We call a subgroup $H$ of a group $G$ normal if and only if $gH = Hg$ for all $g \in G$.
The normal subgroup test is the following theorem:
A subgroup $H$ of a group $G$ is normal in $G$ if and only if $gHg^{-1}\subseteq H$ for all $g \in G$.
The proof for $\implies$ given in the book is the following:
If $H$ is normal in $G$ then for any $g \in G$ and $h \in H$ there is $h'$ in $H$ such that $gh = h' g$. Thus $ghg^{-1} = h'$ and therefore $gHg^{-1}\subseteq H$.
I tried to prove this and I wrote this proof:
If $H$ is normal then $gH = Hg$. Multiplying by $g^{-1}$ on both sides yields $gHg^{-1} = H$ hence in particular, $gHg^{-1} \subseteq H$.
Now I'm wondering whether this is a false proof because it is shorter than what they do in the book and therefore would be my first choice.
Is my proof false?