P-automorphism-invariant not implies characteristic

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This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., p-automorphism-invariant subgroup) need not satisfy the second subgroup property (i.e., characteristic subgroup)
View a complete list of subgroup property non-implications | View a complete list of subgroup property implications
Get more facts about p-automorphism-invariant subgroup|Get more facts about characteristic subgroup
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Let p be a prime number. Then, we can construct a finite p-group G with a subgroup H such that H is invariant under all the p-automorphisms of G, but H is not characteristic in G.

Facts used

  1. Bryant-Kovacs theorem
  2. Burnside's theorem on coprime automorphisms and Frattini subgroup


An example

This example works for odd primes p.

Let K be a non-Abelian group of order p^3, and let K_1 and K_2 be isomorphic copies of K. Define G = K_1 \times K_2, and define H_1 = K_1Z(G), H_2 = K_2Z(G). Then, H_1 and H_2 are the only two subgroups in their automorphism class, and both are invariant under all p-automorphisms.

A more general example

This example works for all primes p, but relies on a sophisticated result called Bryant-Kovacs theorem (fact (1)).

Let r be a number distinct from 1 and relatively prime to p, and let V be the vector space of dimension r over p. Consider G to be the cyclic group of order r acting on V by cyclic permutation on coordinates. The Bryant-Kovacs theorem asserts that there exists a p-group P such that P/\Phi(P) \cong V and the image of the natural homomorphism:

\alpha: \operatorname{Aut}(P) \to \operatorname{Aut}(P/\Phi(P)) = GL(V)

is precisely G.

By fact (2), the kernel of the homomorphism \alpha is a p-group.Notice that since G has order relatively prime to p, the kernel of \alpha must be the p-Sylow subgroup, and thus the p-Sylow subgroup is normal.

Now, let H be a subgroup of P containing \Phi(P) such that H/\Phi(P) is not G-invariant. For instance, H/\Phi(P) could be the vector space spanned by the first coordinate. Then, since the p-Sylow subgroup acts trivially on P/\Phi(P), every p-automorphism fixes H. On the other hand, since there are elements of G that do not preserve H/\Phi(P), there are elements of \operatorname{Aut}(P) that do not preserve H.