Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
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
add comment

1 Answer

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. $■$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.