Proving Abelianity in Groups with $g^{2} = 1$: A Simple Proof

[GreenGoblin]

I need to show a group with $g^{2} = 1$ for all g is Abelian. This is all the information given, I do know what Abelian is, and I know that this group is but I don't know how to 'show' it. Can someone help?

Gracias,
GreenGoblin

 

[Evgeny.Makarov]

For a start, see this page from The Math Forum.

 

[GreenGoblin]

I did the proof and it doesn't seem to be relevant that $g^{2}=1$. Just that it is the group of $g^{2}$. Is this right?

 

[hmmmmm]

For $$ g,h\in G $$ consider $$ (gh)^2=1 $$ then expand out and see what happens

 

[Evgeny.Makarov]

I did the proof and it doesn't seem to be relevant that $g^{2}=1$. Just that it is the group of $g^{2}$. Is this right?

I am not sure what "the group of $g^{2}$" means. Could you post your proof?

 

[GreenGoblin]

I just showed algebraically gh = hg, without using the fact $(gh)^{2}=1$.

 

[Deveno]

perhaps you should show your proof.

it is not the case that all groups are abelian, only some groups are.

a typical proof starts out like this:

suppose $g^2 = 1$ for all g in G. let g, h be any two elements of G (which perhaps might be the same element, if |G| = 1).

then gh is an element of G, so

$(gh)^2 = 1$, that is:

ghgh = 1. then,

g(ghgh) = g(1) = g, and...?