Let $G = H \times K$ be a finite group (direct product), $P$ a Sylow $p$-subgroup of $G$. Prove that there exist Sylow $p$-subgroups $H'$, $K'$ of $H$ and $K$ respectively so that $P$ = $H' \times K'$

I am very new to the group theory, so can you explain solution properly? Thank you for your help.

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    $\begingroup$ Have you tried anything? $\endgroup$ Dec 17, 2013 at 8:47
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    $\begingroup$ It's not true. Let $G=S_3$, $H=\langle (1,2) \rangle$, $K=\langle (1,2,3) \rangle$, $P=\langle (1,3) \rangle$ with $p=2$. $\endgroup$
    – Derek Holt
    Dec 17, 2013 at 8:57
  • $\begingroup$ Was the product $HK$ supposed to be direct? $\endgroup$ Dec 17, 2013 at 9:13
  • $\begingroup$ @TobiasKildetoft yes, it supposed to be direct. $\endgroup$ Dec 17, 2013 at 9:14
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    $\begingroup$ Presumably the problem should be to prove that there exists a Sylow $p$-subgroup of $G$ which is the product of Sylow $p$-subgroups of $H$ and $K$. I am not seeing immediately how to do that! $\endgroup$
    – Derek Holt
    Dec 17, 2013 at 9:15

4 Answers 4


However, the following is true. Let $G=HK$, $H$, $K$ subgroups and let $p$ a prime dividing the order of $G$. Then there exists a $P \in Syl_p(G)$ such that $P=(P\cap H)(P \cap K)$, with $P \cap H \in Syl_p(H)$ and $P \cap K \in Syl_p(K)$.

Proof Let us first find a Sylow $p$-subgroup $P$ of $G$ such that $P\cap H$ is a Sylow $p$-subgroup of $H$ and $P\cap K$ is a Sylow $p$-subgroup of $K$. Let $Q$ be a Sylow $p$-subgroup of $H$ and let $R$ be a Sylow $p$-subgroup of $K$. Choose a Sylow $p$-subgroup $S$ of $G$ such that $Q\subseteq S$. By Sylow theory, there is a $g\in G$ such that $R\subseteq S^g$. In particular, $S\cap H=Q$ and $S^g\cap K=R$.
But $g=hk$ for some $h\in H$ and $k\in K$. Then $S^g\cap K=R=S^{hk} \cap K=(S^h \cap K)^k$, hence $R^{k^{-1}}=S^h \cap K$ and this is a Sylow $p$-subgroup of $K$, being a conjugate of $R$. On the other hand, $S^h \cap H=(S \cap H)^h=Q^h \in Syl_p(H)$, since it is a conjugate of $Q$. So $P=S^h$ is the Sylow $p$-subgroup we were looking for.

Finally we use a counting argument to show that indeed $(P \cap H)(P \cap K)=P$. Observe that $$|(P \cap H)(P \cap K)|=\frac{|P \cap H| \cdot |P \cap K|}{|P \cap H \cap K|}=\frac{|H|_p \cdot |K|_p}{|P \cap H \cap K|}$$ where the $p$-subscript denotes the largest $p$-power dividing a positive integer (which is understood to be $1$ if the integer in question is not divisible by $p$).

Since $P \cap H \cap K$ is a $p$-subgroup of $H \cap K$, note that $|P \cap H \cap K| \leq |H \cap K|_p$. Combining this: $$|(P \cap H)(P \cap K)| \geq \frac{|H|_p \cdot |K|_p}{|H \cap K|_p}=[\frac{|H| \cdot |K|}{|H \cap K|}]_p=|G|_p=|P|$$ since $G=HK$ and $P \in Syl_p(G)$. As a set $(P \cap H)(P \cap K) \subseteq P$, so we conclude $P=(P \cap H)(P \cap K)$.$\square$

  • $\begingroup$ I need to proof that every Sylow p-subgroup of $G$ is the direct product of Sylow p-subgroups of $H$ and $K$ $\endgroup$ Dec 17, 2013 at 10:27
  • $\begingroup$ @Nicky Hekster: is that easy to prove? $\endgroup$
    – Derek Holt
    Dec 17, 2013 at 12:51
  • $\begingroup$ @Derek, I edited with a proof. $\endgroup$ Dec 17, 2013 at 22:43
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    $\begingroup$ Counterexample: $G = S_3$, $H = \{(1),(12)\}$, and $P = \{(1),(13)\}$. Then $P \in {\rm Syl}_2(G)$ and $H \cap P = \{(1)\}$, so $[H:H \cap P] = |H| = 2$, which is divisible by 2. More generally, if $G$ is divisible by $p$ just once and $H$ and $P$ are distinct $p$-Sylow subgroups of $G$ then $P \in {\rm Syl}_p(G)$ and $H \cap P$ is trivial, so $[H:H \cap P] = |H| = p$, which is divisible by $p$. Your error is that you need to require $HP$ to be a subgroup of $G$. Otherwise $|HP|$ need not divide $|G|$, and without that you can't conclude that $|HP|/|P|$ is not divisible by $p$. $\endgroup$
    – user1728
    May 6, 2019 at 0:33
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    $\begingroup$ The reason I knew something had to be wrong is that if $G$ is a finite group and $H$ is a subgroup, there's no general theorem that every $p$-Sylow subgroup $P$ of $G$ intersects $H$ in a $p$-Sylow subgroup of $H$. There is a theorem that this is true if $H$ is a normal subgroup of $G$, and that's useful because the result is just not true for intersections of Sylow subgroups with arbitrary subgroups. In general a conjugate of $P$ will intersect $H$ in a $p$-Sylow subgroup of $H$, but that conjugate might not be $P$ itself. $\endgroup$
    – user1728
    May 6, 2019 at 0:36

Realized that we (with all the comments above) did not answer the OP’s original question fully. So assume $G=MN$, $M \unlhd G$, $N \unlhd G$, with $M \cap N=1$, that is, $G$ is an (internal) direct product of $M$ and $N$. We use the well-known fact that for any $K \unlhd G$, and $P \in Syl_p(G)$, $P \cap K \in Syl_p(K)$, and all Sylow $p$-subgroups of $K$ arise in this way.

The key thing to show here is that for every $P \in Syl_p(G)$ we have $P=(P \cap M)(P \cap N)$. Note that this implies that $P \cong (P \cap M) \times (P \cap N)$.

Observe that $|G|=|MN|=|M|\dot|N|$ and $|(P \cap M)(P \cap N)|=|P \cap M| \dot |P \cap N|$ (remember $P \cap M \cap N \subseteq M \cap N = 1$).

So $|G|$=$ \color{darkblue}{\frac{|M|}{|P \cap M|}\cdot\frac{|N|}{|P \cap N|}}\cdot|(P \cap M)(P \cap N)|$, where the darkblue numbers are not divisible by $p$. It follows that $|(P\cap M)(P \cap N)|$ $|$ is divisible by $|P|$, and hence $P=(P \cap M)(P \cap N)$. The line of proof is similar to my other answer here on this page.

Note that if one would start from an external direct product, the proof would be somewhat simpler, by noting that the direct product of the Sylow $p$-subgroups of each of the factors, gives a Sylow $p$-subgroup of the whole group.


I believe that we can prove everything using the fact that Sylow $p$-subgroups are maximal $p$-subgroups (by definition).

Let $G = H \times K$ and let $P$ be a Sylow $p$-subgroup of $G$. Consider the projections onto each coordinate of the direct product:

\begin{align} \pi_1 : H \times K &\rightarrow H, \quad \pi_2 : H \times K \rightarrow K \\ (h,k) &\mapsto h, \quad (h,k) \mapsto k \end{align}

Notice that $P \leq \pi_1(P) \times \pi_2(P) \leq G$. But $\pi_1(P) \times \pi_2(P)$ is a $p$-group (see note below) and so since $P$ is a Sylow $p$-subgroup (and thus is maximal), we must have $P = \pi_1(P) \times \pi_2(P)$.

Now let's show that $\pi_1(P)$ is a Sylow $p$-subgroup of $H$ and $\pi_2(P)$ is a Sylow $p$-subgroup of $K$.

Without loss of generality, suppose $\pi_1(P)$ is not a Sylow $p$-subgroup of $H$. Since $\pi_1(P)$ is a $p$-subgroup (see note below), then in particular our assumption implies that it is not maximal in $H$. Then there exists a $p$-subgroup $Q \leq H$ with $|\pi_1(P)| < |Q|$. But then $Q \times \pi_2(P)$ is a $p$-subgroup of $G$ with

\begin{align} |P| = |\pi_1(P) \times \pi_2(P)| < |Q \times \pi_2(P)| \end{align}

But this is impossible, because $P$ is maximal.

Note: I have used the following general property of elements of direct products:

\begin{align} |(h,k)| = \text{lcm}(|h|,|k|) \end{align}

So in particular, if $|(h,k)| = p^n$ for some $n$, then we know that both $h$ and $k$ must have order a power of $p$.


I will be using your notation. Since $G$ has direct factors $H$ and $K$ , the elements of $H$ and $K$ commute, which particularly means elements of $H'$ and $K'$ since they are subgroups of $H$ and $K$ respectively. Therefore, $ H'K'=K'H'$ which implies that $H'K'$ is a subgroup (see Prove that $HK$ is a subgroup iff $HK=KH$.). Since $|H'K'|=|P|$ , $H'K'$ is a Sylow-$p$ subgroup.


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