Let $G$ be a finite group of order $n$ and let $\psi:G\to S_n$ be the homomorphism defined by the right multiplication action. I have to prove that:
$\text{(a)}$ Let $x\in G$ such that $o(x)=2$. Show that $\psi(x)\in A_n\iff4\mid n$.
$\textbf{My attempt:}$ Since $o(x)=2$, $o(\psi(x))|2\implies o(\psi(x))\in\{1,2\}$. If $o(\psi(x))=1$, the image of $x$ is the identity element and we have nothing to show. If $o(\psi(x))=2=\text{lcm}(type(\psi(x)))$ and we want that the condition $\psi(x)\in A_n$ is satisfied, the element $\psi(x)$ will be an element of order $2$ composed by the multiplication of an even number of $2$-cycles. The smallest $2$-cycle that satisfies this condition has to be in form $(ab)(cd)$, so we need at least $4$ elements in $G$.
The operation that we do if we want $\psi(x)$ composed by a greater number of $2$-cycle is adding $2\cdot k$ $2$-cycles, for example:$$n=4\implies\psi(x)=(ab)(cd)$$$$n=8\implies\psi(x)=(ab)(cd)(ef)(gh)$$$$\vdots$$Conversely is really quick.
$\text{(b)}$ Assuming $|G|=2\cdot s$, where $s\in 2\mathbb Z+1$, show that $\psi^{-1}(A_n)\le G$ , concluding that $\exists K\le G$ such that $|G:K|=2$.
Here is my difficulty but I suppose that the subgroup $K$, for the fact that it's normal, is the kernel of the map.
$\text{(c)}$ Show that $K$ is characteristic in $G$.
$\textbf{My attempt:}$ This request follows easily from point $\text{(b)}$ because $K$ has index $2$ in $G\implies K\trianglelefteq G$ and
$$\begin{align} |G:K|=2&\implies |K|=\dfrac{|G|}{2}=\dfrac{\not 2\cdot s}{\not2}=s\\ &\implies \gcd(|K|,|G:K|)=1\\ &\implies K\text{ is characteristic}. \end{align}$$
Could you suggest to me a method to attack point $\text{(b)}$?
Thank for you help.
The crucial idea for the entire answer, not just part (b), is that the action is right multiplication. Consequently, the kernel of $\psi$ is the identity element. (Note that $\psi (g)$ maps each element $x$ to $xg$ and that $x$ only equals $xg$ if $g$ is the identity.)
Part (b)
Let $I$ be the image of $G$. Then $|I|=|G|$.
Now consider the subgroup $I\bigcap A_n$ of $S_n$. By the argument of part (a), $I$ is not contained in $A_n$. Then $|IA_n||I\bigcap A_n|=|I||A_n|$ and so $|I|=2|I\bigcap A_n|$.
Therefore $\psi ^{-1}(|I\bigcap A_n|)$ is a subgroup of $G$ of index $2$.