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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've been trying this problem for quite a while but to no avail. What I can't understand is, how do you relate the subgroup being normal/abnormal to its order?

This question is from I.N.Herstein's book Topics in Algebra Page 53, Problem no. 9. This is NOT a homework problem!! I'm studying this book on my own.

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    A stronger result is th$a$t if a group has a unique su$b$group of a given order, then that subgroup is characteristic.2014-03-04

3 Answers 3

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This is a small contribution, but in response to a comment above, you can also solve this without using any homomorphism properties. Assume the subgroup $H$ has order $n$ and pick $g\in G$. Then, for any $h\in H$: \begin{align} \left(ghg^{-1}\right)^n=\underbrace{ghg^{-1}\cdot ghg^{-1}\cdot\dotsm\cdot ghg^{-1}}_{\text{n times}}=gh^ng^{-1}=geg^{-1}=gg^{-1}=e \end{align} Since $H$ has order n. The above holds for all $h\in H$, so the subgroup $gHg^{-1}$ has order n, so it is equal to $H$ by assumption. Then every $h\in H$ has some $h'\in H$ such that $ghg^{-1}=h'$, implying $gh=h'g$, so $gH=Hg$.

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    @3SAT Hello, then how do you show that $|gNg^{-1}|=n$ hence $H=gNg^{-1}$?2018-03-08
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(sorry didn't see the comments - this is a spoiler :-) )

You need to prove that $\sigma_x(h) \in H \ \forall x \in G \ $ and $\ \forall h \in H$, where $\sigma_x(h)=x^{-1}hx$. You know that $\forall x \in G$, $\sigma_x$ is an automorphism of $G$, which implies that if $K$ is a subgroup of $G$ then also $\sigma_x(K)$ is a subgroup of $G$, of the same order of $K$ (because $\sigma_x$ is bijective). Therefore, since $H$ is the only subgroup of $G$ of its order, $\sigma_x(H)=H \ \ \forall x \in G$ and you are done.

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    @Vishnu in that case you can think of it this way: prove that $\forall x \in G$ the function $\sigma_x$ defined above is bijective, and has the property that it maps subgroups into subgroups. Also: in case you were used to the definition $H \unlhd G$ iff $xH = Hx \forall x \in G$, you can also prove it is equivalent to $\sigma_x(h) \in H$ $\forall x \in G $. Hope this helps ;-)2012-01-06
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Consider the set $xHx^{-1}$. Prove that $xHx^{-1}$ is a subgroup of $G$. (show $ab^{-1} \in xHx^{-1} ;\forall a,b \in xHx^{-1} $ )

Now If we prove that there exist one-one and onto relation between $H$ and $xHx^{-1}$ i.e, $ \phi : H \rightarrow xHx^{-1} $ we are done because H is unique subgroup of G and establishing a one one relation will mean both have same order and thus thus meaning both are the same.

i.e, $ |xHx^{-1}| = |H| $

thus $ xHx^{-1} = H $

Q.E.D