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Suppose $H$ is the only subgroup of order $o(H)$ in the finite group $G$. Prove that $H$ is a normal subgroup of $G$.

I have to solve the following problem. It's an exercise from Herstein's Topics in Algebra book.

Suppose $G$ is a finite group and let $H$ be a subgroup of $G$. Suppose that $H$ is the only subgroup of $G$ of order $o(H)$. Then prove that $H$ is normal in $G$.

Any hints?

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    Is there concrete examples for such groups ?2012-06-09
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    Why do you make a question if it is written "your answer"? And what do you mean "concrete examples for such groups"? *What* groups do you mean?2012-06-09
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    In the abelian case, take the cyclic group of order 4, which has only one subgroup of order 2. The symmetric group on three letters $S_3$ is the smallest non-abelian example. $H$ would be the alternating subgroup here, $A_3$.2012-06-09
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    @Mohamed: hmm, this raises possibly the question: what if $G$ is a finite group with the property that *all of its proper* subgroups $H$ have a unique order? Well, then $G$ must be *cyclic*. Proof (sketch): every p-Sylow subgroup of $G$ is normal and hence $G$ is nilpotent. So we can assume that $G$ is a p-group for some prime p. Note that $Z(G)$ is non-trivial and hence by induction $G/Z(G)$ is cyclic, whence $G$ is abelian. If $G$ would not be cyclic then we can spit of a non-trivial direct factor of $G$, giving rise to two different subgroups of order p. Contradiction.2012-06-09

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