# 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) need not satisfy the second subgroup property (i.e., characteristic subgroup)
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## Statement

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 $G$.

## Proof

### 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 $H$.