A proof

This is a proof of which I'm rather proud.
Claim: |U(n)| is even when n>2.
Pf: It is known that Aut(Z(n)) is isomorphic to U(n). Let a,b be elements if Z(n). Then, a+b=b+a, so Z(n) is Abelian. If a group G is Abelian, then there exists an automorphism P(g)=g^-1. As proof, note that inverses of elements are unique, so P is one-to-one and onto. Further, letting x,y be elements of G, P(ab)=(ab)^-1=(b^-1)(a^-1)=(a^-1)(b^-1)=P(a)P(b), showing that P preserves the operation. This suffices to show that P(g)=g^-1 is an automorphism on a group G. Then, P is an element of Aut(Z(n)). Finally, observe that P(P(g))=g; that is to say, the inverse of the inverse of g is g. This shows that the inverse of P is P, or, more precisely, that P is of order 2. Then, Aut(Z(n)) has at lease one element of order 2. By the properties of isomorphisms acting on elements, if G is a finite group, then G and P(G) have exactly the same number of elements of every order. Then, since U(n) is isomorphic to Aut(Z(n)) and Aut(Z(n)) has an element of order 2, U(n) has an element of order 2. Then, by a corollary of Lagrange's Theorem (specifically, the corollary states |a| divides |G|, or, the order of a group's element a divides the order of the group G), because U(n) contains an element of order 2, the order of U(n) must be a multiple of 2; that is, |U(n)| must be even. QED.
I bet there is an easier way to do this, but the problem specifically required the use of the noted corollary. And, really, this proof is pretty sexy -- it basically just compiles a bunch of other proofs and says, look, the claim's true. That's the definition of a sexy proof right?