# Could this be the subgroup test described by one axiom?

Proposition: Let $$H$$ be a non-empty subset of a group $$G$$ w.r.t. some binary operation, $$*$$. We have:

$$H$$ is a subgroup of $$G$$ $$\iff$$ $$(\forall h_1, h_2 \in H) (h_1 * h_2 ^{-1} \in H)$$.

I start the first part of the proof by the supposition that $$H$$ is a subgroup of $$G$$ w.r.t. the binary operation $$*$$. We already know that if $$H$$ is a subgroup of $$G$$, then for all $$h_1, h_2 \in H$$, $$h_1 * h_2 \in H$$, and $$h_1 \in H \implies h_1 ^{-1} \in H$$ and $$h_2 \in H \implies h_2 ^{-1} \in H$$ so that $$h_1 * h_1 ^{-1} = e \in H$$ and $$h_2 * h_2 ^{-1} = e \in H$$, where $$e$$ is the unique identity element in $$H$$.

How could one relate this information to the fact that $$(\forall h_1, h_2 \in H) (h_1 * h_2 ^{-1} \in H)$$?

Kind regards.

• This is worth a look. Commented Nov 17, 2020 at 15:57
• @Shaun Thank you for the information, Shaun. Commented Nov 17, 2020 at 16:04

## 1 Answer

Hint: By definition, a nonempty subset $$H$$ of $$G$$ forms a subgroup of $$G$$ iff (1) for all $$h,h'\in H$$, we have $$h\cdot h'\in H$$, and (2) for each $$h\in H$$, we have $$h^{-1}\in H$$.

These two conditions are equivalent to the subgroup criterion: for all $$h,h'\in H$$, we have $$h\cdot h'^{-1}\in H$$.

This is easy to validate.

• Perhaps your hint also is rather a criterion than a definition, since (1) and (2) imply $e\in H$ and associativity holds throughout any subset of a group.
– user810157
Commented Nov 17, 2020 at 16:24