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.

Related facts

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

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.