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Let G be a group and assume that every nonidentity element of G has order two. Prove that G is abelian.
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Well, if you take a,b ∈ G
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Alternate proof:
a^2 = 1, so a = a^-1. Hence, ab = a^-1b^-1 = (ba)^-1 = ba.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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Thank you JaneFairfax. You were a great help!
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That is another good one, Ricky. Thanks.
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