If $|G/H|=4$ then $G$ is union of three proper subgroups

Question

Let $G$ be a finite group and $N$ a normal subgroup of $G$ of index $4$. Prove that $G$ has a subgroup of index $2$ (hence normal) and if $G/H$ is not cyclic then $G$ is equal to the union of three proper subgroups.

I understand that $G/H$ is isomorphic to $\mathbb{Z}_4$ or $\mathbb{Z}_2\times \mathbb{Z}_2$. Let $K^*$ be a subgroup of $G/H$ of order $2$. Then $K^*\cong K/H$ for some $K$ such that $H$ is normal in $K$ and hence $[G:K]=[G/H:K/H]=4/2=2$.

If $G/H$ is not cyclic then $G/H\cong \mathbb{Z}_2\times \mathbb{Z}_2$. I don't know how to prove $G$ is the union of three subgroups. Could it be true that as $G/H$ is the union of three proper subgroups $K_1^*,K_2^*,K_3^*$ then the corresponding groups $K_1,K_2,K_3$ have union equal to $G$? Would anyone give me a hint to prove this?

Thank you.

Answer 1

Hint: If you use, for Klein group $\Bbb Z_2\times\Bbb Z_2$, the presentation $$\langle a,b\mid a^2=e,b^2=e,ab=ba\rangle$$ then there are three subgroups for $(a)=\{e,a\}$, $(b)=\{e,b\}$ and $(ab)=\{e,ab\}$.

Now you could use the natural epimorphism $G\to G/H$.