= Solution
For subgroups $H,P\leq G$, define $x\sim y$ when $y\in HxP$. The identity elements show reflexivity, inverses show symmetry, and multiplication shows transitivity. Thus the <double coset>[double cosets] $HxP$ partition $G$.
Let $H$ act by left multiplication on the set of left <coset>[cosets] $G/P$. The orbit of $xP$ consists of the cosets contained in $HxP$, so its size is $|HxP|/|P|$. Its stabilizer is
$$
\begin{aligned}
\operatorname{Stab}_H(xP)
&=\{h\in H:hxP=xP\}\\
&=H\cap xPx^{-1}.
\end{aligned}
$$
The <orbit-stabilizer theorem> therefore gives
$$
\boxed{\frac{|HxP|}{|P|}
=\frac{|H|}{|H\cap xPx^{-1}|}}.
$$
Suppose $P$ is a <Sylow subgroup> of $G$, with $|P|=p^a$, and write the largest power of $p$ dividing $|H|$ as $p^b$. If every $H\cap xPx^{-1}$ had order at most $p^{b-1}$, the displayed formula would make every double-coset size divisible by $p^{a+1}$. Their sum $|G|$ would then also be divisible by $p^{a+1}$, contradicting the choice of $P$. Hence some $H\cap xPx^{-1}$ has order $p^b$ and is a Sylow $p$-subgroup of $H$. This is the <Sylow subgroup of a subgroup from double cosets> argument.
To count the <general linear group over a finite field> $GL_n(\mathbb F_p)$, choose its columns successively. There are $p^n-1$ choices for the first, $p^n-p$ for the second, and $p^n-p^k$ for column $k+1$. Consequently
$$
\boxed{|GL_n(\mathbb F_p)|
=\prod_{k=0}^{n-1}(p^n-p^k)
=p^{n(n-1)/2}\prod_{j=1}^n(p^j-1)}.
$$
None of the factors $p^j-1$ is divisible by $p$, so the <upper unitriangular group> is a Sylow $p$-subgroup: it has one arbitrary field entry in each of the $n(n-1)/2$ positions above the diagonal and hence order $p^{n(n-1)/2}$. The <permutation matrix>[permutation matrices] form a subgroup isomorphic to the <symmetric group> $S_n$.
By <Cayley theorem>, every finite group $G$ embeds in $S_{|G|}$, and permutation matrices embed this symmetric group in $GL_{|G|}(\mathbb F_p)$. The latter has the explicit Sylow $p$-subgroup just described, so the result proved in the first part, applied to the embedded copy of $G$, proves that every finite group has a Sylow $p$-subgroup.
The counting part of the <Sylow theorems> says that if $p^a$ is the largest power of $p$ dividing $|G|$, then the number $n_p$ of Sylow $p$-subgroups satisfies
$$
\boxed{n_p\equiv1\pmod p,\qquad n_p\mid |G|/p^a}.
$$
Finally, let $|G|=pq$ with prime numbers $p>q$. The Sylow counts give
$$
n_p\mid q,\quad n_p\equiv1\pmod p,
$$
so $n_p=1$ and the Sylow $p$-subgroup is a <normal subgroup>. Also $n_q\mid p$ and $n_q\equiv1\pmod q$. If $n_q=1$, both Sylow subgroups are normal; their elements commute, so $G$ is their <direct product of groups>[direct product] and is <abelian group>[abelian]. In the nonabelian case one must therefore have $n_q=p$, whence
$$
\boxed{p\equiv1\pmod q},
$$
or equivalently $q\mid p-1$, as recorded by <nonabelian group of order pq>.
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