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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 $(|K|,|G:K|) = 1$.2012-12-17
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    What is $(H,[G:H])$? Thi is a pair of a subgroup with an index...2012-12-17
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    @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.2012-12-17
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    @DonAntonio But only the trivial group has a unique such subgroup.2012-12-17
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    @Tobias, the group $\,S_3\,$ has *a unique (non-trivial) normal subgroup* which is both normal and of order coprime to its index, namely $\,A_3\,$: $$(3=|A_3|\,,\,2=|S_3/A_3|)=1$$2012-12-17
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    @DonAntonio: No, it also has the entire group and the trivial subgroup.2012-12-17
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    Ahh, I missed the non-trivial part. But still, most abelian grouups will be counter examples.2012-12-17
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    Yes, something's fishy in this. I must be misunderstanding Rotman's intention. I'll read from the book again.2012-12-17
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    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.2012-12-17

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As stated, the question is a bit vague, and I am having trouble interpreting it in a way that makes it correct.

Here is something that does hold and which might be what is being asked:

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 $|KH|$.