Show that $\phi(g)$ is an even permutation

Question

Let $G$ be a group of order $n$ then $G$ is isomorphic to a subgroup of $S_n$,Denote it by $\phi$, $\phi:G\to S_n$ be an monomorphism of Cayleys Theorem

Let $g\in G$ has order $k$.Show that $\phi(g)$ is a product of disjoint cycles of length $k$.Moreovver Show that $\phi(g)$ is an even permutation unless $g$ has even order and $\langle g\rangle $ has odd index in $G$.

My Try:

Since $g$ has order $k$ so $\phi(g)$ also has order $k$ and hence is either a $k-$ cycle or a product of $k$ cycles.

Given that $g$ has even order and so $\phi(g)$ has even order and hence it is a product of odd number of transpositions.

But I need to prove that it is even permutation?

Where am I missing the point?

Please help.

Answer 1

As pointed out in the comments, the homomorphism $\phi:G\to Sym(G)$ is the one arising from the action of $G$ on itself by left multiplication. The cycle decomposition of $\phi(g)$ is obtained from the orbits of the action of $g$ on $G$. Let $\mathcal{X}$ be a set of representatives of the orbits of $g$. Then, as $g$ has order $k$, $$\phi(g)=\prod_{x\in\mathcal{X}}(x,gx,g^2x,\ldots,g^{k-1}x)$$ which is a product of disjoint cycles.

Now, cycles of odd length are even. Therefore, if $k$ is odd, $\phi(g)\in A_n$. If $k$ is even, then $\phi(g)\in A_n$ if, and only if $|X|$ is even (why?). The result now follows from the following lemma:

Lemma: $|X|=[G:\langle g\rangle]$.

Proof: For $x,y\in G$, $g^rx=y$ if, and only if $yx^{-1}\in\langle g\rangle$. Therefore, the map $x\mapsto \langle g\rangle x$ defines a bijection between $\mathcal{X}$ and the set of right cosets $\langle g\rangle\backslash G$.