Sylow $p$-subgroup and Sylow $q$-subgroup both normal

Question

Let $G$ be a finite group and $p\neq q$ prime numbers that divide $\|G\|$. Let $P$ and $Q$ be a Sylow $p$‐subgroup and a Sylow $q$‐subgroup of $G$, respectively. If $n_p=1$ and $n_q=1$, show that

$\forall p\in P, \forall q\in Q,pq=qp$.

By 2nd Sylow theorem, $n_q=n_p=1$ means that there is a single Sylow $p/q$-subgroup, meaning that $P$ and $Q$ are normal in $G$.

I tried to use the normality:

$q_1p_1=p_2q_1=q_2p_2$,

but I got nowhere since I don't see how to show either $p_2=p_1$ or $q_2=q_1$.

I also see that $G/P$ and $G/Q$ are groups, but coset operations didn't lead me anywhere different.

Any help is appreciated!

Answer 1

Continuing from "By 2nd Sylow theorem, $n_q=n_p=1$ means that there is a single Sylow $p/q$-subgroup, meaning that $P$ and $Q$ are normal in $G$."

Then by normality, for $p\in P, pQp^{-1}=Q$ meaning that for $q\in Q, pqp^{-1}\in Q$. Similarly, $qpq^{-1}\in P$, so $pqp^{-1}q^{-1}$ is an element of both $Q$ and $P$. By definition of $p/q$-group, the only element that can be in both is $e$, which has order $p^0=1=q^0$. So $pqp^{-1}q^{-1}=e$, which is equivalent to $pq=qp$. As $p,q$ were taken arbitrarily, the exercise is solved.

Answer 2

Hint: in general (chipping away the superfluous information of Sylow subgroups) if $M,N \unlhd G$ and $M \cap N=1$, then $[M,N]=1$, that is $M$ and $N$ commute element wise with each other. In the comments you find the hint for the proof (look at $m^{-1}n^{-1}mn$, $m \in M, n\in N$).