Subgroup element composition Please validate the following proof.

Let $H$ be a subgroup of the Abelian group $G$ with elements $h \in H$ and $g \in G$ but $g \notin H$.
Theorem
The product $gh \notin H$.
Proof
Let $x = gh$. Assume $x \in H$, then by multiplying by $h^{-1} \in H$ we have that $xh^{-1} = gh(h^{-1}) = g(h h^{-1}) = g$. So we have composed two elements of $H$ and obtained an element $g \notin H$ and this violates the closure of the subgroup, hence $gh \notin H$. $\square$

Hypothesis: $H\leq G, g\in G-H, h\in H$.
Suppose that $gh\in H$ and we will have a contradiction/
Since $h\in H$ and $H$ is a subgroup we have $h^{-1}\in H$.
Since $h^{-1}, gh\in H \Rightarrow (gh)\cdot h^{-1}=g\in H$ contradiction.
I used that fact that "if $a,b\in H\leq G$ then $ab,ba\in H$"
