Let $G$ be a finite group. My problem is: $$|\hom(G, \Bbb Z/2)|>1 \iff |G:[G,G]| \mbox{ is even}.$$
I know that $\hom(G, \Bbb Z/2)=\hom(G^{ab}, \Bbb Z/2)$, where $G^{ab}:=G/[G,G]$. If $|\hom(G^{ab}, \Bbb Z/2)|>1$, there exists $f \in \hom(G^{ab}, \Bbb Z/2)$ and $|G^{ab}|= |\ker f||\rm{im}~f|$. Since $f$ is nontrivial, $f$ needs to be onto i.e. $|\rm{im}~f|=2$. Hence $|G:[G,G]|$ is even.
Conversely, if $|G^{ab}|$ is even how can I prove that $|\hom(G^{ab}, \Bbb Z/2)|>1$? I know $G^{ab}$ is abelian and contains an element of order $2$, say $x$. From this can we find any nontrivial element of $\hom(G^{ab}, \Bbb Z/2)$?
Any help would be much appreciated.
Let $G^{\mathrm{ab}}$ be the abelianization and suppose that it contains an element of order $2$. It follows that the map $x\in G^{\mathrm{ab}}\mapsto 2x\in G^{\mathrm{ab}}$ has a non trivial kernel, so —since $G^{\mathrm{ab}}$ is finite— it is not surjective. The subgroup $2G^{\mathrm{ab}}$ of $G^{\mathrm{ab}}$ is therefore a proper subgroup. Now $H=G^{\mathrm{ab}}/2G^{\mathrm{ab}}$ is non-trivial abelian group in which every element has order $2$. Such a thing is, in a canonical way, a vector space over the field $\mathbb Z/2\mathbb Z$. Since $H\neq0$, from linear algebra we know that there is a non-zero linear map $H\to \mathbb Z/2\mathbb Z$. The composition of that with the canonical projection $G^{\mathrm{ab}}\to H$ is non-zero, clearly.