Let $G$ and $H$ be finite groups and let $p$ be a positive prime number that divides the order of $G$ and that divides the order of $H.$ Let $S$ be a Sylow $p$-subgroup of $G$ and let $T$ be a Sylow $p$-subgroup of $H$. Prove that
$S×T$ is a Sylow $p$-subgroup of $G× H$.
Suppose $S \times T$ is a Sylow $p$-subgroup. Then $S \times T = p^k.$ Which implies $|S| = p^k$ and $|T|= p^k.$ Since  $p^k$ is the largest power of $p$ that divides an integer $n$ if and only if $n = p^k \ell$ and $\gcd ( p, \ell )  = 1,$ and since $S$ is a Sylow $p$-subgroup of $G$ and $T$ is a Sylow $p$-subgroup of $H,$ then $|G|= p^k \ell$ and $|H|= p^k \ell.$ This implies $G \times H = p^k\ell.$ Therefore $S \times T$  is a Sylow $p$-subgroup of $G \times H$.
This is all I could come up with. I think that it is wrong but makes some good points...maybe. Help?

Oops. It looks like you started off assuming what you wanted to prove, if i'm not mistaken.
Anyway, as I said, it suffices to prove that $S\times T$ has the right order.
We know $|S|=p^k$, where $|G|=p^kl$, and $|T|=p^r$, where $|H|=p^rs$ and $(p,l)=(p,s)=1$.
Now $|S\times T|=|S||T|=p^{k+r}$.  But $|G\times H|=|G||H|=p^{k+r}ls$.
So, to finish, you just need that $(p,ls)=1$.  Use Euclid's lemma.
