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

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}{=}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 $n\geq5$.