# Groups: b^5a = ab^5

1. Feb 9, 2006

### sat

Would it be possible to infer that $b^5 = e$ (where $e$ is the group's identity element) from

$$b^{5} a = ab^{5}$$
given that $a^{2}=e$?

(Basically we are given $b^{2}a=ab^{3}$ and $a^{2}=e$ and asked to show that $b^{5}=e$, though I've managed to infer the "equation" above and I can't quite see how we'd move to inferring what is needed. Maybe it's either very simple and I'm missing it or there's a bit of reasoning that I need.)

Thanks.

2. Feb 9, 2006

### matt grime

No, you can't conclude that. All you can conclude is that two elements commute, and there are elements not equal to the identity that commute with other elements in some groups.

Try showing that b^2=b^3, that's the only thing that will work and show b=e

3. Feb 9, 2006

### sat

Thanks

Thanks for that. I do agree with what you've said though I thought perhaps there might be some way of manipulating it so that in this special case you could show it.

$b^{2}=b^{3}$ certainly sounds like a way forward.

4. Feb 9, 2006

### matt grime

well that is vacuously true since it can be shown 'from this special case' in the first place.

5. Feb 9, 2006

### StatusX

As matt grime has said, it doesn't follow from ab5=b5a and a2=e that b5=e. For example, consider D10, the group of symmetries of a decagon. If a is a reflection and b is a rotation by 36o, then the above two equalities hold. But b5 is not the identity, but a rotation through 180o. In other words, you can't start from these equations and prove your result; you'll have to go back to the original equations.

6. Feb 10, 2006

### matt grime

It's not going out on a limb to suggest that you got the deduction you made from playing around with (ab^3)^2 and (b^2a)^2 and so on. If you play around with those alone you can prove the result. I got to the point of finding that abab^6=b^5 from these manipulations which will give you what you want.

Note I got it wrong when i said show b^3=b^2 cos I misread the question, I thought you wanted to show b=e, rather than b^5=e.

a and b generate the dihedral group of the pentagon.