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Fixed point free automorphism of order 2
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[QUOTE="andrewkirk, post: 6868440, member: 265790"] Your last equals sign is wrong: $$\sigma^{-1}(x) = \sigma(x^{-1})$$ I think what you meant is the property we get from automorphism being a homomorphism, which is that: $$(\sigma(x))^{-1}= \sigma(x^{-1})$$ The second is correct. The first is not. Note that in the second one an exponent of -1 always means "take the inverse in this group", whereas in the first one, the first -1 exponent means "take the inverse function", which is a completely different thing: nothing to do with group inverses. [/QUOTE]
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Fixed point free automorphism of order 2
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