# Structure lemma for p-group with coprime automorphism group having automorphism trivial on invariant subgroups

## Statement

Suppose $G$ is a Group of prime power order (?), and $A$ is a subgroup of $\operatorname{Aut}(G)$, of order relatively prime to $G$. Suppose there exists a non-identity element $\varphi \in A$ such that $\varphi$ acts as the identity on every proper $A$-invariant subgroup of $G$. Then, we have the following conclusions about the structure of $G$:

1. $[G,G] = \Phi(G) \le Z(G)$. In other words, the commutator subgroup is contained in the center, and the quotient by the commutator subgroup is elementary Abelian.
2. Either of two cases can occur: $\Phi(G) = Z(G)$, in which case $G$ is a Special group (?), or $\Phi(G)$ is trivial, in which case $G$ is an Elementary Abelian group (?).

We also have the following conclusions about the action of $A$ on $G$:

1. $\varphi$ acts nontrivially on $G/G'$, and the action of $A$ on $G/G'$ is irreducible (in other words, there are no proper nontrivial $A$-invariant subgroups).
2. In case $G$ is a special group, $\varphi$ also acts trivially on $G'$.

## Proof

Given: A $p$-group $G$, a subgroup $A$ of $\operatorname{Aut}(G)$ such that $A$ has order relatively prime to $G$. A non-identity element $\varphi \in A$ such that $\varphi$ acts as the identity on every proper $A$-invariant subgroup.

To prove: $\varphi$ acts nontrivially on $G/G'$ and trivially on $G'$.$G/G'$ is elementary Abelian and $A$ acts irreducibly on $G/G'$. $G$ is either elementary Abelian or special.

### Proof that $\varphi$ acts nontrivially on $G/G'$

Since $G'$ is a characteristic subgroup of $G$, it is in particular $A$-invariant. Thus, $\varphi$ acts trivially on $G'$. If $\varphi$ acted trivially on $G/G'$, fact (1) would yield that $\varphi$ is the identity map, a contradiction. Thus, $\varphi$ acts nontrivially on $G/G'$. In particular, $A$ acts nontrivially on $G/G'$.

### Proof that $G/G'$ is elementary Abelian and that the action of $A$ is irreducible

Let $\overline{G} = G/G'$. Suppose the action of $A$ on $G$ is decomposable, so $\overline{G} = \overline{G_1} \times \overline{G_2}$, where both $\overline{G_1}$ and $\overline{G_2}$ are proper $A$-invariant. Their inverse images in $G$ are proper $A$-invariant subgroups $G_1, G_2 \le G$, and $\varphi$ acts trivially on both $G_1$ and $G_2$. Thus, $\varphi$ acts trivially on $G_1G_2 = G$, a contradiction.

Thus, the action of $A$ on $\overline{G}$ is indecomposable.

Next, consider the induced action of $A$ on $\Omega_1(\overline{G})$. Suppose $\Omega_1(\overline{G})$ is a proper subgroup of $\overline{G}$. Then, the inverse image of this in $G$ is a proper $A$-invariant subgroup, so $\varphi$ acts trivially on it, forcing $\varphi$ to act trivially on $\Omega_1(\overline{G})$. By fact (2), we see that this forces $\varphi$ to be the identity on $\overline{G}$, a contradiction.

Thus, $\Omega_1(\overline{G}) = \overline{G}$, so $\overline{G}$ is elementary Abelian. The action of $A$ on $\overline{G}$ is indecomposable, so by fact (3), it is irreducible.

### Proof that $G$ is elementary Abelian or special

We first show that $G'$ centralizes $G$. PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]

Thus, $G' \le Z(G)$. So, $Z(G)$ is an $A$-invariant subgroup containing $G'$. Thus, $Z(G)/G'$ is an $A$-invariant subgroup of $\overline{G}$. By the irreducibility of the action of $A$ on $\overline{G}$, either $Z(G)/G'$ is trivial or $Z(G)/G' = \overline{G}$. Thus, either $Z(G) = G$ (in which case $G'$ is trivial and $G$ is elementary Abelian) or $Z(G) = G'$ (in which case $G' = Z(G) = \Phi(G)$, and so $G$ is a special group).

## References

### Textbook references

• Finite Groups by Daniel Gorenstein, ISBN 0821843427, More info, Theorem 3.7, Page 181-183, Section 5.3 ($p'$-automorphisms of $p$-groups)