# IA equals inner in extraspecial

## 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).

## 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 $|I| = |G/Z(G)|$.