Is a product of $p$-subgroups still a $p$-subgroup?

A $p$-group is a group in which every element has order a power of $p$.

Let $G$ be a group, and let $P$ and $Q$ be $p$-subgroups of $G$. Suppose that the product $PQ$ is a subgroup of $G$ (equivalently, $PQ=QP$). Is $PQ$ necessarily a $p$-subgroup of $G$?

If $P$ and $Q$ are finite then the answer is yes, because $\lvert PQ\rvert=\lvert P\rvert\cdot\lvert Q\rvert/\lvert P\cap Q\rvert$ is a power of $p$.

If $P\leq N_G(Q)$ (or vice versa), then the answer is yes.

Proof: Let $a\in P$ and $b\in Q$. If $a^{p^k}=1$, then \begin{align*} (ab)^{p^k}&=a^{p^k}(a^{-(p^k-1)}ba^{p^k-1})(a^{-(p^k-2)}ba^{p^k-2})\cdots(a^{-2}ba^2)(a^{-1}ba)b\\ &=(a^{-(p^k-1)}ba^{p^k-1})(a^{-(p^k-2)}ba^{p^k-2})\cdots(a^{-2}ba^2)(a^{-1}ba)b\in B, \end{align*} so $(ab)^{p^k}$ has order a power of $p$, which shows that $ab$ has order a power of $p$.

Alternative Proof (thanks to David Craven): Note that $Q\trianglelefteq PQ$, where the quotient group $PQ/Q\cong P/(P\cap Q)$ is a $p$-group. Let $g\in PQ$. Then $gQ\in PQ/Q$, so $Q=(gQ)^{p^k}=g^{p^k}Q$ for some $k$. Then $g^{p^k}\in Q$, so $1=(g^{p^k})^{p^j}=g^{p^{k+j}}$ for some $j$.

Solution 1:

It turns out that this is Question 1.36 in the Kourovka Notebook (the first issue, in 1965), and the item about it in the "Archive of Solved Problems" says that it was shown that the answer is negative in general (i.e., $PQ$ need not be a $p$-group) in:

Sysak, Ya. P., Products of infinite groups, Math. Inst. Akad. Nauk Ukrain. SSR, Kiev, 1982 (Russian).

However, I've not been able to access a copy of this paper, and probably wouldn't be able to read it if I did.