TITLE: If unique subgroups of order $m$ and $n$ exist, is subgroup of order $nm$ unique if it exists?
QUESTION [6 upvotes]: Suppose that $G$ is a group and has two unique subgroups of order $m$ and $n$ respectively, where $m$ and $n$ are different and greater than $1$.  If there exists a subgroup of order $mn$, then does it has to be unique?

It seems that the statement is true when $G$ is abelian: $G$ can be represented by the direct product of $\mathbb{Z}_n$ and uniqueness of group of order $mn$ can be shown if it exists.
And I thought it would be false when $G$ is not abelian so tried to find a counterexample, but couldn't have found any. 
Is there any counterexample for this statement? Or could you help me with proving this?

REPLY [6 votes]: When you say "2 unique subgroups of order $n$ and $m$ respectively", then I guess you mean a unique subgroup of order $m$ and a unique subgroup of order $n$.
If you had allowed $m=n$, then we could have taken the quaternion group $Q_8$ is a counterexample with $m=n=2$.
Otherwise, we can take $G = Q_8 \times C_3$, with $m=2$ and $n=6$. There are unique subgroups of order $2$ and $6$, but three subgroups of order $12$.