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.

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

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 $gH=Hg$.

(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.