Homomorphism more than 3/4 the inverse

I think the point of this whole $3/4$ business is the following. If $G_1$ is the set of elements such that $f(x) = x^{-1}$, then if we look at left multiplication on $G$ by an element of $G_1$, more than half the elements have to make back into $G_1$.

Combining this with what we know about $f$ it should follow that any $g \in G_1$ commutes with more than $1/2$ the elements of $G$, so if you say Lagrange's theorem enough times it should follow that $G$ is abelian and $G_1$ generates $G$, which together imply the result.