IA equals inner in extraspecial

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Statement

In an Extraspecial group (?) (a group of prime power order whose commutator subgroup, center and Frattini subgroup are all equal and cyclic) any IA-automorphism (?) is inner.

Definitions used

Extraspecial group

Further information: extraspecial group

A group of prime power order is termed an extraspecial group if its commutator subgroup, center, and Frattini subgroup are all equal, and this subgroup is a cyclic group of prime order.

IA-automorphism

Further information: IA-automorphism

An IA-automorphism of a group is an automorphism that induces the identity automorphism on the Abelianization of the group (the quotient by its commutator subgroup).

Related facts

An equivalent fact

Proof

Given: An extraspecial p-group G. Let A denote the group of IA-automorphisms of G, and I denote the group of inner automorphisms of G

To prove: I = A

Proof: Note that I \le A (i.e., any inner automorphisms is IA). Also, I \cong G/Z(G) (again, a standard fact). Since both groups are finite, it suffices to show that |A| \le |I|.

The first step in this is to show that elements of A act as the identity. not only on the quotient G/G', but also on G' = Z(G). Thus, A can be viewed as the stability group of the series 1 \le Z(G) \le G. So, what we need to prove is that the cardinality of this stability group is at most |I| = |G/Z(G)|.