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Let $G$ be a finite group and let $\pi : G\rightarrow S_G$ be the left regular representation.

Question is to :

Prove that if $x$ is an element of order $n$ and $|G|=mn$ then $\pi(x)$ is a product of $m$ cycles of length $n$.Deduce that $\pi(x)$ is an odd permutation iff $|x|$ is even and $\frac{|G|}{|x|}$ is odd.

I do not have much to say about what i have tried on the way to solution.

what i am thinking is to consider $H=\big<x\big>$ and consider restricted action on cosets of $H$ in $G$.

I am not going anywhere in this way.

please help ,e by givinig some hint for this.

i would be thankful if some one can make me writeup the solution (by giving some hints) than just write an answer in the answer block.

Thank You

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  • $\begingroup$ The property of the regular representation that you need is that, for any nonidentity element $g \in G$, $\pi(g)$ fixes no point of $G$. If $\pi(x)$ were anything other than $m$ cycles of length $n$, then it would have a cycle of length $t<n$, and then $\pi(x^t) = \pi(x)^t$ woul fix a point of $G$. $\endgroup$ – Derek Holt Sep 5 '13 at 8:20
  • $\begingroup$ I am sorry, I am unable to connect to your point. :( $\endgroup$ – user87543 Sep 5 '13 at 12:41
  • $\begingroup$ What is the first step in the argument that you do not understand? $\endgroup$ – Derek Holt Sep 5 '13 at 13:04
  • $\begingroup$ I do not know that for any non identity element $g\in G$ $\pi(g)$ fixes no point of $G$. But, it seems to be natural. I will try proving it. $\endgroup$ – user87543 Sep 5 '13 at 13:06
  • $\begingroup$ if $\pi(x)$ have a cycle of length $t$, I did not understand after this. I do not want to say just like that, but, some jow i could not connect it. $\endgroup$ – user87543 Sep 5 '13 at 13:08
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Let's break the question into two parts. First you need to show that if $x$ has order $n$, then $\pi(x)$ is a product of $m$ cycles of length $n$. So how does $x$ act on $G$? If you pick $g \in G$, then we get the elements $g, xg, x^2g, ..., x^{n-1}g$. Show that these are all distinct since otherwise the order of $x$ is smaller than $n$. Thus, the action of $x$ splits $G$ into $m$ sets of $n$ elements. These are nothing more than the cosets of $\langle x \rangle$, and these cosets are your $n$-cycles. In particular, $(g, xg, x^2g, ..., x^{n-1}g)$ is an $n$-cycle.

Second part. Show that $\pi(x)$ is an odd permutation if and only if $|x|$ is even and $\frac{|G|}{|x|}$ is odd. So in terms of $n$ and $m$ this means $n$ is even and $m$ is odd. Ok, $\pi(x)$ is the product of $m$ cycles of length $n$. Each cycle of length $n$ gives a permutation of $G$ (where you fix everything not in the cycle). We have $sgn(\sigma_1 \circ \sigma_2)=sgn(\sigma_1)\cdot sgn(\sigma_2)$ for permutations $\sigma_1, \sigma_2$. Thus, the sign of $\pi(x)$ is the product of the signs of the $n$-cycles. Show that an $n$-cycle has sign $-(-1)^n$. So the sign of $\pi(x)$ is just $[-(-1)^n]^m$. Show this is $-1$ precisely when $m$ is odd and $n$ is even.

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