If $H$ and $K$ are cyclic subgroup of $G$ then $HK$ is subgroup of $G$?

Question

If $H=\langle a \rangle$ and $K=\langle b \rangle$ are cyclic subgroup of $G$ then $HK = \{ a^ib^j \; | \; 0 \leq i,j \leq p-1 \}$ is subgroup of $G$, given that the $|H| = |K| = p$ where $p$ is prime.

Proof: It’s clear that $e \in HK$ Now we have to show that set is closed . But I am not able to prove that since $G$ is not Abelian group. Since if G is Abelian then $(a^i \cdot b^j) \cdot (a^p \cdot b^q)=a^{i+p} \cdot b^{j+q}$. I also have a feeling that it’s not subgroup but I can’t think of any counter example.

Answer 1

The simplest counterexample is $S_3$. The product of cyclic subgroups generated by $(1,2)$ and $(2,3)$ resp. is not a subgroup.

Answer 2

First note that $|HK|=\frac{|H|\cdot|K|}{|H\cap K|}$. Since $|H|=|K|=p$, $H\cap K\ne 1$ if and only if $H=K$, in which case $HK=H$ is certainly a subgroup of $G$.

Now if $H\cap K=1$ (which means $H\ne K$), then $|HK|=p^2$. It is well-known that a group of order $p^2$ is abelian. So if $x,y$ are elements of order $p$ in $G$ and $x,y$ are not commutative, then $H=\langle x\rangle$ and $K=\langle y\rangle$ is certainly a counterexample.

There are a lot of such examples for $(G,x,y)$: such as $(S_3,(12),(13))$, $(A_4,(123),(134))$ and so on.