If I need to prove that something is a subgroup and I say that $\displaystyle a,b$ are elements in the subgroup, will constructing a Cayley table with $\displaystyle a,b,1$ show that the subgroup is closed?

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- Oct 22nd 2009, 05:34 PMsfspitfire23Cayley table and subgroup
If I need to prove that something is a subgroup and I say that $\displaystyle a,b$ are elements in the subgroup, will constructing a Cayley table with $\displaystyle a,b,1$ show that the subgroup is closed?

- Oct 22nd 2009, 05:58 PMaliceinwonderland
- Oct 22nd 2009, 06:34 PMsfspitfire23
so it will also show associativity right?

- Oct 22nd 2009, 08:52 PMaliceinwonderland
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), $\displaystyle a * b^{-1} \in H$. 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.