Show that the union of a nonempty directed family of subgroup of a group G is a subgroup of G.

Proof so far:

Let G be a group, and suppose that $\displaystyle \{ S _{a} \} _{a \in A} $ be a directed family of subgroups of G.

Then $\displaystyle \bigcup \{ S _{a} \} _{a \in A} $ is the union.

Since it is nonempty, the subgroup of {e}, act as the identity of this subgroup.

Now, suppose that x and y are elements of this union, then there exist i,j in A such that $\displaystyle x \in S_{i}$ and $\displaystyle y \in S_{j} $. And since this family is directed, there exist some k in A such that $\displaystyle S_{i} \subseteq S_{k} $ and $\displaystyle S_{j} \subseteq S_{k} $. Implies that both x,y are in $\displaystyle S_{k} $ And since the union is consist of all subgroups, that means $\displaystyle S_{k} $ is a subgroup, implies that $\displaystyle xy^{-1} \in S_{k} $, thus proves the claim.

Is this right? Thank you.