Let \(A, B\) be finite groups and let \(p\) be a prime. Prove that any Sylow \(p\)-subgroup of \(A \times B\) is of the form \(P \times Q\), where \(P \in Syl_p(A)\) and \(Q \in Syl_p(B)\). Prove that \(n_p(A \times B) = n_p(A)n_p(B)\). Generalize both of these results to a direct product of any finite number of finite groups.
Let \(\left|A\right| = p^\alpha m\) and \(\left|B\right| = p^\beta n\), where \(p \nmid m\) and \(p \nmid n\). Suppose \(C \in Syl_p(A \times B)\). Consider now the restriction of the projection homomorphism \(\pi_A : A \times B \rightarrow A\) to \(C\) (which we shall still denote as \(\pi_A\), and whose kernel and image we shall denote as \(K\) and \(I\)). \(I\) is a subgroup of \(A\) and \(K\) can be identified with a subgroup \(K'\) of \(B\), i.e. \(\left|I\right| \mid p^\alpha m\) and \(\left|K'\right| \mid p^\beta n\). This in addition with the fact that \(\left|C\right| = p^{\alpha+\beta}\) and \(C/K \cong I\) force both \(I \in Syl_p(A)\) and \(K' \in Syl_p(B)\). In particular, if \(a \in I\) and \(b \in K'\), then it is easily seen that \((a,1), (1,b) \in C\), so \((a,b) \in C\). Thus \(I \times K' \subseteq C\), and equality holds since their order is equal.
Clearly \(n_p(A \times B)\) is the number of ways to form direct products of Sylow \(p\)-subgroups of \(A\) and \(B\), which is simply \(n_p(A)n_p(B)\).
The last statement is trivial.
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