Is the hypothesis of $G$ being a finite group necessary in this exercise?

Question

I've been trying to solve an exercise for my Algebra class and even found this question asked already:

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

Although, is the finiteness hypothesis a must for this exercise?

Answer 1

Suppose $H$ is the only subgroup of $G$ of order $|H|$. Fix an element $g \in G$ and let $K = gHg^{-1}$. Take $a,b \in K$. There exist $a' , b' \in H$ such that $a = ga'g^{-1}$ and $b = gb'g^{-1}$. Thus $$ ab = ga'g^{-1} gb' g^{-1} = ga'b'g^{-1} $$ which is an element of $K$ since $a'b' \in H$. Moreover, $e \in H$ since $H$ is a subgroup, so $$ e = geg^{-1} \in K $$ Thus, we have shown $K \leq G$. Now, consider the maps \begin{align} f &: H \to K \\ &h \mapsto ghg^{-1} \\ g &: K \to H\\ &k \mapsto g^{-1}k g \end{align} I leave it to you to check that $g = f^{-1}$, which shows that $|H| = |K|$. By assumption, this means $H = K = gHg^{-1}$. Since $g \in G$ was arbitrary, this proves that $H$ is a normal subgroup of $G$.

Notice that nowhere did we need to assume that $G$ was finite.