Let $H$ and $K$ be subgroups of a group $G$. We know that $HK$ is a subgroup of $G$ if and only if $HK=KH$. Is it true that $HK$ is a subgroup of $G$ if and only if either $H$ or $K$ is a normal subgroup of $G$? Clearly if $H$ or $K$ is a normal subgroup of $G$, then $HK=KH$ which implies $HK$ is a subgroup of $G$. So really is it true that if $HK$ is a subgroup of $G$ (or really $HK=G$) does it imply that either $H$ or $K$ normal? I am mainly interested in finite groups and when $G$ factors over $H$ and $K$ so that $G=HK$.
Is it true that $HK$ is a subgroup of $G$ iff either $H$ or $K$ is a normal subgroup of $G$?
2
$\begingroup$
abstract-algebra
group-theory
finite-groups
-
2See https://en.wikipedia.org/wiki/Zappa%E2%80%93Sz%C3%A9p_product. – 2017-02-22
-
0@QiaochuYuan So good to see you 'round!!! – 2017-02-22
1 Answers
2
This is not the case. For example neither $H=\langle(1,2)\rangle$ or $K=\langle(3,4)\rangle$ is normal in $G=S_5$, but $HK$ is a subgroup of $G$.
A weaker sufficent condition for $HK$ to be a subgroup of $G$ than $H$ or $K$ normal is $K\le N_G(H)$ (that is for any $k\in K$ $Hk=kH$) or $H\le N_G(K)$.
Unfortunately, this is not necessary. For example if $G=S_4$ then set $H=\langle(1,2,3,4)\rangle$ and $K=G_1$ (the stabiliser in $G$ of $1$). Neither $K\le N_G(H)$ or $H\le N_G(K)$, but $HK=G$.
-
2That can't be right because $|HK| \le |H||K| = 8$. It would work with $H = \langle (1,2),(1,2,3) \rangle$. – 2017-02-22
-
0Of course you're right, I'll remove that part of the answer, thank you – 2017-02-22
-
0What would be the case is $HK=G$? – 2017-02-22
-
0If $HK=G$ then $HK$ is obviously a subgroup of $G$. I've added a correct example to my answer so you can see $H$ and $K$ don't need to satisfy the sufficient condition I'd stated – 2017-02-23