Thread: Cayley table and subgroup

If I need to prove that something is a subgroup and I say that are elements in the subgroup, will constructing a Cayley table with show that the subgroup is closed?

Originally Posted by sfspitfire23

If I need to prove that something is a subgroup and I say that are elements in the subgroup, will constructing a Cayley table with show that the subgroup is closed?

Yes. If H is a subgroup of a group (G, *), then H should be closed with respect to '*'. It should also satisfy the group axioms with respect to '*' of course.

so it will also show associativity right?

Originally Posted by sfspitfire23

so it will also show associativity right?

To show that H is a subgroup of (G, *), it suffice to show that for all a and b in H (as a subset of G), . This basically checks the closure of an operation '*' and its existence of an inverse in H. If H satifies this subgroup condition, it means H also satisfies the remaining group axioms (identity, associativity). You can think that "associativity" w.r.t '*' is simply inherited from G.