If $G$ has exactly one subgroup $H$ of order $k$, prove that $H$ is normal in $G$. I am doing a presentation in my class, and I have to write the proof to this theorem. Could someone please verify whether it is correct or not? It is based on (https://math.stackexchange.com/q/156008)'s proof.

If $G$ has exactly one subgroup $H$ of order $k$, prove that $H$ is normal in $G$.

Proof: Let $H=\left \{ e,h_{1},h_{2},...,h_{k-1} \right \}\leq G$. Then $|H|=k$.
Claim: The set $gHg^{-1}=\left \{ghg^{-1}:g\in G  \right \}$ is a subgroup of $G$.
Proof: (one-step subgroup test)
Non-empty: Since $H\leq G$, $e\in H$. Then $geg^{-1}=gg^{-1}=e\in gHg^{-1}$. Therefore, $gHg^{-1}\neq \varnothing$.
Closure Inverse: Let $gh_{1}g^{-1},gh_{2}g^{-1}\in gHg^{-1}$. Then $(gh_{1}g^{-1})(gh_{2}g^{-1})^{-1}=gh_{1}g^{-1}(g^{-1})^{-1}h_{2}^{-1}g^{-1}$ (since $G$ and $H$ have inverses) $=g(h_{1}h_{2})g^{-1}\in gHg^{-1}$ (since $H$ is a subgroup hence a group) $\square$
Then $gHg^{-1}=\left \{ e,gh_{1}g^{-1},gh_{2}g^{-1},...,gh_{k-1}g^{-1} \right \}$ and so $|gHg^{-1}|=k$. Since $G$ has exactly one subgroup of order $k$, $H=gHg^{-1}$. Therefore, $H$ is normal in $G$. $\square$
 A: The proof has a couple of glitches.

(Closure) Let $gh_1g^{-1},gh_2g^{-1}\in gHg^{-1}$. Then
  $$
(gh_1g^{-1})(gh_2g^{-1})^{-1}=
gh_1g^{-1}(g^{-1})^{-1}h_2^{-1}g^{-1}=
g(h_1h_2^{-1})g^{-1}\in gHg^{-1}
$$
  because $H$ is a subgroup.

You have $h_2$ instead of $h_2^{-1}$. Besides, the parenthetical remarks are irrelevant: you're working in a group using known properties of the operations.
You haven't really proved that $|gHg^{-1}|=|H|$, though, which is however easy.

There is a much simpler proof, however.

For every $g\in G$, the map $\varphi_g\colon G\to G$, $\varphi_g(x)=gxg^{-1}$ is a group homomorphism, since
  $$
\varphi(x)\varphi(y)=(gxg^{-1})(gyg^{-1})=g(xy)g^{-1}=\varphi(xy)
$$
  Moreover $\varphi_g$ is bijective since $\varphi_{g^{-1}}$ is the inverse map of $\varphi_g$. Indeed
  $$
\varphi_{g^{-1}}(\varphi_g(x))=
\varphi_{g^{-1}}(gxg^{-1})=g^{-1}gxg^{-1}(g^{-1})^{-1}=x
$$
  and similarly $\varphi_g(\varphi_{g^{-1}}(x))=x$.

Thus $\varphi_g(H)=gHg^{-1}$ is a subgroup of $G$ (by homomorphism) having the same cardinality as $H$ (by bijectivity).
A: It's almost there.
You're missing a step here:

Then $gHg^{-1}=\left \{ e,gh_{1}g^{-1},gh_{2}g^{-1},...,gh_{k-1}g^{-1} \right \}$ and so $|gHg^{-1}|=k$.

You need to prove that the elements of $gHg^{-1}$ are distinct, but that's easy, since if $gh_ig^{-1}=gh_jg^{-1}$, then, cancelling the $g$ and the $g^{-1},$ we have $h_i=h_j$.
NB: Let $h_0=e$. Then $e=gh_0g^{-1}$.
A: You should put 'let $g\in G$' at the beginning. Also at line 2, it should be $gHg^{-1}=\left \{ghg^{-1}:h\in H  \right \}$. And at line -3, it should be $=g(h_{1}h_{2}^{-1})g^{-1}\in gHg^{-1}$.
