Few But Ripe

Few but ripe March 6, 2009 Theorem 1. An automorphism T on a finite group G that takes more than three fourth of the elements to their inverses, must take all the elements to their inverses. The group will be Abelian. Proof. Let S = {x ∈ G : T x = x−1 } If G is abelian, T takes the product of elements in S to their inverses. Since S has more than half the elements of G, it must generate the group, and so T takes all the elements of G to their inverses. If G is non-abelian at least one element most have a proper normalizer. Observe that if T (gx) = (gx)−1 , then g and x must commute. In other words, if x ∈ S does not commute with g ∈ S, T (gx) ∈ G\S. So for a g whose normalizer N (g) is proper, the map θg : S \ N (g) → G \ S defined as θg (x) = gx is well defined. This map is easily seen to be injective. But since S has more than three fourth the elements of the groups, the order of the domain is greater than the order of the range, contradicting the injectivity of the map. So G must be abelian.

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