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 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.
 A: 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 $gH=Hg$.
A: (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.
A: 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
A: For every $g\in G$, the map $h\mapsto ghg^{-1}$ is a bijection from $H$ to $gHg^{-1}$. If $H$ is the only subgroup of order $o(H)$, then $\forall g\in G, gHg^{-1}=H$, i.e. $H\unlhd G$.
A: I am reading "Topics in Algebra 2nd Edition" by I. N. Herstein.
This problem is Problem 9 on p.53 in Herstein's book.
I solved this problem as follows:

If $H$ is a subgroup of $G$, then $aHa^{-1}$ is also a subgroup of $G$.
And if $H$ is finite, then $o(aHa^{-1})=o(H)$.
(Please see Problem 4 on p.47 in this book.)
So, by the assumption of Problem 9, $H=aHa^{-1}$ must hold for any $a\in G$.
Therefore $H$ must be a normal subgroup of $G$.

