If $G$ is finite and $H$ is the only subgroup of a given order, then $H$ is normal.
I have a proof idea that I'm not sure works or not:
By Lagrange's theorem, the order of $G$ is equal to the order of $H$ times the number of distinct left cosets of $H$. But every left coset of $H$ has the same number of elements as $H$, so the order of $H$ must be equivalent to the order of $G$. Hence, $H = G$, so $H$ is normal.
Does this work? I'm aware of the proof using the conjugation map, but was curious to know if this works or not. Thank you!
Best Answer
This does not work, since cosets are not subgroups (except for the trivial cosets), and your conclusion is not true. For instance, if $G$ is a cyclic group of order $2n$, then it has a unique subgroup of order $n$ (which is of course normal, since the group is abelian, but it is also definitely not all of $G$ if $n>0$).
Instead, recall that $H\leq G$ is normal iff $H^g=H$ for all $g$.