Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $A,B$ be finite groups, such that a prime $p$ divides $|A|$ and $|B|$. Show that $$n_p(A )\cdot n_p(B) = n_p(A \times B).$$

Any ideas on how to proceed? I was thinking that the idea was something along the lines of $n_p(A) = n_p(A \times \{e\})$, and using that fact, but clearly no prime divides the order of $\{e\}$, so that's not helpful.

share|cite|improve this question
Where $n_p(A)$ is the number of $p$-Sylows of $A$. – 1015 Apr 26 '13 at 4:02
up vote 4 down vote accepted

If $P$ is a Sylow subgroup of $A$ and $Q$ is a Sylow subgroup of $B$, then $P \times Q$ is a Sylow subgroup of $A \times B$. We will show that every Sylow subgroup of $A \times B$ is of this form, which proves the claim.

Suppose that $H$ is a Sylow subgroup of $A \times B$. Let $\pi_A: A \times B \rightarrow A$ and $\pi_B: A \times B \rightarrow B$ be the projection homomorphisms. Since $\pi_A(H)$ and $\pi_B(H)$ are both $p$-groups and $H$ is contained in the product $\pi_A(H) \times \pi_B(H)$, we get $H = \pi_A(H) \times \pi_B(H)$. From this it follows that $\pi_A(H)$ is a $p$-Sylow subgroup of $A$, because otherwise the order of $H$ would be smaller than the largest power of $p$ dividing $|A \times B|$. By the same argument $\pi_B(H)$ is a $p$-Sylow subgroup of $B$.

share|cite|improve this answer

Let $P$ be a Sylow $p$-subgroup of $G$ and $Q$ be a Sylow $p$-subgroup of $H$.

Is $P\times Q$ a Sylow subgroup of $G\times H$?

share|cite|improve this answer
Thus $n_p(A)n_p(B) \leq n_p(A \times B)$, but this is not enough for equality. – Mikko Korhonen Dec 12 '12 at 20:06
@m.k. Yeah, I'm just giving OP a hint on how to proceed. He still must make a bijection. – Alexander Gruber Dec 13 '12 at 1:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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