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
Given a group \((G, \ast)\) and subset of \(H \subseteq G\), \(H\) is a subgroup of \(G\) if and only if for all \(a, b \in H\),
- \(H \neq \varnothing\)
- \(a \ast b^{-1} \in H\)
Proof
Given that \(H \neq \varnothing\), 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 \(h, h' \in H\), \(h'^{-1} \in H\) and hence by property 2, \(h \ast (h'^{-1})^{-1} = h \ast h' \in H\), 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.
Given a finite group \((G, \ast)\) and subset of \(H \subseteq G\), \(H\) is a subgroup of \(G\) if and only if for all \(a, b \in H\),
- \(H \neq \varnothing\)
- \(a \ast b \in H\)
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,