$A$ and $B$ are finite groups. Prove that any Sylow $p$-subgroup of $A \times B$ is of the form $P \times Q$.

Question

We let $A$ and $B$ be two finite groups. We want to show that any Sylow $p$-subgroup of $A \times B$ is of the form $P \times Q$, where $P$ is a Sylow $p$-subgroup of $A$ and $Q$ is a Sylow $p$-subgroup of $B$. \ We first let $|A| = p^rm$ with $p \nmid m$. And we let $|B| = p^s n$ with $p \nmid n$. Then we have $|A \times B| = |A| \cdot |B| = p^{r+s}mn$, with $p \nmid mn$. And if $P$ is a Sylow $p$-subgroup of $A$ and $Q$ is a Sylow $p$-subgroup of $B$, then $|P| = p^r$, $|Q| = p^s$. Then $|P \times Q| = |P| \cdot |Q| = p^{r+s}$, which implies that $P \times Q$ is a Sylow subgroup of $A \times B$. \ Then we suppose that $R \leq A\times B$ is a Sylow $p$-subgroup of $A \times B$. We define $$X = =\{x \in A: (x,y) \in R \text{ for some } y \in B\}$$ and $$Y = \{y \in B: (x,y) \in R \text{ for some } x \in A\}$$ We claim that $X \leq A$, $Y \leq B$. And since $R$ is a Sylow subgroup of $A \times B$, $|R| = p^{k}$ for some $k \in \mathbb{N}$. And we also know that $|(x,y)| = \text{lcm} (|x|,|y|)$.This means that $|x|$ divides $p^k$ and $|y|$ also divides $p^k$, which means that both orders of $x$ and $y$ are powers of $p$. And this means that both $X$ and $Y$ are $p$-groups. And then this means that we could find $P$ and $Q$ such that $P$ is a Sylow $p$-subgroup of $A$ and $Q$ is a Sylow $p$-subgroup of $B$ and $X \leq P \leq A$ and $Y \leq Q \leq B$. And it follows that $R \leq X \times Y \leq A \times B$. And I searched the solution online. It says that $|R| = p^{a+b} = |P \times Q|$ for some $a$ and $b$. But I don't know how to get this conclusion from the previous steps. Can anyone help me please? Thanks!