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

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.

• If $H, K$ are these subgroups, then clearly $G \cong H \times K$ by your observations. But $G/H = \mathbb{Z}_2$ and $G/K = \mathbb{Z}_2$. Can you conclude that $H = \mathbb{Z}_2$ and $K = \mathbb{Z}_2$? Commented Jul 19, 2023 at 1:16

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.

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, $$H\cap K = \{e\}$$.
• I'm not quite sure how you conclude that $HK = G$. I understand that $HK$ is a subgroup since $HK = KH$. I am also not sure how one of the subgroups being simple comes into play here. Commented Jul 19, 2023 at 22:07
• @Important_man74 $HK$ is a subgroup strictly larger than $H$, and smaller or equal to $G$; by Lagrange theorem, this forces the index of $HK$ to be $1$. As for your second question, we used the condition of simple group when proving $H\cap K = \{e\}$: $H\cap K$ is a normal subgroup of $H$ by the second isomorphism theorem, so it is either trivial or $H$, but $H\cap K = H$ means $H\subseteq K$, but they have the same index so that gives $H = K$ which is a contradiction, hence $H\cap K = \{e\}$. Commented Jul 20, 2023 at 1:21
• thanks so much for the clarification! Commented Jul 21, 2023 at 1:22

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., $$H, and there's no room for proper subgroups of $$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$$.