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

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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If $g\in G$, what is the order of the subgroup $gHg^{-1}$? What can you conclude from that? – Mariano Suárez-Alvarez Jan 6 '12 at 22:38
i dont know :( ..what i can't understand is how do u link order with the group being normal/abnormal. – Vishnu Jan 6 '12 at 22:45
Do not think what is the relation between normality and the order for a while: try to see what is the order of the subgroup $gHg^{-1}$? – Mariano Suárez-Alvarez Jan 6 '12 at 22:47
Notice that you are studying the book on your own does not change the fact that this is homework (you can assign homework to yourself!)... The point of knowing if a question is homework or not is to know if the best option to answer the question is to actually answer it or to give hints for you to answer it yourself. – Mariano Suárez-Alvarez Jan 6 '12 at 22:50
Well said!..The reason i was specific that i was self-studying and this was not a 'homework' problem, was that i am BRAND NEW to this site (I joined 15 minutes back), and i read in some passing comment something like " is this a homework prob"..so i thought , maybe your aren't allowed to pose 'hw probs' here. That's all.Not that i m trying to evade making efforts on my own. ThankYou for your help! – Vishnu Jan 6 '12 at 22:56

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 $gH=Hg$.
If $(ghg^{-1})^n=e$ it dos't says that ord$(ghg^{-1})=n$ it's says that $n\big |_{ord(gng^{-1})}$ – RedMushroom Jul 9 '15 at 16:10
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.
@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 ;-) – Emilio Ferrucci Jan 6 '12 at 23:12