Abelian

[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 :)

[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).

[hedlund]

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).

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?

[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).

[jcsd]

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).

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

[Muzza]

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


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.

[jcsd]

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.

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