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
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.