# P-automorphism-invariant not implies characteristic

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) neednotsatisfy the second subgroup property (i.e., characteristic subgroup)

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## Statement

Let be a prime number. Then, we can construct a finite -group with a subgroup such that is invariant under all the -automorphisms of , but is not characteristic in .

## Facts used

## Proof

### An example

This example works for odd primes .

Let be a non-Abelian group of order , and let and be isomorphic copies of . Define , and define . Then, and are the only two subgroups in their automorphism class, and both are invariant under all -automorphisms.

### A more general example

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

Let be a number distinct from and relatively prime to , and let be the vector space of dimension over . Consider to be the cyclic group of order acting on by cyclic permutation on coordinates. The Bryant-Kovacs theorem asserts that there exists a -group such that and the image of the natural homomorphism:

is precisely .

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

Now, let be a subgroup of containing such that is *not* -invariant. For instance, could be the vector space spanned by the first coordinate. Then, since the -Sylow subgroup acts trivially on , every -automorphism fixes . On the other hand, since there are elements of that do not preserve , there are elements of that do not preserve .