Show that Aut(G) = 1 if and only if |G| = 1 or 2.
I have a proof (literally found it in my sleep), though it uses the axiom of choice. I'm curious about other approaches and if that can be avoided. Utterly unimportant.
Eugene
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Hi Josh, here's my proof (which uses AC, sorry :)). Suppose to the contrary that |G| > 2. Take distinct elements g, h in G which are different from the identity. Since conjugation by an element of G is an automorphism of G and Aut(G) = 1, G is abelian. Then inversion in G is an automorphism of G, so G has exponent two. Hence G is an F2 -vector space. Let S be a basis for G containing g and h. Define a linear transformation T : G -> G such that T(g) = h, T(h) = g, and T(v) = v for all v in S - {g,h}. It belongs to Aut(G), but it differs from the identity automorphism. This is a contradiction.