Fraction of Elements in a group mapped to own inverse by automorphism

deluks917

Q: Given a fraction A/B when does there exist a finite group G and an automorphism f s.t. exactly A/B elements of G are mapped to their own inverses (f(a) = a-1? If so how can we find the group? Does anything change if we allow infinite groups?

I have a friend who was preparing for an intro to abstract algebra class. So we were going over questions that the professor had asked him. On one exam he had been asked if there was a group with automorphism that sent 3/4 of the group to their own inverses. He found by considering small groups that D4 the symmetries of the circles worked with the identity automorphism. We could not think of a way to solve this question in general.

if 1 > A/B > 1/2 then G can't be abelian because in abelian group the elemnts mapped homomorphically to their own inverses is a subgroup.

If A/B = C/D * E/F then we can find the solution for the RHS groups we can take a direct product and get A/B.

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fresh_42

Mentor
2018 Award
I would consider Galois groups of field extensions. It should be possible to find extension towers with the required properties, depending a bit on $A$ and $B$.

"Fraction of Elements in a group mapped to own inverse by automorphism"

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