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 $j$.

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    $\begingroup$ For your proof, it's notationally simpler to note that $PQ$ is a group with $Q$ a normal subgroup. Then if $g\in PQ$, then $Qg\in P/Q$ a $p$-group, so $(Qg)^{p^n}=Q$ for some $n$. Then $g^{p^n}$ has $p$-power order, and done. This is the same proof as yours, of course. $\endgroup$ Sep 1, 2021 at 17:50
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    $\begingroup$ About your question, my general position on such things is that if it isn't obviously true it's probably false... Groups can be pretty pathological. $\endgroup$ Sep 1, 2021 at 17:52
  • $\begingroup$ Thanks for the cleaner proof. And your guess is my position too. $\endgroup$ Sep 1, 2021 at 17:53
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    $\begingroup$ @DavidA.Craven I can give you an infinite group $G=PQ$, with $P$ and $Q$ $p$-groups but $G$ not a $p$-group, if you can give me a family of finite $p$-groups and subgroups $G_i=P_iQ_i$ where the exponents of $P_i$, $Q_i$ are bounded but those of $G_i$ are not. But maybe there’s a reason that’s not possible? $\endgroup$ Sep 2, 2021 at 13:30

1 Answer 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.

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    $\begingroup$ The survey article "Factorizations of groups and related topics" (Amberg & Kazarin) cites the paper of Sysak, mentioning results on "triply factorized groups" constructed from "radical rings". $\endgroup$ Sep 4, 2021 at 21:25
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    $\begingroup$ I contacted Sysak, and he sent me a copy of the paper "Some Examples of Factorized Groups and their Relation to Ring Theory", which is in English, and also contains a construction of counterexample. $\endgroup$ Sep 5, 2021 at 15:54
  • $\begingroup$ @ThomasBrowning That paper also seems to be available online. My university seems to have a suitable institutional subscription, but it's probably not available for free without that. $\endgroup$ Sep 5, 2021 at 17:18

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