# How can I show that $n_p(A )\cdot n_p(B) = n_p(A \times B)$?

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.

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Where $n_p(A)$ is the number of $p$-Sylows of $A$. –  1015 Apr 26 '13 at 4:02

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 $B$.

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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 $G\times H$?

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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