Abstract Algebra (Modern) Automorphisms Question?

Question

Show that Aut (Z,+) has only two elements.

So I've figured out what the two elements are, f(x) = x and g(x) = -x. But I can't figure out how to prove that there are no other elements.

Answer 1

First, we prove a lemma -- if f is any homomorphism from Z → Z, then f(n) = n*f(1)

Proof: Since f is a homomorphism, f(0) = f(0+0) = f(0) + f(0), and subtracting f(0) from both sides reveals f(0) = 0 = 0*f(1), so this holds for n=0. Suppose this holds for some n. Then f(n+1) = f(n) + f(1) = n*f(1) + f(1) = (n+1)*f(1), so it holds for n+1, and by induction, it holds for any nonnegative integer n. Finally, note that since f(-n) + f(n) = f((-n)+n) = f(0) = 0, we have that f(-n) = -f(n). So if n is a negative integer, then -n is positive, so from the previous sentence and the fact that the formula holds for nonnegative integers, we have that f(n) = -f(-n) = -(-n)*f(1) = n*f(1). So the formula f(n) = n*f(1) holds for all integers n.

Now, proving that Aut(Z, +) has only two elements is easy -- note that for f to be an automorphism, it must be surjective, so there must be some k such that f(k)=1. But since f(k)=k*f(1), this means k=1/f(1), so 1/f(1) must be an integer. The only way this can happen is if f(1) = 1 or f(1) = -1, which gives you the two functions you already found. Q.E.D.

Answer 2

Let A be an autom on (Z, +)
Let A(1) = k. Then A(j) = k + . . . +k (j times) = jk. It's not an automorph since nothing goes to 1, unless k = + or - 1.
So the two automorphs are the ones you've already found;
f(x) = x sends 1 to 1 and g(x) = -x sends 1 to -1.
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