1. abelian groups

Is there a group that satisfies a^3=1 for each a in G but isn't abelian?

2. Originally Posted by MathBird
Is there a group that satisfies a^3=1 for each a in G but isn't abelian?
I can't find any reason why not - but if it is finite then it is at least 3-generated (because it is regular so it can't be 2-generated).

What about $\displaystyle <x,y,z|x^3=y^3=z^3=1, xyx^{-1}=y^3, xzx^{-1}=z^3, yzy^{-1}=z>$, the semidirect product of $\displaystyle C_3 \times C_3$ with $\displaystyle C_3$? I'm not sure if that is actually properly defined - I'm doing this in a bit of a hurry...

3. Thanks a lot. But may I ask for a little more details.

4. Originally Posted by MathBird
Thanks a lot. But may I ask for a little more details.
About the 3-generated bit? I was making it too complicated, and I think my logic was wrong. However, the group must be at least 3-generated if it is finite as otherwise it will have order 3 or 9, and there are no non-abelian groups of order 3 or 9. Also,
$\displaystyle <x,y,z; x^3 = y^3 = z^3 = 1, yz = zyx, xy = yx, xz = zx>$

is a group that works. I believe the group I gave earlier is nonsense...

EDIT: Of course it's nonsense - $\displaystyle y$ and $\displaystyle z$ are just the identity element...