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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"?

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    $\begingroup$ What are $G$ and $S$? $\endgroup$ – fkraiem Dec 4 '14 at 22:14
  • $\begingroup$ I added it above! Sorry about that! $\endgroup$ – Fmonkey2001 Dec 4 '14 at 22:18
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In general your proposed theorem doesn't hold. For example: The dihedral groups are generated by a pair of cyclic (hence abelian) groups. But dihedral groups are not abelian.

Specifically, take the symmetries of an equilateral triangle. This is a group of order 6 which is generated by a 120-degree rotation and (any) reflection. The rotation itself generates a cyclic group of order 3. The reflection generates a cyclic group of order 2. But the symmetries of an equilateral triangle isn't an abelian group.

On the other hand, if you have two cyclic groups whose generators commute, then your resulting group will be abelian.

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.

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