1. ## group theory proof

Let G be a group

if the order(g) = 2 for all g then prove G is abelian

My proof:

order(g) = 2

g^2 = e
g.g(^-1) = e

g.g = g.g(^-1)

therefore g= g^-1

g.g(^-1) = g^2 = e = g(^-1).g

therefore G is abelian

my proof is dont look right

2. Originally Posted by Dreamer78692
Let G be a group

if the order(g) = 2 for all g then prove G is abelian

My proof:

order(g) = 2

g^2 = e
g.g(^-1) = e

g.g = g.g(^-1)

therefore g= g^-1

g.g(^-1) = g^2 = e = g(^-1).g

therefore G is abelian

my proof is dont look right

Your "proof" doesn't look right because you didn't prove anything close to what you were asked: you

only "proved" that $\displaystyle g=g^{-1}\,\,and\,\,gg^{-1}=e\,,\,\,\forall g\in G$ , the former being

a rather obvious property in this kind of groups, and the latter being an axiom of group theory.

What you have to prove is: $\displaystyle \forall x,y \in G\,,\,xy=yx$ . Hint: use that $\displaystyle (xy)^2=e\,\,and\,\,also\,\,x=x^{-1}$

Tonio

3. Thanks... i get how to do it