Here is a problem from "An introduction to the Theory of Groups" by J.J.Rotman:
Let $G$ be a finite group, and let $H$ be a normal subgroup with $(|H|,[G:H])=1$. Prove that $H$ is the unique such subgroup in $G$.
I assumed there was another normal subgroup like $H$, say $K$, such that $(|K|,[G:K])=1$. My aim was to show that $[K: K\cap H]=1 $ that was not held if I didn’t suppose $|H|=|K|$ . My question is if my last assumption about two subgroups is right? If it isn’t, please guide me. Thanks.
Best Answer
Alternatively, you might note that if $K$ is any other subgroup of order $|H|$, whether or not $K$ is normal (but assuming $H$ is normal), then $HK$ is a subgroup of $G$, so its order must divide $|G|$.