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.


1 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 $G\to G/H$.

  • $\begingroup$ Then if $A_1,A_2,A_3$ are those three subgroups of $G/H$, then $G=\cup \pi^{-1}(A_i)$. Is that what you mean? Thank you. $\endgroup$
    – Talexius
    Jun 11, 2016 at 14:28
  • $\begingroup$ @Talexius, yes you are thru $\endgroup$
    – janmarqz
    Jun 11, 2016 at 15:38

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