# Abelian math help

1. Sep 29, 2004

### hedlund

I want to show that if G is a group where there exists an $$x$$ which is it's own inverse then G is abelian. Ie $$x * x = e$$. I get the hint that let $$x = ab$$. So we have abab=e, I'm not sure how to continue from this. But I think I should try something like this

$$(1) \quad a*abab = a$$
$$(2) \quad abab*a = a$$
$$(3) \quad b*abab = b$$
$$(4) \quad abab*b = b$$

But I'm not sure how to continue ... please give me help but don't spoil it :)

2. Sep 29, 2004

### jcsd

I'm slightly puzzled as all groups contain an idenityt so, every group has an elemnt such that x*x = e, namely e.

I can also think of Abelian and non-Abelian groups that contain elements such that x*x = e were x is not equal to e (for example muplication in the rationals and matrix mutplication).

3. Sep 29, 2004

### hedlund

Hmm I might have understood it wrong. But it say

Prove that in a group G where $$x \ast x = e$$ for all $$x \in G$$ is abelian. Hint: Look at $$(a \ast b) \ast (a \ast b)$$. Maybe I misunderstood the question ...

Edit: I think I can prove it now, I MUST have misunderstood. We have
$$x \ast x = e$$ let the x:s be $$x_1$$ and $$x_2$$ with $$x_1 = x_2$$ (Just to make it easier to see the difference).

$$x_1 \ast x_2 = e$$
$$x_1 \ast x_2 \ast x_2^{-1} = x_2^{-1}$$
$$x_1 \ast e = x_2^{-1}$$
$$x_2 \ast x_1 = x_2 \ast x_2^{-1}$$
$$x_2 \ast x_1 = e$$
And since $$x_1 \ast x_2 = x_2 \ast x_1$$ the group G must be abelian?

Last edited: Sep 29, 2004
4. Sep 29, 2004

### Muzza

Yes, there is a difference between "there exists an x such that x*x = e" (which you said in your first post) and "for all x, x*x = e".

Another hint is to consider (ab)^-1 (there is a formula for that expression in terms of the inverses of a and b).

5. Sep 29, 2004

### jcsd

I'm still a little confused tho', does that also implicit that x is not equal to e?

6. Sep 29, 2004

### Muzza

I'm not sure what you mean. Are you asking if x^2 = e for all x in G => x != e...? x could of course be e.

7. Sep 29, 2004

### jcsd

Right now, sorry I relaize now, the orginal quetsion wasn't the the question that he wanted answered.