I was trying to solve a problem on Sylow-p subgroups that read
If $G$ is such that $|G| = p^nq$ with $p>q$ primes, show that $G$ has only one normal subgroup of index $q$.
Answering this requires only one of the Sylow theorems and is quite easy to prove. I was distracted, however, and was trying to prove
If $G$ is such that $|G| = p^nq$ with $p>q$ primes, show that $G$ has only one normal subgroup of order $q$.
I tried several things, mainly pointing towards trying to prove that said normal subgroup was the Sylow-q subgroup of $G$. I tried assuming $H, K$ were normal subgroups of $G$ with order $q$ and tried to show $H = K$; I tried induction on the exponent $n$ in $|G| = p^nq$; I tried taking the quotient $G/P$ where $P$ is the only Sylow-p subgroup...
I didn't make any of that work, but I also failed to provide a counter-example, hence I am looking for hints in the right direction.
As the comments point out, a counter-example arises easily by considering $S_3$.
Indeed, $|S_3| = 3\cdot2$ so we make $p=3,q=2$. Now we show that there are $3$ Sylow-2 subgroups of order $2$, namely
$$\langle(1\ 2)\rangle, \langle(1\ 3)\rangle, \langle(2\ 3)\rangle$$
which are as many as there could be, for if $n$ is the number of Sylow-2 subgroups, we have $n|3 \wedge n \equiv 1\ \mod2$ implying we have the case $n = 3$.