# Proving $H$ is normal in $G$. [duplicate]

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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## marked as duplicate by Jack Schmidt, Asaf Karagila, Dylan Moreland, Jason DeVito, Zev ChonolesJun 9 '12 at 19:43

Is there concrete examples for such groups ? – Mohamed Jun 9 '12 at 11:28
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? – DonAntonio Jun 9 '12 at 12:08
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, $A_3$. – Nicky Hekster Jun 9 '12 at 13:03
@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. – Nicky Hekster Jun 9 '12 at 13:28

Hint. If $g\in G$, how many elements does $gHg^{-1}$ have?

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Although this may be clear, one should also show that $gHg^{-1}$ is indeed a subgroup of $G$ but I forget why the cardinality of $gHg^{-1}$ must equal the cardinality of the subgroup above. Why is it that $|gHg^{-1}| < |H|$ cannot happen? – math-visitor Jun 9 '12 at 5:55
$gHg^{-1}$ contains the identity, is closed, and is closed under taking inverses, so is a subgroup (it is an instructive exercise to check these facts). It is also an instructive exercise to check that the clear candidate for a bijection between $H$ and $gHg^{-1}$ is indeed a bijection. – Geoff Robinson Jun 9 '12 at 6:11
@math-visitor, since in the question $G$ is a finite group, it's enough to note $gxg^{-1}=gyg^{-1}$ implies $x=y$. – Gerry Myerson Jun 9 '12 at 6:24
@GerryMyerson but this is true for any group... – Belgi Jun 9 '12 at 7:32
@theron: Great and enough Hint. +1 – Babak S. Jun 9 '12 at 10:52

@theron: $H$ is even a characteristic subgroup: it is invariant under any automorphism of $G$. Can you see why? The normality comes from the so-called inner automorphisms, that is conjugation by any element $g \in G$.
Observe that the same holds true if your $H$ is the only subgroup of $index[G:H]$.

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For any $a∈G$, the set $a^{−1}Ha$ is also a subgroup of $G$ ($∵$ If $H$ is a subgroup of$G$, then $x^{−1}Hx$ is also a subgroup of $G$ for$x∈G$. Also if $$H=\{e,h_{1},...,h_k\}$$ then$$a^{−1}Ha=\{e,a^{−1}h_1a,...,a^{−1}h_ka\}$$ and these are all distinct elements of $a^{−1}Ha$, i.e. $o(H)=o(a^{−1}Ha)$. But if $H$ is the only subgroup of a given order. Hence $a^{−1}Ha=H$. In other words $H$ is a normal subgroup of $G$.

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