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Let $G$ a finite group. And let $\varphi : G→S(G):g↦λ_g$. If $|G|=nm$ and $|g|=n$, then $\varphi(g)$ is the product of $m$ disjoint $n$ cycles.

When writing this question, I think I've proven the question myself :) Could anybody check if this is correct ? If there are other ways to prove this, I'd be glad to hear them as well!

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There is a problem with your proof that for $x \neq y$, the cycles mentioned are disjoint. It is not a true statement. For example, they are not disjoint if $y=gx$. – user3533 Apr 19 '13 at 10:47
@user3533 You are right, I changed my proof a little bit. – Kasper Apr 19 '13 at 10:52
Now it looks good. – user3533 Apr 19 '13 at 11:08
@user3533 Thanks for looking at it ! – Kasper Apr 19 '13 at 11:28

Proof $\;$ We have $\varphi(g)=λ_g$. Note that for any $x\in G$:

$$x\overset{λ_g}{\mapsto}gx \overset{λ_g}{\mapsto}g^2x\overset{λ_g}{\mapsto}...\overset{λ_g}{\mapsto}g^nx=x$$

So we find a cycle of length $n$: $(x \; gx \;g^2x\;...\;g^{n-1}x)$.

Note that $\langle g\rangle x$ corresponds with this cycle. Since $\langle g \rangle≤G$, then this is a right coset. So $\{\langle g \rangle x: x\in G \}$ gives a partition of $G$ and every right coset corresponds with a $n$ cycle. Since $|G|=nm$, then there are $m$ different right cosets of order $n$. Since the right cosets give a partition of $G$, then all of them are disjoint. So $λ_g$ a product of $m$ disjoint $n$ cycles. $■$

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