Suppose that $G$ is a group with the property that $g^{2}= 1$ for all $g \in G$.Prove that $G$ is a commutactive group.
Abelian group $ab = ba$. I think like this $g*g^{-1} = 1$ after that i get stuck any hints?
How to prove that a group is abelian if $a^2=1$ for all $a \in A$.
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abelian-groups
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1also http://math.stackexchange.com/questions/238171/prove-that-if-g2-e-for-all-g-in-g-then-g-is-abelian – 2017-01-15
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0@ArnaudD. : the question I linked is also a possible duplicate (of another duplicate!)... you might have a look if you wish ;-) – 2017-01-15
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0Thanks Arnaud D for linking a page. – 2017-01-15
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1@Watson I have seen the question you linked (I followed your vote) but later realised the other one was even older, and had more views, so I wanted to mention it (even though I couldn't change my vote). – 2017-01-15
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We have that $g^2 = 1$ implies $g = g^{-1}$. Thus for any $x,y \in G$ we have $$xy = (xy)^{-1} = y^{-1}x^{-1} = yx$$
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0One question can i say that x and y is g?? – 2017-01-15
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0@JonasBo You have $g^2 = 1$ for **ANY** $g \in G$. So if $x,y \in G$ you have $x^2 = 1 = y^2$. If you want this more formally, this is called substitution in first order predicate logic, I think. – 2017-01-15