When x^2=indetity in a group,then show group is commutative?

Answer 1

Let e denote the identity of our given group G. Observe that for x, y ∈ G

e = xx = xex = x(yy)x = (xy)(yx),

and so yx is the inverse of xy. But inverses are unique, and we know that xy is its own inverse (since (xy)(xy) = e). Hence,

xy = yx,

Since x and y are arbitrary, the group is commutative.

Answer 2

Let G be a group such that for all x in G we have x^2 = 1, where 1 denotes the identity of G. Let * denote the operation on G.
Hence, x^(-1) = x. That is, x is it's own inverse, for all x in G. We use this property.
Let a,b be in G.
We know that (a*b)^(-1) = b^(-1) * a^(-1), right? You can show this by:
(a*b) * ( b^(-1) * a^(-1) ) = a * ( b * b^(-1) ) * a^(-1)
= a * 1 * a^(-1) = a * a^(-1) = 1.

We can now write (a*b) = (a*b)^(-1) = b^(-1) * a^(-1) = b * a.
That is, (ab) = (ba).
QED (Quite Entertainingly Deceptive)