut(C_{\infty })=\{\pm1\}\). The homomorphism \(S_{3}\rightarrow\Aut(S_{3})\) is injective because \(Z(S_{3})=1\), but \(S_{3}\) has exactly \(3\) elements \(a_{1},a_{2},a_{3}\) of order \(2\) and \(2\) elements \(b,b^{2}\) of order \(3\). The elements \(a_{1},b\) generate \(S_{3}\), and there are only \(6\) possibilities for \(\alpha(a_{1})\), \(\alpha(b)\), and so \(S_{3}\rightarrow\Aut(S_{3})\) is also onto.

3-7 (a) The element \(g^{o(q)}\in N\), and so has order dividing \(|N|\). (c) The element \(g=(1,4,3)(2,5)\), and so this is obvious. (d) By the first part, \(\left( (1,0,\ldots,0),q\right) ^{p}=((1,\ldots,1),1)\), and \((1,\ldots,1)\) has order \(p\) in \((C_{p})^{p}\). (e) We have \((n,q)(n,q)=(nn^{-1},qq)=(1,1).\)

3-8 Let \(n\cdot q\in Z(G)\). Then

\[\left. \begin{array} [c]{ccc}% (n\cdot q)(1\cdot q^{\prime}) & = & n\cdot qq^{\prime}\\ (1\cdot q^{\prime})(n\cdot q) & = & q^{\prime}nq^{\prime-1}\cdot q^{\prime}q \end{array} \quad\text{all }q^{\prime}\in Q\right\} \implies% \begin{array} [c]{c}% n\in C_{N}(Q)\\ q\in Z(Q) \end{array} \nonumber \]

and

\[\left. \begin{array} [c]{ccc}% (n\cdot q)(n^{\prime}\cdot1) & = & nqn^{\prime}q^{-1}\cdot q\\ (n^{\prime}\cdot1)(n\cdot q) & = & n^{\prime}n\cdot q \end{array} \quad n^{\prime}\in N\quad\right\} \implies n^{-1}n^{\prime}n=qn^{\prime }q^{-1}. \nonumber \]

The converse and the remaining statements are easy.

4-1 Let \(\varphi\colon G/H_{1}\rightarrow G/H_{2}\) be a \(G\)-map, and let \(\varphi(H_{1})=gH_{2}\). For \(a\in G\), \(\varphi(aH_{1})=a\varphi (H_{1})=agH_{2}\). When \(a\in H_{1}\), \(\varphi(aH_{1})=gH_{2}\), and so \(agH_{2}=gH_{2}\); hence \(g^{-1}ag\in H_{2}\), and so \(a\in gH_{2}g^{-1}\). We have shown \(H_{1}\subset gH_{2}g^{-1}\). Conversely, if \(g\) satisfies this condition, the \(aH_{1}\mapsto agH_{2}\) is a well-defined map of \(G\)-sets.

4-2 (a) Let \(H\) be a proper subgroup of \(G\), and let \(N=N_{G}(H)\). The number of conjugates of \(H\) is \((G:N)\leq(G:H)\) (see [ga08]). Since each conjugate of \(H\) has \((H:1)\) elements and the conjugates overlap (at least) in \(\{1\}\), we see that

\[\left\vert \bigcup gHg^{-1}\right\vert <(G:H)(H:1)=(G:1). \nonumber \]

(b) Use that the action of \(G\) on the left cosets of \(H\) defines a homomorphism \(\varphi\colon G\to S_n\), and look at the finite group \(G/\Ker(\varphi)\).

(c) Let \(G=\GL_n(k)\) with \(k\) an algebraically closed field. Every element of \(G\) is conjugate to an upper triangular matrix (its Jordan form). Therefore \(G\) is equal to the union of the conjugates of the subgroup of upper triangular matrices.

(d) Choose \(S\) to be a set of representatives for the conjugacy classes.

4-3 Let \(H\) be a subgroup of a finite group \(G\), and assume that \(H\) contains at least one element from each conjugacy class of \(G\). Then \(G\) is the union of the conjugates of \(H\), and so we can apply Exercise [x20]. (According to , this result goes back to Jordan in the 1870s.)

4-4 According to [ga15], [ga16], there is a normal subgroup \(N\) of order \(p^{2}\), which is commutative. Now show that \(G\) has an element \(c\) of order \(p\) not in \(N\), and deduce that \(G=N\rtimes\langle c\rangle\), etc..

4-5 Let \(H\) be a subgroup of index \(p\), and let \(N\) be the kernel of \(G\rightarrow\Sym(G/H)\) — it is the largest normal subgroup of \(G\) contained in \(H\) (see [ga19]). If \(N\neq H\), then \((H:N)\) is divisible by a prime \(q\geq p\), and \((G:N)\) is divisible by \(pq\). But \(pq\) doesn’t divide \(p!\) — contradiction.

4-6 Embed \(G\) into \(S_{2m}\), and let \(N=A_{2m}\cap G\). Then \(G/N\hookrightarrow S_{2m}/A_{2m}=C_{2}\), and so \((G:N)\leq2\). Let \(a\) be an element of order \(2\) in \(G\), and let \(b_{1},\ldots,b_{m}\) be a set of right coset representatives for \(\langle a\rangle\) in \(G\), so that \(G=\{b_{1},ab_{1},\ldots,b_{m}% ,ab_{m}\}\). The image of \(a\) in \(S_{2m}\) is the product of the \(m\) transpositions \((b_{1},ab_{1}),\ldots,(b_{m},ab_{m})\), and since \(m\) is odd, this implies that \(a\notin N\).

4-7 The set \(X\) of \(k\)-cycles in \(S_{n}\) is normal, and so the group it generates is normal ([bd25l]). But, when \(n\geq5\), the only nontrivial normal subgroups of \(S_{n}\) are \(A_{n}\) and \(S_{n}\) itself. If \(k\) is odd, then \(X\) is contained in \(A_{n}\), and if \(k\) is even, then it isn’t.

4-8 (a) The number of possible first rows is \(2^{3}-1\); of second rows \(2^{3}-2\); of third rows \(2^{3}-2^{2}\); whence \((G:1)=7\times6\times4=168\). (b) Let \(V=\mathbb{F}_{2}^{3}\). Then \(|V|=2^{3}=8\). Each line through the origin contains exactly one point \(\neq\) origin, and so \(|X|=7\). (c) We make a list of possible characteristic and minimal polynomials:

\[% \begin{array} [c]{lllll} & \text{Characteristic poly.} & \text{Min'l poly.} & \text{Size} & \text{Order of element in class}\\ 1 & X^{3}+X^{2}+X+1 & X+1 & 1 & 1\\ 2 & X^{3}+X^{2}+X+1 & (X+1)^{2} & 21 & 2\\ 3 & X^{3}+X^{2}+X+1 & (X+1)^{3} & 42 & 4\\ 4 & X^{3}+1=(X+1)(X^{2}+X+1) & \text{Same} & 56 & 3\\ 5 & X^{3}+X+1\text{ (irreducible)} & \text{Same} & 24 & 7\\ 6 & X^{3}+X^{2}+1\text{ (irreducible)} & \text{Same} & 24 & 7 \end{array} \nonumber \]

Here size denotes the number of elements in the conjugacy class. Case 5: Let \(\alpha\) be an endomorphism with characteristic polynomial \(X^{3}+X+1\). Check from its minimal polynomial that \(\alpha^{7}=1\), and so \(\alpha\) has order \(7\). Note that \(V\) is a free \(\mathbb{F}_{2}[\alpha]\)-module of rank one, and so the centralizer of \(\alpha\) in \(G\) is \(\mathbb{F}_{2}[\alpha]\cap G=\langle\alpha\rangle\). Thus \(|C_{G}(\alpha)|=7\), and the number of elements in the conjugacy class of \(\alpha\) is \(168/7=24\). Case 6: Exactly the same as Case 5. Case 4: Here \(V=V_{1}\oplus V_{2}\) as an \(\mathbb{F}% _{2}[\alpha]\)-module, and

\[\End_{\mathbb{F}_{2}[\alpha]}(V)=\End_{\mathbb{F}_{2}[\alpha]}(V_{1}% )\oplus\End_{\mathbb{F}_{2}[\alpha]}(V_{2}). \nonumber \]

Deduce that \(|C_{G}(\alpha)|=3\), and so the number of conjugates of \(\alpha\) is \(\frac{168}{3}=56\). Case 3: Here \(C_{G}(\alpha )=\mathbb{F}_{2}[\alpha]\cap G=\langle\alpha\rangle\), which has order \(4\). Case 1: Here \(\alpha\) is the identity element. Case 2: Here \(V=V_{1}\oplus V_{2}\) as an \(\mathbb{F}% _{2}[\alpha]\)-module, where \(\alpha\) acts as \(1\) on \(V_{1}\) and has minimal polynomial \(X^{2}+1\) on \(V_{2}\). Either analyse, or simply note that this conjugacy class contains all the remaining elements. (d) Since \(168=2^{3}\times3\times7\), a proper nontrivial subgroup \(H\) of \(G\) will have order

\[2,4,8,3,6,12,24,7,14,28,56,21,24\text{, or }84. \nonumber \]

If \(H\) is normal, it will be a disjoint union of \(\{1\}\) and some other conjugacy classes, and so \((N:1)=1+\sum c_{i}\) with \(c_{i}\) equal to 21, 24, 42, or 56, but this doesn’t happen.

4-9 Since \(G/Z(G)\hookrightarrow\Aut(G)\), we see that \(G/Z(G)\) is cyclic, and so by ([ga17]) that \(G\) is commutative. If \(G\) is finite and not cyclic, it has a factor \(C_{p^{r}}\times C_{p^{s}}\) etc..

4-10 Clearly \((ij)=(1j)(1i)(1j)\). Hence any subgroup containing \((12),(13),\ldots\) contains all transpositions, and we know \(S_{n}\) is generated by transpositions.

4-11 Note that \(C_{G}(x)\cap H=C_{H}(x)\), and so \(H/C_{H}(x)\approx H\cdot C_{G}(x)/C_{G}(x))\). Prove each class has the same number \(c\) of elements. Then

\[|K|=(G:C_{G}(x))=(G:H\cdot C_{G}(x))(H\cdot C_{G}(x):C_{G}(x))=kc. \nonumber \]

4-12 (a) The first equivalence follows from the preceding problem. For the second, note that \(\sigma\) commutes with all cycles in its decomposition, and so they must be even (i.e., have odd length); if two cycles have the same odd length \(k\), one can find a product of \(k\) transpositions which interchanges them, and commutes with \(\sigma\); conversely, show that if the partition of \(n\) defined by \(\sigma\) consists of distinct integers, then \(\sigma\) commutes only with the group generated by the cycles in its cycle decomposition. (b) List of conjugacy classes in \(S_{7}\), their size, parity, and (when the parity is even) whether it splits in \(A_{7}\).

\[% \begin{array} [c]{cccccc} & \text{Cycle} & \text{Size} & \text{Parity} & \text{Splits in }A_{7}? & C_{7}(\sigma)\text{ contains}\\ 1 & (1) & 1 & E & N & \\ 2 & (12) & 21 & O & & \\ 3 & (123) & 70 & E & N & (67)\\ 4 & (1234) & 210 & O & & \\ 5 & (12345) & 504 & E & N & (67)\\ 6 & (123456) & 840 & O & & \\ 7 & (1234567) & 720 & E & Y & \text{720 doesn't divide 2520}\\ 8 & (12)(34) & 105 & E & N & (67)\\ 9 & (12)(345) & 420 & O & & \\ 10 & (12)(3456) & 630 & E & N & (12)\\ 11 & (12)(3456) & 504 & O & & \\ 12 & (123)(456) & 280 & E & N & (14)(25)(36)\\ 13 & (123)(4567) & 420 & O & & \\ 14 & (12)(34)(56) & 105 & O & & \\ 15 & (12)(34)(567) & 210 & E & N & (12) \end{array} \nonumber \]

4-13 According to GAP, \(n=6\), \(a\mapsto(13)(26)(45)\), \(b\mapsto(12)(34)(56)\).

4-14 Since \(\Stab(gx_{0})=g\Stab(x_{0})g^{-1}\), if \(H\subset\Stab(x_{0})\) then \(H\subset\Stab(x)\) for all \(x\), and so \(H=1\), contrary to hypothesis. Now \(\Stab(x_{0})\) is maximal, and so \(H\cdot\Stab(x_{0})=G\), which shows that \(H\) acts transitively.

5-1 Let \(p\) be a prime dividing \(|G|\) and let \(P\) be a Sylow \(p\)-subgroup, of order \(p^{m}\) say. The elements of \(P\) all have order dividing \(p^{m}\), and \(P\) (even \(G\)) has at most \(p^{m-1}\) elements of order dividing \(p^{m-1}\); therefore \(P\) must have an element of order \(p^{m}\), and so it is cyclic. Each Sylow \(p\)-subgroup has exactly \(p^{m}\) elements of order dividing \(p^{m}\), and so there can be only one. Now ([st10]) shows that \(G\) is a product of its Sylow subgroups.

6-2 No, \(D_{4}\) and the quaternion group have isomorphic commutator subgroups and quotient groups but are not isomorphic. Similarly, \(S_{n}\) and \(A_{n}\times C_{2}\) are not isomorphic when \(n\geq5\).