Subgroup Test (two-step)

Subgroup Test (two-step): Let $G$ be a group and let $H$ be a nonempty subset of $G$. Then $H$ is a subgroup of $G$ if and only if:

1. for all $x,y\in H$ one has $xy\in H$ (closure under the group operation), and
2. for all $x\in H$ one has $x^{-1}\in H$ (closure under inverses).

This criterion is equivalent to the one-step subgroup test ; use whichever closure condition is easier to verify in a given situation.

Proof sketch: A subgroup satisfies both closure properties by definition. Conversely, if $H$ is nonempty and closed under products and inverses, then $H$ contains the identity (take $xx^{-1}$ for any $x\in H$) and satisfies the subgroup axioms inside $G$.