If f: Z--> Z is an isomorphism, prove that f is the identity map. [ Hint: what a

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Since f is an isomorphism f(1) = 1. f(-1)*f(-1) = f((-1)(-1)) = f(1) = 1. So f(-1)^2 = 1 => f(-1) = ±1. f(-1) can't be 1 since f(1) = 1 and f is injective. So f(-1) = -1. f(0) = f(1+(-1)) = f(1) + f(-1) = 1 + (-1) = 0. Suppose n is positive and the smallest positive integer (exists by the Well-Ordering property) such that f(n) ? n. f(n) = f((n-1)+1) = f(n-1) + f(1) = f(n-1) + 1. Since f(n) ? n, so f(n-1) + 1 ? n => f(n-1) ? n-1, contradicting either our assumption that n is the smallest integer to fail the identity or contradicting f(0) = 0. So for n positive f(n) = n. For -n just note that f(0) = f(n + (-n)) = f(n) + f(-n) = n + f(-n). So f(-n) = f(0) - n = 0 - n = -n. So we have shown that f is the identity on all integers.

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