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.

  • 5
    $\begingroup$ Either you have a different meaning for "isomorphism" than the usual one (perhaps a $\;1-1\;$ homomorphism=a monomorphism...?), or else there is a mistake in the question, since a group of order $\;n\;$ cannot be isomorphic with a group of order $\;n!\;$ , for $\;n\ge3\;$ ... $\endgroup$
    – DonAntonio
    Dec 6, 2018 at 15:09
  • $\begingroup$ @DonAntonio;question edited $\endgroup$
    – user596656
    Dec 6, 2018 at 15:14
  • $\begingroup$ Still, where you mention Cayley's theorem it should be $\;\phi:G\to K\le S_n\;$ or something similar, and not what you wrote: $\;G\;$ cannot be isomorphic with $\;S_n\;$ . $\endgroup$
    – DonAntonio
    Dec 6, 2018 at 15:20
  • $\begingroup$ @Jean-ClaudeArbaut;why is that not true ,isomorphic image of an element has the same order as the element $\endgroup$
    – user596656
    Dec 6, 2018 at 15:22
  • 2
    $\begingroup$ It seems that the question is meant to be about the specific homomorphism coming from the usual proof of Cayley's Theorem. That is the one arising from the group acting on itself by left multiplication $\endgroup$ Dec 6, 2018 at 15:52

1 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 $\langle g\rangle\backslash G$.


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