# Centerless implies inner automorphism group is centralizer-free in automorphism group

## Statement

Suppose $G$ is a Centerless group (?): the center of $G$ is trivial. Then, the Inner automorphism group (?) $\operatorname{Inn}(G)$ (which is naturally isomorphic to $G$) is a Centralizer-free subgroup (?) inside $\operatorname{Aut}(G)$: there is no non-identity automorphism of $G$ that commutes with every inner automorphism.

Equivalently, under the identification of $G$ with $\operatorname{Inn}(G)$, we say that $G$ is centralizer-free in $\operatorname{Aut}(G)$. Note that this also implies that $\operatorname{Aut}(G)$ is a centerless group.

## Facts used

1. Group acts as automorphisms by conjugation: In particular, $c_{gh} = c_g \circ c_h$, where $c_g$ is the map $x \mapsto gxg^{-1}$.

## Proof

### Hands-on proof

Given: A group $G$, with automorphism group $\operatorname{Aut}(G)$ and inner automorphism group $\operatorname{Inn}(G)$. Here, for $g \in G$, the inner automorphism $c_g$ is defined as $x \mapsto gxg^{-1}$.

To prove: If $\sigma \in \operatorname{Aut}(G)$ commutes with $c_g$ for all $g \in G$, then $\sigma$ is the identity map.

Proof:

1. For all $g \in G$, $\sigma \circ c_g \circ \sigma^{-1} = c_{\sigma(g)}$: For any $x \in G$, the left side is $\sigma(g\sigma^{-1}(x)g^{-1}) = \sigma(g)\sigma(\sigma^{-1}(x))\sigma(g)^{-1} = \sigma(g)x\sigma(g)^{-1} = c_{\sigma(g)}(x)$. Thus, the two sides are equal for all $x \in G$, and are hence equal as functions.
2. If $\sigma$ and $c_g$ commute, we have $c_g = c_{\sigma(g)}$: This follows directly from step (1).
3. If $\sigma$ and $c_g$ commute, we have that $c_{g^{-1}\sigma(g)}$ is the identity map: Since $c_g = c_{\sigma(g)}$, we get $(c_g)^{-1}c_{\sigma(g)}$ is the identity map. Using the fact that conjugation is a group action, we obtain that $c_{g^{-1}\sigma(g)}$ is the identity map.
4. (Given data used: $G$ is centerless): If $\sigma$ and $c_g$ commute, $\sigma(g) = g$: Step (3) shows that $g^{-1}\sigma(g)$ acts trivially by conjugation, and is hence in the center. Since we know that $G$ is centerless, $g^{-1}\sigma(g)$ must be the identity element, yielding $g = \sigma(g)$.
5. If $\sigma$ commutes with $c_g$ for every $g \in G$, $\sigma$ is the identity map: This is a direct consequence of step (4), applied to all $g \in G$.

### Proof using the commutator language

Given: A group $G$, with automorphism group $\operatorname{Aut}(G)$ and inner automorphism group $\operatorname{Inn}(G)$. Here, for $g \in G$, the inner automorphism $c_g$ is defined as $x \mapsto gxg^{-1}$.

To prove: If $\sigma \in \operatorname{Aut}(G)$ commutes with $c_g$ for all $g \in G$, then $\sigma$ is the identity map.

Proof: For convenience, we identify $G$ with the subgroup $\operatorname{Inn}(G)$ of $\operatorname{Aut}(G)$. Note that $[G,\sigma]$ now has two interpretations:

• It is the subgroup of $\operatorname{Aut}(G)$ generated by commutators between elements of $G$ and the automorphism $\sigma$, viewed as automorphisms of $G$.
• It is the subgroup of $G$ generated by elements of the form $g^{-1}\sigma(g)$, where $g \in G$.

The former interpretation tells us that $[G,\sigma]$ is trivial, which, viewed using the latter interpretation, yields that $\sigma$ is the identity automorphism.