# Cayley table and subgroup

• Oct 22nd 2009, 05:34 PM
sfspitfire23
Cayley table and subgroup
If I need to prove that something is a subgroup and I say that $a,b$ are elements in the subgroup, will constructing a Cayley table with $a,b,1$ show that the subgroup is closed?
• Oct 22nd 2009, 05:58 PM
aliceinwonderland
Quote:

Originally Posted by sfspitfire23
If I need to prove that something is a subgroup and I say that $a,b$ are elements in the subgroup, will constructing a Cayley table with $a,b,1$ 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.
• Oct 22nd 2009, 06:34 PM
sfspitfire23
so it will also show associativity right?
• Oct 22nd 2009, 08:52 PM
aliceinwonderland
Quote:

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), $a * b^{-1} \in H$. 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.