Guys. Let $G$ be a finite group en let $a\in G$. Then define $\lambda_a\colon G\to G$ by $\lambda_a(x)=ax$ for $x\in G$. So I’ve shown that $\lambda_a$ is the product of $[G:\langle a\rangle]$ disjoint cyckles of length $\operatorname{ord}(a)$. Now my book is asking the following: Show that $\lambda_a$ is an odd permutation iff the order of $G$ is even, and the order of $a$ is divisible by the highest power of 2 that divides the order of $G$.

So I’m guessing what we’re looking for is for the order of $a$ to be even in any case. And then we want $[G:\langle a\rangle]$ to be odd. I’m not rea…

Say $\vert G\vert=n$. And say $\operatorname{ord}(a)=k$. I wrote out the elements in the following way:
\begin{align}
ax_1,a^2x_1,&\dots,a^kx_1\\
&\vdots\\
ax_n,a^2x_n,&\dots,a^kx_n.
\end{align}

We know that $a,a^2,\dots,a^k$ are all different elements. So we know that for each $i$, $ax_i,\dots,a^kx_i$ are different elements. Now I basically have to show that those sets (the rows) form a partition. So if $a^mx_i=a^nx_j$, I need to show that the two sets are equal. Obviously $a^{m+1}x_i=a^{n+1}x_j$, and so on. It might be useful to use modular arithmetic here (even though, technically, it's…