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 $gh \notin H$. $\square$

In fact

If $H$ is a subgroup of $G$ and $h \in H$, then for all $g \in G$ we have $gh \in H \iff g \in H$.

Indeed, $gh \in H \implies g \in Hh^{-1} \subseteq H$.
Conversely, $g \in H \implies gh \in H$. Both claims follow because $H$ is a subgroup and so is closed under products and inverses.
