Exercise 13.2.3

Proof.

(a)

Let $\sigma \in G$ and $\tau \in S_5\setminus G$. If $\mathit{\sigma \tau }{\sigma }^{-1}\in G$, then $\tau \in G$, in contradiction with the hypothesis. So, if $\sigma \in G$, $$\tau \in C\Rightarrow \sigma \cdot \tau \in C.$$

Moreover, if $\sigma ,{\sigma }^{\prime }\in G$, and $\tau \in C$, then $e\cdot \tau =\mathit{e\tau }{e}^{-1}=\tau$, and

$$\sigma \cdot ({\sigma }^{\prime }\cdot \tau )=\sigma \cdot ({\sigma }^{\prime }\tau {\sigma {}^{\prime }}^{-1})=\sigma {\sigma }^{\prime }\tau {\sigma {}^{\prime }}^{-1}{\sigma }^{-1}=(\sigma {\sigma }^{\prime })\cdot \tau .$$

Therefore $\sigma \cdot \tau =\mathit{\sigma \tau }{\sigma }^{-1}$ defines an action of $G$ on $C$.

(b)

Let $\tau =(a_1{a}_2{a}_3{a}_4{a}_5)\in S_5$ be a $5$-cycle.

$\text{∙}$

Suppose that $\sigma \in S_5$ satisfies $\mathit{\sigma \tau }{\sigma }^{-1}=\tau$. Then $(\sigma (a_1)\sigma (a_2)\sigma (a_3)\sigma (a_4)\sigma (a_5))=(a_1{a}_2{a}_3{a}_4{a}_5)$.

Therefore $\sigma (a_1)\in \{a_1,a_2,a_3,a_4,a_5\}$, so $\sigma (a_1)=a_i$ for some $i\in \text{[[}1,5\text{]]}$.

Then, for all $j\in \text{[[}1,5\text{]]}$, since $\mathit{\sigma \tau }=\mathit{\tau \sigma }$,

$$\begin{array}{llll}\hfill \sigma (a_j)& =(\sigma {\tau }^{j-1})(a_1)& \hfill & \\ \hfill & =({\tau }^{j-1}\sigma )(a_1)& \hfill & \\ \hfill & ={\tau }^{j-1}(a_i)& \hfill & \\ \hfill & ={\tau }^{j-1}({\tau }^{i-1}(a_1))& \hfill & \\ \hfill & ={\tau }^{i-1+j-1}(a_1)& \hfill & \\ \hfill & ={\tau }^{i-1}(a_j),& \hfill & \end{array}$$

Since $\{a_1,a_2,a_3,a_4,a_5\}=\{1,2,3,4,5\}$, $\sigma ={\tau }^{i-1}\in ⟨\tau ⟩$.

$\text{∙}$

Conversely, suppose that $\sigma \in ⟨\tau ⟩$. Since $⟨\tau ⟩$ is cyclic, it is an Abelian subgroup, therefore $\mathit{\sigma \tau }=\mathit{\tau \sigma }$, so $\mathit{\sigma \tau }{\sigma }^{-1}=\tau$.

Conclusion: If $\tau \in S_5$ is a 5-cycle,

$$\forall <!--\; FUNCTION\; APPLICATION\; -->\sigma \in S_5,\mathit{\sigma \tau }{\sigma }^{-1}=\tau ⟺\sigma \in ⟨\tau ⟩.$$

(c)

By part (b), the stabilizer of $\tau \in C$ in $G$ is $${\mathrm{Stab}}_G(\tau )=⟨\tau ⟩\cap G.$$

Since $\tau \in C$, $\tau \notin G$. If ${\tau }^k\in G$ for some $k\in \{2,3,4\}$, since $k\wedge 5=1$, $\mathit{uk}+5v=1$ for some integers $u,v$, so $\tau ={\tau }^{\mathit{uk}}{\tau }^{5v}={({\tau }^k)}^u\in G$, which is a contradiction, so $\tau ,{\tau }^2,{\tau }^3,{\tau }^4$ are not in $G$, so $⟨\tau ⟩\cap G=\{e\}$. Therefore

$$G_{\tau }={\mathrm{Stab}}_G(\tau )=\{e\}.$$

If $\mathcal{}{O}_{\tau }$ is the orbit of $\tau$ for the action defined in part (a),

$$|\mathcal{}{O}_{\tau }|=(G:G_{\tau })=|G|.$$

As $C={\coprod <!--\; FUNCTION\; APPLICATION\; -->}_{\tau \in S}\mathcal{}{O}_{\tau }$, where $S$ is a complete system of the representatives of the orbits, if $m=|S|$ is the number of orbits, $|C|=m|G|$, so

$$|G|\text{ divides }|C|.$$

(d)

By Exercise 2, $S_5$ has $24$ 5-cycles. $$\begin{array}{llll}\hfill \{\tau \in S_5|\tau \text{ is a 5-cycle}\}& =\{\tau \in G|\tau \text{ is a 5-cycle}\}\cup \{\tau \in S_5\setminus G|\tau \text{ is a 5-cycle}\}& \hfill & \\ \hfill & =\{\tau \in G|\tau \text{ is a 5-cycle}\}\cup C,& \hfill & \end{array}$$

where the union is a disjoint union.

Moreover, G has $N$ subgroups of order 5, and the intersection of two such subgroups is $\{e\}$, so $G$ has $N\times 4$ 5-cycles. Therefore

$$24=4N+|C|.$$

(e)

Since $G\subset S_5$ is a transitive subgroup, by Lemma 13.2.1, $5\mid |G|$, and by part (c), $|G|\mid |C|$, so part (d) implies $$5\mid 24-4N.$$

Therefore $4N\equiv 24\equiv 4(\mathit{mod}5)$, so $N\equiv 1(\mathit{mod}5)$, and since $4N\le 24$, $N\le 6$, so

$$N=1\text{ or }N=6.$$