Let G be a finite group and φ: G—>G be an automorphism that fixes more than half of the elements of G. Prove that φ is the identity of G.

Let S={g ∈ G : φ(g)=g}. We know that by hypothesis |S|>|G|/2 and, by what was said previously, S is a subgroup of G (it is a non-empty subset of G and if x,y∈S we have that φ(x⁻¹y)=φ(x⁻¹)φ(y)=φ(x)⁻¹φ(y) )therefore
|G|=|S|k for non-zero natural k. If k=1 I am done. In fact:
|G|=|S|k>(|G|/2)k. Therefore if k>1 then k≥2 from which |G|>|G| absurd, therefore k=1. ■