Let $G$ be a finite group such that $|G| = nm$. Suppose $x\in G$ has order $n$ and let $\sigma_x\in S_G$ be the permutation such that $\sigma_x(g)=xg$ for every $g\in G$. Note that $\sigma_x$ is a product of $m$ disjoint n-cycles.
a) Show that if $n=2$ and $m$ is odd, show that there is a homomorphism $f : G \to S_{nm}$ such that $f(G) \not \subseteq A_{nm}$.
b) Show that if $n=2$ and $m$ is odd, conclude that $G$ has a subgroup of index $2$.
The homomorphism $h$ is given by $x\to\sigma _x$, where $\sigma _xg=xg\,,\forall g\in G$. When $\mid x\mid=n$, since $\sigma _x$ is a product of $m$ disjoint $n$ cycles, $n=2$ and $m$ odd implies $\sigma_x\in S_{nm}$ is odd.
For the second part, let $H\subset G$ be given by $H=h^{-1}(A_{nm})$. Then it is easy to see that $[G:H]=2$.