Show $G$ is isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$

Question

Let $G$ be a finite group with at least two (distinct) subgroups of index $2$, and suppose that at least one of the index-$2$ subgroups of $G$ is simple. Prove that $G\cong \mathbb{Z}_2 \times \mathbb{Z}_2$.

I am pretty stuck on this problem. I know that subgroups of index $2$ are normal. I also know that since they intersect trivially, that their elements commute with one another, but I'm not sure where to go in terms of proving the isomorphism.

Answer 1

Suppose $H$ and $K$ subgroups of $G$ with index $2$, and $H$ is simple. As you observed, $H\cap K = \{e\}$ by the second isomorphism theorem. The second isomorphism theorem also tells us that

• $H\subsetneq HK\subseteq G$ is a subgroup of $G$, and in particular $HK = G$.
• $HK/K\cong H/(H\cap K)$ is a group isomorphism. We now have $G/K\cong H$. But the group $G/K$ has order $2$ and this forces $H\cong \mathbb Z/2\mathbb Z$. Exchanging the order of $H$ and $K$ tells us that $K\cong \mathbb Z/2\mathbb Z$ in an analogous fashion.

In particular, we have $G = HK$ and $H\cap K = \{e\}$, so $G = H\times K$ is a direct product.

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Edit: Elaboration on OP's questions

• Since $H$ and $K$ are distinct, then $HK$ must be strictly bigger than $H$, so it has index less than $2$, therefore it must have index $1$, but by definition this means $G = HK$.
• We used the condition of a simple group when showing $H\cap K = \{e\}$. By the second isomorphism theorem, we have $H\cap K \lhd H$ is a normal subgroup, but $H$ is a simple group, so $H\cap K$ is either $H$ or the trivial subgroup $\{e\}$. But $H\cap K\neq H$, otherwise $H\subseteq K$, and since they have the same index this means $H = K$, contradiction. Therefore, $H\cap K = \{e\}$.

Answer 2

Call $H$ and $K$ the two subgroups. They are both normal (as of index $2$) and intersect trivially (as the intersection of normal subgroups is normal, and either $H$ or $K$ is simple). Therefore, $G\cong H\times K$ (because, e.g., $H<H\times K$, and there's no room for proper subgroups of $G$ of order greater than $|H|$). Say $n:=|H|=|K|$, it must then be $|G|=n^2=2n$, whence $n=2$ and hence $G\cong C_2\times C_2$.