# On a group having a simple subgroup of index $2$

This is an exercise from Rotman's An Introduction to the Theory of Groups (2.72) that is giving me some headaches. I would really just like a hint if at all possible. The question asks

Let $G$ be a group having a simple subgroup $H$ of index $2$. Prove that either $H$ is the unique proper normal subgroup of $G$ or that $G$ contains a normal subgroup $K$ of order $2$ with $G \cong H \times K$. (Hint. Use the second isomorphism theorem.)

So suppose $H$ is not the unique proper normal subgroup of $G$. That is, there exists another proper normal subgroup of $G$, call it $K$. Now $H \cap K \lhd H$, and so $H \cap K = 1$ because $H$ is simple. Now the second isomorphism theorem states $HK / H \cong K / H \cap K \cong K$. That is, $[HK : H] = |K|$. Now $2 = [G : H] = [G : HK] [HK : H] = [G : HK] |K|$. I'm stuck here. If I can conclude that either $[G : HK] = 1$ or $|K| = 2$ then I'll be done. Am I at a dead end here? Hints would be greatly appreciated!

EDIT: I believe we have to put $K$ nontrivial in the assumptions. Consider $G = \mathbb{Z}$ and $H = 2 \mathbb{Z}$. Then clearly $H \lhd G$ has index $2$. But $H$ is not the unique proper normal subgroup of $G$, and there does not exist a subgroup of $G$ of order $2$, much less a normal one. With the assumption now that $K$ is nontrivial, it is clear that $|K| = 2$, which completes the proof.

Assume $H$ is not the unique proper normal subgroup of $G$ and let $1 \lt K \lhd G$, $K \neq H$. Note that $HK$ is a subgroup containing $H$. Since index $[G:H]=2$, $G=HK$ or $HK=H$. In the latter case $K \subseteq H$, which is impossible, since $H$ is simple and $H \neq K$. Hence $G=HK$. But clearly $H \cap K$ is a normal proper subgroup of the simple $H$, so $H \cap K =1$. This means that $[H,K]=1$ (all elements of $H$ commute with those of $K$). But then the map $\phi: G=HK \rightarrow H \times K$ defined by $\phi(hk)=(h,k)$, is in fact a well-defined isomorphism. Hence $G \cong H \times K$. Finally, $|K|=|G/H|=2$. Note that this proof does not depend on any finiteness conditions.
• Why $[G:H]=2$ implies $G=HK$ or $HK=H$? – Twink Jun 10 '14 at 1:24
• @Twink: $[G:H]=[G:HK]\cdot[HK:H]$. – Nicky Hekster Jun 10 '14 at 5:17
Note that either $[HK:H]=1$ (in which case $K$ is trivial), or $[HK:H]=2,$ in which case $|HK| = 2 |H| = |G|.$ So, what could $HK$ be, given that it is a subgroup of $G?$
• We can say that $[HK : H] = 1, 2$ because otherwise $[G : HK] < 1$, correct? In answer to your question, of course $HK = G$ if $G$ is finite, but aren't there issues when $G$ is infinite? Also, is there anything stopping $K$ from being trivial? – tylerc0816 Dec 9 '13 at 2:09
• Well, $HK$ is a subgroup of $G,$ therefore (by some other isomorphism theorem), $HK/N$ is a subgroup of $G/N,$ and $G/HK = (G/N)/(HK/N).$ Since $G/N$ has order $2\dots.$ – Igor Rivin Dec 9 '13 at 2:11