Proving that $gHg^{-1}$ is a subgroup of $G$ Let $G$ be a group and $H$ subgroup of $G$ with $\operatorname{ord}(H)=k$

I need to prove that $gHg^{-1}$ is a subgroup of $G$ 

$\color{grey}{(gHg^{-1}=\{ghg^{-1}\mid g\in G, h\in H\})}$
My attempt:
Let $a,b \in H$
$1)  $  To check that $(gag^{-1})^{-1}\in G$
$2)$  $(gag^{-1})\cdot (gbg^{-1}) \in G$

$1) $$  (gag^{-1})^{-1}=g^{-1^{-1}}a^{-1}g^{-1}=ga^{-1}g^{-1}$
$2)$ $ ga(g^{-1}g)bg^{-1}=g(ab)g^{-1}$

I'm stuck at this point, Is it correct so far? is there other methods to solve this? any hints please? 

 A: You're way sounds good. I'm just gonna do it the standard way
$gHg^{-1}$ is a subgroup of $G \Longleftrightarrow $
$$$$
$gHg^{-1}$ Nonempty ie
$$gHg^{-1}\neq \emptyset
$$
and (the closure inverse condition)
$$\forall x,y \in gHg^{-1} , \quad xy^{-1} \in gHg^{-1}
$$
To show $gHg^{-1}$ nonempty, we consider the fact that $H \leq G \Longrightarrow e \in H$
but $$e = geg^{-1} $$ so this means
$$e=geg^{-1} \in gHg^{-1}$$
So the identity is in $gHg^{-1}$ meaning $$gHg^{-1}\neq \emptyset
$$ as needed.
Now the second part. Let
$$x=gag^{-1}, y=gbg^{-1}: x,y \in gHg^{-1}
$$
Then consider 
$$xy^{-1}=gag^{-1} \circ \left( gbg^{-1} \right)^{-1}
$$
where $a,b \in H$. Since $H$ has inverses, we find,
$$\Longrightarrow xy^{-1}=gag^{-1} \circ \left( g^{-1} \right)^{-1} bg^{-1}
$$
$$\Longrightarrow xy^{-1}=ga\circ e \circ bg^{-1}
$$
$$ \Longrightarrow xy^{-1}=g(ab)g^{-1}
$$
Now by closure and inverses of $H$, we have that $ab \in H$
Thus overall, we have shown $xy^{-1} \in gHg^{-1}$.
A: If you can use homomorphisms, then this is clear because $gHg^{-1}$ is the image of $H$ under the homomorphism $G \to G$ given by $x \mapsto gxg^{-1}$.
A: The proof that I am familiar with : 
We first note that as $ H \leq G$, we have that $e \in H$. Then resultantly $geg^{-1} \in H$ which implies that H is non-empty. 
Then we let $gh_{a}g^{-1},gh_{b}g^{-1},gh_{c}g^{-1} \in gHg^{-1}$.
Now 
$$ 
gh_{a}g^{-1}gh_{b}g^{-1} = gh_{a}h_{b}g^{-1}
$$
which is in $gHg^{-1}$ as $H$ is closed. Now using the fact that $H$ is a subgroup of G and contains inverses, 
\begin{align*} 
(gh_{c}g^{-1})^{-1} &= (g^{-1})^{-1}h_{c}^{-1}g^{-1} \\
&= gh_{c}^{-1}g^{-1} \in gHg^{-1}
\end{align*}
We have thus shown that $gHg^{-1}$ is a subgroup of $G$. 
