Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 5.1 Exercise 5.1.4
Solution: 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 $k$.
