# Rotman introduction to theory of groups exercise

From Rotman "Introduction to the Theory of Groups", ex. 2:54:

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.

That exercise was introduced here before: link

It's easy that $HK$ is a subgroup of $G$, thus $|K| / |H\cap K|$ divides $|G|/|H|$. But I don't see what to do in next step anyway, in spite of reading link above.

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Could you clarify what is meant by "such"? Clearly there might be another (normal) subgroup $K$ with $(|K|,|G:K|) = 1$. –  Tobias Kildetoft Dec 17 '12 at 17:28
What is $(H,[G:H])$? Thi is a pair of a subgroup with an index... –  alex.jordan Dec 17 '12 at 17:30
@Tobias, that's exactly the wording in Rotman's book...a rather poor wording, in fact. The meaning is: prove $\,H\,$ is the only normal subgroup in $\,G\,$ s.t. its order is coprime to its index. –  DonAntonio Dec 17 '12 at 17:33
@DonAntonio But only the trivial group has a unique such subgroup. –  Tobias Kildetoft Dec 17 '12 at 17:35
It's probably trying to get you to prove that if a Hall $\pi$-subgroup is normal then it's the only Hall $\pi$-subgroup. –  Alexander Gruber Dec 17 '12 at 18:05

Let $G$ be a finite group and $H$ a normal subgroup of $G$ with $\rm{gcd}(|H|,|G/H|) =1$. Let $K$ be any subgroup of $G$ such that $|K|$ divides $|H|$. Then $K\subseteq H$ (as a special case of this, if $|K| = |H|$ then $K =H$ and I think this might be what is asked).
Proof: Since $H$ is normal, $KH$ is a subgroup of $G$ of order $\frac{|K||H|}{|K\cap H|}$. If $K$ is not contained in $H$ then $\frac{|K|}{|K\cap H|} > 1$. Let $p$ be a prime dividing $\frac{|K|}{|K\cap H|}$. Since $|K|$ divides $|H|$ we know that $p$ divides $|H|$, so since $\rm{gcd}(|H|,|G/H|) =1$ we know that $p$ does not divide $|G/H|$. But then $p|H|$ does not divide $|G|$ which contradicts the fact that it should divide $|KH|$.