Algebra, Problems For Fun (37)

**NonCommAlg** · Jul 29th 2009, 07:14 PM

Let $G$ be a finite non-abelian group and $f \in \text{Aut}(G).$ Let $S=\{x \in G: \ f(x)=x^{-1} \}.$ Prove that: $|S|\leq \frac{3}{4}|G|.$

**siclar** · Jul 29th 2009, 09:32 PM

What about this approach: Suppose $|S|>\dfrac{3|G|}{4}$

Though $S$ is not necessarily a subgroup, it does contain the identity and inverses by the automorphism properties, i.e. if $e$ is the identity and $x\in S$ , $f(e)=e;\, f(x^{-1})=f(x)^{-1}=(x^{-1})^{-1}=x$ . Then in particular $f(f(x))=x$ , and the set of fixed points of $f\circ f$ does form a subgroup. Since this subgroup contains more than half of the elements it follows that $f(f(x))=x$ for all elements of the group.

If we can somehow show that in fact $S=G$ , we are done.

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