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

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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

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.