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.

  • 1
    $\begingroup$ Have you tried anything? $\endgroup$ – Nicky Hekster Dec 17 '13 at 8:47
  • 1
    $\begingroup$ 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 '13 at 8:57
  • $\begingroup$ Was the product $HK$ supposed to be direct? $\endgroup$ – Tobias Kildetoft Dec 17 '13 at 9:13
  • $\begingroup$ @TobiasKildetoft yes, it supposed to be direct. $\endgroup$ – Evgeny Eltishev Dec 17 '13 at 9:14
  • 1
    $\begingroup$ 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 '13 at 9:15

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 $K$ $\endgroup$ – Evgeny Eltishev Dec 17 '13 at 10:27
  • $\begingroup$ @Nicky Hekster: is that easy to prove? $\endgroup$ – Derek Holt Dec 17 '13 at 12:51
  • $\begingroup$ @Derek, I edited with a proof. $\endgroup$ – Nicky Hekster Dec 17 '13 at 22:43
  • 1
    $\begingroup$ 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 at 0:33
  • 1
    $\begingroup$ 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 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 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.


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 $p$.


Let $P_H=\{h\in H:(h,k)\in P\ \text{for some}\ k\in K\}$ and $P_K=\{k\in K: (h,k)\in P\ \text{for some}\ h\in H\}$. It should be clear that $P_H$ and $P_K$ are subgroups of $H$ and $K$ respectively.

Suppose $|H|=p^n q$ and $|K|=p^m q'$, then $|G|=p^{n+m}qq'$.

Note that the orders of element of $P$ are divisors of $p^{n+m}$, so are the orders of element of $P_H$ and $P_K$. Thus $P_H$ and $P_K$ are $p$-subgrouops of $H$ and $K$ respectively.

On the other hand, $|P|=p^{n+m}\le|P_H|\cdot |P_K|\le p^n\cdot p^m$, so $|P_H|=p^n$, $|P_K|=p^m$ and $P=P_H\times P_K$ where $P_H$ and $P_K$ are Sylow $p$-subgroups of $H$ and $K$ respectively. Otherwise, the equality in $|P|\le |P_H|\cdot |P_K|$ cannot be achieved. And we are done.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.