The Sylow numbers of a direct product are products of the Sylow numbers

Let A and B be finite groups and let p be a prime. Prove that any Sylow p-subgroup of A \times B has the form P \times Q, where P and Q are Sylow p-subgroups of A and B, respectively. Deduce that n_p(A \times B) = n_p(A) \cdot n_p(B). Generalize these results to a direct product of finitely many finite groups (so that the number of Sylow p-subgroups of a product is the product of the numbers of Sylow p-subgroups of the factors).


Let A and B be finite groups and p a prime, and write |A| = p^am and |B| = p^bn, where p does not divide either m or n. Note that |A \times B| = p^{a+b}mn, and that p does not divide mn.

Suppose P \leq A and Q \leq B are Sylow p-subgroups. Then |P| = p^a and |Q| = p^b, so that |P \times Q| = p^{a+b}. Thus P \times Q \leq A \times B is a Sylow p-subgroup.

Now suppose R \leq A \times B is a Sylow p-subgroup.

Define X = \{ x \in A \ |\ (x,y) \in R\ \mathrm{for\ some}\ y \in B \} and Y = \{ y \in B \ |\ (x,y) \in R\ \mathrm{for\ some}\ x \in A \}. We claim that X and Y are subgroups of A and B, respectively; we prove this for X only. Suppose x_1,x_2 \in X; then (x_1,y_1), (x_2,y_2) \in R for some y_1,y_2 \in B. Then (x_1x_2^{-1},y_1y_2^{-1}) \in R, so that x_1x_2^{-1} \in X. By the Subgroup Criterion, X \leq A. Similarly Y \leq B.

Note that if (x,y) \in R, then |(x,y)| = p^k for some k. We also have |(x,y)| = \mathsf{lcm}(|x|,|y|), so that |x| and |y| have p-power order. Thus X and Y are p-subgroups, as otherwise some nonidentity element does not have p-power order. By Sylow’s Theorem, there exist Sylow p-subgroups P \leq A and Q \leq B with X \leq P and Y \leq Q.

Note that R \leq X \times Y \leq P \times Q, but that |R| = p^{a+b} = |P \times Q|. Because these sets are finite, R = X \times Y = P \times Q.

Thus we have proved that \mathsf{Syl}_p(A \times B) = \{ P \times Q \ |\ P \in \mathsf{Syl}_p(A), Q \in \mathsf{Syl}_p(B) \}; hence n_p(A \times B) = n_p(A) \cdot n_p(B).

The generalization to arbitrary finite direct products proceeds by induction as follows.

Suppose that for some k \geq 2, for an arbitrary direct product of groups G = \times_{i=1}^k G_i, every Sylow p-subgroup of G is a product of Sylow p-subgroups of the G_i (and vice versa). Let G = \times_{i=1}^{k+1} G_i be arbitrary. Then every Sylow p-subgroup of G is of the form P \times P_{k+1} where P \leq \times_{i=1}^k G_i and Q \leq G_{k+1} are Sylow p-subgroups (and vice versa); by the induction hypothesis, P = \times_{i=1}^k P_i for Sylow p-subgroups P_i \leq G_i. Thus every Sylow p-subgroup of G has the form \times_{i=1}^k P_i for some Sylow p-subgroups P_i \leq G_i (and vice versa).

Thus we have n_p(\times_{i=1}^k G_i) = \prod_{i=1}^k n_p(G_i) for all groups G_i and positive integers k.

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