Let $G$ be a finite group of order $n$. Let $H$ be a unique subgroup of $G$ of index $k$. Show that $H$ is a normal subgroup of $G$.
My attempt: Since, $H$ is a subgroup of $G$; and assume that $|H|=m$,
Now define a mapping $\phi: H \rightarrow gHg^{-1}$ by $\phi(h)=ghg^{-1}$, for some $h \in H$, and $g\in G$.
Well defined:
Let $h_1,h_2 \in H$, and $h_1=h_2$ then, $gh_1g^{-1}=gh_2g^{-1}$. This implies $\phi(h_1)=\phi(h_2)$.
Homomorphism:
Let $h_1,h_2 \in H$, then $$\phi(h_1h_2)=g(h_1h_2)g^{-1}=gh_1g^{-1}gh_2g^{-1}=\phi(h_1)\phi(h_2)$$
One-to-One:
Let $\phi(h_1)=\phi(h_2)$ then $$gh_1g^{-1}=gh_2g^{-1}\implies h_1=h_2$$
Onto:
Let $ghg^{-1} \in gHg^{-1}$, then there is $h \in H$, such that $\phi(h)=ghg^{-1}$.
This implies $H$ and $gHg^{-1}$ are isomorphic to each other. So $|gHg^{-1}|=m$; By the uniqueness of the subgroup, so $H=gHg^{-1}$. This implies $Hg=gH$. Hence, H is a normal subgroup of G.
Is this proof correct for this question? I am confused how do I use index $k$ property.
Anyone can suggest me some improvements or directions?
For any finite group $G$, element $g \in G$, and subgroup
$H \subseteq G, \tag 1$
the mapping
$H \to G, \; H \ni h \mapsto ghg^{-1} \in gHg^{-1} \tag 2$
is an isomorphism 'twixt $H$ and $gHg^{-1}$, for this map is clearly injective, since if
$gh_1g^{-1} = gh_2g^{-1}, \tag 3$
then
$h_1g^{-1} = eh_1g^{-1} = g^{-1}gh_1g^{-1} = g^{-1}gh_2g^{-1} = eh_2g^{-1} = h_2g^{-1}, \tag 4$
whence
$h_1 = h_1 e = h_1g^{-1}g = h_2g^{-1}g = h_2e = h_2; \tag 5$
here $e \in G$ is the identity element; furthermore, the map (2) is manifestly surjective by definition; therefore
$\vert H \vert = \vert gHg^{-1} \vert, \tag 6$
that is, the two subgroups $H$ and $gHg^{-1}$ are of the same cardinaliity. By Lagrange's theorem,
$[G:H] = \vert G \vert / \vert H \vert = \vert G \vert / \vert gHg^{-1} \vert = [G:gHg^{-1}], \tag 7$
that is, $H$ and $gHg^{-1}$ are of the same index. Thus
$H = gHg^{-1}; \tag 8$
since this holds for any $g \in G$,
$H \lhd G, \tag 9$
$OE\Delta$.