TITLE: If $H\unlhd G$ with $(|H|,[G:H])=1$ then $H$ is the unique such subgroup in $G$.
QUESTION [7 upvotes]: 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.

REPLY [2 votes]: To make things explicit: Assume $\lvert H \rvert = \lvert K \rvert$. 
As a consequence of the 2nd isomorphism theorem, $HK \le G$ and 
$$\lvert HK \rvert 
= \frac{\lvert H \rvert \lvert K \rvert}{\lvert H \cap K \rvert} 
= \frac{\lvert H \rvert^2}{\lvert H \cap K \rvert}$$
By Lagrange's Theorem:
$$\lvert G \rvert = [G:H] \lvert H \rvert 
= [G:HK] \lvert HK \rvert $$
We derive 
$$[G:H] = [G:HK] \frac{\lvert H \rvert}{\lvert H \cap K \rvert}$$
If $\lvert H \cap K \rvert < \lvert H \rvert$ then there is some factor $n \ne 1$ of $\lvert H \rvert$ on the RHS (note: $\lvert H \vert /\lvert H \cap K \rvert 
 = n \in \mathbb N$). Thus $n \mid [G:H] $, a contradiction that $\operatorname{gcd}(\lvert H \rvert, [G:H]) = 1$. 
We conclude $\lvert H \cap K \rvert = \lvert H \rvert$ and since $\lvert H \rvert = \lvert K \rvert$ we can conclude $H = K$.