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

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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'.

Facts used

  1. Stability group of subnormal series of p-group is p-group
  2. Omega-1 of Abelian p-group is coprime automorphism-faithful
  3. Maschke's averaging lemma


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


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).


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)