Example of two subgroups $H$ and $K$ of a non-abelian group $G$ such that $HK$ is not a subgroup of $G$?

Question

I need to disprove that the set $HK$ is not a subgroup of $G$ if $G$ is non-abelian. Does anyone have a trivial or easy counterexample? I cannot think of one...

Answer 1

Consider $G=S_3,H=\langle (12)\rangle$ and $K=\langle (13)\rangle$.
By product formula, $$|HK|=\frac{|H||K|}{|H\cap K|}=4$$ If $HK$ is a subgroup of $G$, then by Lagrange's Theorem,$|HK|$ divides $|G|$, which is a contradiction.
Hence $HK$ is not a subgroup of $G$.

Answer 2

In the symmetric group $S_4$, you can take $H = \langle (1,2,3)\rangle$ and $K=\langle (1,4)(2,3)\rangle$ (among many other choices). Then it's easy to check that $HK \neq KH$, so $HK$ is not a subgroup of $S_4$. (Or, just directly check that $(1,2,4), (1,2,3)\in HK$, but $(1,2,4)\cdot(1,2,3) = (1,3)(2,4)\not\in HK$.)

Answer 3

Take $G=S_3,H=\left<(1,2)\right>$ and $K=\left<(2,3)\right>$. Then $H$ and $K$ are subgroups of $G($each containing two elements$)$ and $$HK=\{1,(12),(23),(132)\}$$a set of size $4$. Therefore, $HK$ is not a subgroup of $G$$($by Lagrange's Theorem$)$ since $4$ does not divide $6$.