Prove that the image of an element $g \in G$ of order $k$ is a product of $n/k$ disjoint $k$-cycles in $S(G)$.

Question

So for context, this is the problem I'm trying to solve.

Let $G$ be a finite group of order $n$ and $G \to S(G) \cong S_n$ the map associated with Caley's theorem. Prove that the image of an element $g \in G$ of order $k$ is a product of $n/k$ disjoint $k$-cycles in $S(G)$. When does the image of $G$ in $S(G)$ contain odd permutations?

What I'm lost about is the idea of the Cayley map. From what I've understood this map is $f: G \to S(G)$ given by $g \mapsto \lambda _g$ where $\lambda_g: G \to G$ is a map given by $x \mapsto gx$.

For a concrete example, I wrote the map as $g \mapsto(x \mapsto gx)$ and took $G=S_3$ so $\#G=6$ and then the elements $g=(1\;2)$ so $\#g=k=2$; and $x=(1 \; 2 \; 3)$. Which gives $$gx=(1 \; 2)(1 \; 2 \; 3)=(2 \; 3)$$ but this isn't a product 3 disjoint $2$-cycles so I think I've misunderstood something. Can someone help explain it?

Answer 1

$G$ acts on the set $\{1,2,3\}$, and $S(G)$ acts on the set $G=\{\sigma_1,\cdots,\sigma_6\}$, where the $\sigma_i$s are the six possible permutations of $\{1,2,3\}$. The cycle type of an element of $G$ is written using elements of $\{1,2,3\}$, but the cycle type of an element of $S(G)$ is written using elements of $\{\sigma_1,\cdots,\sigma_6\}$.

Let's be explicit. Label the permutations

$$ \begin{array}{lll} \sigma_1=(1)(2)(3), & \sigma_3=(1)(23), & \sigma_5=(31)(2), \\ \sigma_2=(12)\sigma_1=(12)(3), & \sigma_4=(12)\sigma_3=(123), & \sigma_6=(12)\sigma_5=(321). \end{array} $$

Then $\lambda_{(12)}=(\sigma_1\sigma_2)(\sigma_3\sigma_4)(\sigma_5\sigma_6)$ is a product of three $2$-cycles. If you want, you can also use the labels $1$-$6$ for the permutations, so $\lambda=(12)(34)(56)$, as long as you're careful to keep track of when the numbers refer to elements of the original set $\{1,2,3\}$ versus when they refer to permutations of that set.

For your problem, suppose $(x~gx~\cdots~g^{m-1}x)$ is an $m$-cycle of $\lambda_g$. Can you continue?

Answer 2

(Your "$g$" is here "$x$".) Consider the action of the subgroup $\langle x\rangle=\{x^i,i=0,\dots,k-1\}$ on $G$ by left multiplication. For any $a\in G$, the orbit of $a$ reads:

$$O(a)=\{a,xa,\dots,x^{k-1}a\} \tag1$$ which has size $k$$^\dagger$. So, $G$ splits into $\frac{|G|}{k}=\frac{n}{k}$ orbits of size $k$ each.

Now, for every $g\in O(a)$, there is $i\in\{0,\dots,k-1\}$ such that, for every $l\in\Bbb N$: $$\lambda_x^l(g)=\lambda_x^l(x^ia)=x^{i+l\pmod k}a \tag2$$ (induction on $l$). Therefore, the restriction of $\lambda_x$ to $O(a)$ is a $k$-cycle of $S_{O(a)}$. In order to turn it into a $k$-cycle of $S_G$, you have to extend ${\lambda_x}_{\mid O(a)}$ to $G\setminus O(a)$ by the identity map, $\lambda_x(g)=g$. So, named after $(1)$ the set $\{a_1,\dots,a_{r:=n/k}\}$ of orbit representatives, let's define the map $\alpha_i$, $i=1,\dots,r$, in this way: \begin{alignat}{1} &{\alpha_i}_{\mid O(a_i)}:={\lambda_x}_{\mid O(a_i)} \\ &{\alpha_i}_{\mid O(a_j)}:=\operatorname{id}_{O(a_j)}, \text{ for every }j=1,\dots,r \text{ such that } j\ne i\\ \tag3 \end{alignat} Accordingly: \begin{alignat}{2} &g \in O(a_i) &&\Longrightarrow (\alpha_i\alpha_j)(g)=\alpha_i(\alpha_j(g))=\alpha_i(g)=\lambda_x(g) \\ &g \in O(a_j) &&\Longrightarrow (\alpha_i\alpha_j)(g)=\alpha_i(\alpha_j(g))=\alpha_i(\lambda_x(g))=\lambda_x(g) \\ &g \in O(a_{l\ne i,j}) &&\Longrightarrow (\alpha_i\alpha_j)(g)=\alpha_i(\alpha_j(g))=\alpha_i(g)=g \\ \tag4 \end{alignat} or, equivalently: \begin{alignat}{2} &g \in O(a_i)\sqcup O(a_j) &&\Longrightarrow (\alpha_i\alpha_j)(g)=\lambda_x(g) \\ &g \in O(a_{l\ne i,j}) &&\Longrightarrow (\alpha_i\alpha_j)(g)=g \\ \tag5 \end{alignat} By induction on $(5)$: \begin{alignat}{2} &g \in O(a_1)\sqcup\dots\sqcup O(a_r)=G &&\Longrightarrow (\alpha_1\dots\alpha_r)(g)=\lambda_x(g) \\ \tag6 \end{alignat} namely: $$\lambda_x=\alpha_1\dots\alpha_r \tag7$$ and $\lambda_x$ is the product of $n/k$ disjoint $k$-cycles (of $S_G$).

---

$^\dagger$In fact, for $0\le j\le i \le k-1$ (and hence $0\le i-j \le k-1$): \begin{alignat}{1} &x^ia=x^ja &&\iff \\ &x^i=x^j &&\iff \\ &x^{i-j}=e &&\iff \\ &i-j=0&&\iff \\ &i=j \\ \end{alignat} whence all the elements listed in $(1)$ are pairwise distinct.