# Finding/Creating a Modern Algebra theorem

The question I'm trying to prove is this one:

The subgroup $<G,S>$ generated by $G$ and $S$ is abelian and of order $9$.

My Work:

$G=(123)(456)(789)\ \text{and} \ S=(147)(258)(369)$

I've shown that $G^3=S^3=\varepsilon\ \ (\text{Identity})$

I've also shown that GS=SG

I've also shown, if you make the table, that there are only $9$ distinct elements i.e. $$\{\varepsilon,G,G^2,S,S^2,GS,GS^2,G^2S,G^2S^2\}$$

My Question:

Is there a a theorem that states "If you have a group generated by two abelian cyclic subgroups then that group is also abelian"?

• What are $G$ and $S$? – fkraiem Dec 4 '14 at 22:14
• I added it above! Sorry about that! – Fmonkey2001 Dec 4 '14 at 22:18

That's why everything worked out in your example. Since $GS=SG$, powers of $G$ commute with each other, and powers of $S$ commute with each other, then everything commutes with everything.
There is something else coming into play here as well. Every group of order $p^2$ (where $p$ is prime) is abelian. In particular, your group of order $3^2=9$ is abelian.