Let G be a finite group, and let f: G --> G be a homomorphism satisfying f² = 1_G. Assume also that f(g) = g if and only if g = e, the identity element of G.

(a) Show that for all g in G, there exists an h in G such that g = h^-1f(h).

(b) Prove that G must be abelian.

In fact, part (a) is asking us to show that G = {h^-1f(h) : h in G}. Since G is finite, it suffices to show that |G| < |{h^-1f(h) : h in G}|. This is accomplished by showing that the map h |--> h^-1f(h) is an injection:

Suppose that a^-1f(a) = b^-1f(b), where a and b are elements of G.

Then f(ab^-1) = ab^-1.

Since f fixes only the identity element of G, it follows that ab^-1 = e; i.e., a = b and this finishes part (a).

Now for part (b). First, we claim that f(g) = g^-1 for all g in G:

Let g be an element of G. Then there exists an element h in G such that

g = h^-1f(h).

Then f(g) = f(h^-1)h = (h^-1 f(h) )^-1 = g^-1.

The proof that G is abelian, is now trivial:

Let a and b be elements of G. Then (ab)^-1 = f(ab) = f(a)f(b) = a^-1b^-1.

Whence, ab = ba.