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

Statement
Suppose $$G$$ is a fact about::centerless group: the center of $$G$$ is trivial. Then, the fact about::inner automorphism group $$\operatorname{Inn}(G)$$ (which is naturally isomorphic to $$G$$) is a fact about::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.

Related facts

 * Centerless and characteristic in automorphism group implies automorphism group is complete

Facts used

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

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.