There are two possibilities to show that a non-empty subset $H$ of a group $(G,\ast)$ is its subgroup. Both possibilities are equivalent. Which possibility you choose is more or less a matter of taste and of simplicity.

### $(1)$ Verify the subgroup properties

- Demonstrate that $e\in H,$ where $e$ is also the neutral element of $G$ with respect to the operation $”\ast”.$
- Demonstrate that if $a\in H,$ then also its inverse $a^{-1}\in H.$
- Demonstrate that $H$ is closed under the operation $”\ast”,$ i.e. $a\ast b\in H$ for all $a,b\in H.$

I.e. show that $a\ast b^{-1}\in H$ for all $a,b\in H.$

| | | | created: 2019-08-12 21:45:55 | modified: 2019-08-12 21:56:28 | by: *bookofproofs*