Question

Let (G,*) be a group with identity element e such that a*a=e for all a in G. prove that G is abelian

Answer 1

Abstract algebra?

Answer 2

yep

Answer 3

Bleargh I don't know how to solve this. I will call upon @TuringTest and @FoolForMath for help. :P

Answer 4

Although presumably if a*a=e, a is an identity element, and a group of identity elements are abelian?

I don't know; I might be talking out of my retricehere.

Answer 5

If I may suggest a better place to ask advanced questions, http://math.stackexchange.com/ .

The average advanced user here is probably a college math student. :P Abstract algebra isn't for everyone.

Answer 6

ok
yea that web is another level.. let me see if they could help

Answer 7

Okay, this wasn't hard,
\( a*a=e \implies a=a^{-1} \) similarly \( b=b^{-1} \)

So, \[ a*b=(a*b)^{-1} =b^{-1} *a^{-1} =b*a. \]

Answer 8

[QED]

Answer 9

M.SE thread: http://math.stackexchange.com/questions/118772/