Subgroup Tests

The following theorems justify the ability to check that a subset of another group is a subgroup without checking all axioms.

Arbitrary Group

Proof

Given that $H\ne \varnothing$, let $h\in H$ be arbitrary. Then using property 2, we have $h\ast h^{-1}=\mathrm{id}\in H$. As such, we can, for arbitrary $h\in H$, deduce that $\mathrm{id}\ast {\mathrm{h}}^{-1}=h^{-1}\in H$. Now, with closure under inverses, we can deduce that for any $h,h^{\prime }\in H$, $h^{\prime -1}\in H$ and hence by property 2, $h\ast (h^{\prime -1}{)}^{-1}=h\ast h^{\prime }\in H$, thus we also have closure under inverses.

This means that $\ast$ is a well defined operation when restricted to $H\ast H\to H$ and associativity follows from the restricted operation. That is, it is inherited from the main group $G$.

Finite Group

For subgroups of a finite group, the above theorem can be simplified since the assumption of finiteness implies existence of all inverses.

Proof

Closure under inverses is held because each element to the group order is identity, and hence for any $a\in H$, we have that $\mathrm{ord}(a)$ is finite since $|H|$ is (since $|G|$ is). Then,