- #1
hedlund
- 34
- 0
I want to show that if G is a group where there exists an [tex] x [/tex] which is it's own inverse then G is abelian. Ie [tex] x * x = e [/tex]. I get the hint that let [tex] x = ab [/tex]. So we have abab=e, I'm not sure how to continue from this. But I think I should try something like this
[tex] (1) \quad a*abab = a [/tex]
[tex] (2) \quad abab*a = a [/tex]
[tex] (3) \quad b*abab = b [/tex]
[tex] (4) \quad abab*b = b [/tex]
But I'm not sure how to continue ... please give me help but don't spoil it :)
[tex] (1) \quad a*abab = a [/tex]
[tex] (2) \quad abab*a = a [/tex]
[tex] (3) \quad b*abab = b [/tex]
[tex] (4) \quad abab*b = b [/tex]
But I'm not sure how to continue ... please give me help but don't spoil it :)