If x^2 = e for all x in G then G is an abelian group proof

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salmanqureshi

you're saving my ass during this pandemic, I love you sir

saltycoins

xy = x^-1y^-1 since every element is it's own inverse. By the proof from the last video, x^-1y^-1 = (yx)^-1. By closure, (yx)^-1 is in G as a single element, which is also it's own inverse, yx, which is equal to x^-1y^-1 which is equal to xy and so, yx = xy meaning that this group is Abelian.

paulcohen

i feel i'm complicating it too much by thinking hard.!!!! keeping it simple from now on.

punyotakung

only 3 steps.
every element is its own inverse = socks-shoes property = again every elements has its own inverse.
hence abelian

arslannazir

Is that all remain same i put "a" insead of "x" ..and if every element "a" in group g????

anjalijaiswal

Why did you multiply x and y even you don't know about operation btw x and y?

amanfeildconvecture

can we extend this to x^3=e implies G is abelian? why ?

vanshikadatta

If x^2 = e for all x in G then G is an abelian group proof

If x^2 = e for all x in G then G is an abelian group proof

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