Inner automorphisms are I-automorphisms in the variety of groups

Statement

In the Variety of groups (?), treated as a variety of algebras, an I-automorphism (?) is the same thing as an Inner automorphism (?).

Definitions used

Inner automorphism

Further information: Inner automorphism

An automorphism $\sigma$ of a group $G$ is termed an inner automorphism if there exists $g \in G$ such that $\sigma = c_g$, where we define:

$c_g(a) = gag^{-1}$

I-automorphism

Further information: I-automorphism

Suppose $\mathcal{V}$ is a variety of algebras, and $A$ is an algebra in $\mathcal{V}$. An I-automorphism of $A$ is an automorphism that can be expressed as:

$x \mapsto \varphi(x,u_1,u_2,\dots,u_n)$

where $u_1, u_2, \dots, u_n \in A$ are fixed, and $\varphi$ is a word in terms of the operations of the algebra,with the property that for any algebra $B$ of $\mathcal{V}$, and any choice of values $v_1,v_2,\dots,v_n \in B$, the map:

$x \mapsto \varphi(x,v_1,v_2,\dots,v_n)$

gives an automorphism of $B$.

In other words $\varphi$ is guaranteed to give an automorphism.

Facts used

• Inner automorphisms actually are automorphisms. For full proof, refer: Group acts as automorphisms by conjugation

Proof

Inner automorphisms are I-automorphisms

$c_g(a)$ can be viewed as a word with input $a$ and parameter $g$. This word gives an automorphism for every group and every choice of parameter $g$. Thus, inner automorphisms are I-automorphisms.

I-automorphisms are inner automorphisms

Given: A word $\varphi(x,t_1,t_2,\dots,t_n)$ with the property that for any group $G$ and any choice of values of $u_i$ in $G$, the map sending $x$ to $\varphi(x,u_1,u_2,\dots,u_n)$ is an automorphism

To prove: This automorphism is always inner.

Proof: Let $F$ be the free group on $n+2$ generators, and let $u_1,u_2,\dots,u_n,u_{n+1},u_{n+2}$ be a freely generating set for $F$. By the given condition, the map:

$x \mapsto \varphi(x,u_1,u_2,\dots,u_n)$

gives an automorphism of $F$. In particular:

$\varphi(u_{n+1},u_1,u_2,\dots,u_n)\varphi(u_{n+2},u_1,u_2,\dots,u_n) = \varphi(u_{n+1}u_{n+2},u_1,u_2,\dots,u_n)$

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A little manipulation of possible expressions shows that $\varphi$ must be of the form:

$\varphi(x,u_1,u_2,\dots,u_n) \equiv \psi(u_1,u_2,\dots,u_n)x\psi(u_1,u_2,\dots,u_n)^{-1}$

Hence, any automorphism obtained using $\varphi$ must be an inner automorphism.