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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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