Centerless and characteristic in automorphism group implies automorphism group is complete

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Statement

Suppose G is a Centerless group (?) such that, with the natural embedding of G in its Automorphism group (?) \operatorname{Aut}(G), G is a Characteristic subgroup (?). Then, the automorphism group \operatorname{Aut}(G) is a Complete group (?): it is centerless and every automorphism of it is inner.

Related facts

Facts used

For the hands-on proof

  1. Centerless implies inner automorphism group is centralizer-free in automorphism group: If G is centerless, then \operatorname{Inn}(G) (identified naturally with G) has trivial centralizer in \operatorname{Aut}(G). In other words, no non-identity automorphism commutes with every inner automorphism.
  2. Inner automorphism group is normal in automorphism group
  3. Centralizer-free and normal implies automorphism-faithful: If A \le B is a normal subgroup and C_B(A) is trivial, then any automorphism of Bthat fixes every element of A is the identity map.
  4. The basic idea that, if c_g denotes conjugation by g and \tau denotes an automorphism of G, we have:

c_{\tau(g)} = \tau \circ c_g \circ \tau^{-1}

In other words, if \tau is an automorphism of G, the action of \tau on \operatorname{Inn}(G) by conjugation in \operatorname{Aut}(G) precisely mimics the action of \tau as an automorphism on G.

For the proof using the automorphism group action lemma

Automorphism group action lemma: Suppose K is a group and N, H \le K are subgroups such that H \le N_G(N) and \sigma is an automorphism of K that restricts to an automorphism of N as well as of H. Let \alpha be the induced automorphism of N and \sigma' be the induced automorphism of H. Then, if

\rho: H \to \operatorname{Aut}(N)

denotes the homomorphism induced by the action by conjugation, we have:

\rho \circ \sigma' = c_\alpha \circ \rho.

Here, c_\alpha denotes conjugation by \alpha in the group \operatorname{Aut}(N).

Proof

Hands-on proof

Given: A centerless group G, embedded naturally as G = \operatorname{Inn}(G) in the automorphism group \operatorname{Aut}(G). Further, G is a characteristic subgroup of \operatorname{Aut}(G) under this embedding.

To prove: \operatorname{Aut}(G) is complete: the center of \operatorname{Aut}(G) is trivial, and any automorphism of \operatorname{Aut}(G) is inner.

Proof:

  1. (Facts used: fact (1)): \operatorname{Aut}(G) is centerless: By fact (1), G = \operatorname{Inn}(G) is centralizer-free in \operatorname{Aut}(G). Thus, \operatorname{Aut}(G) is centerless.
  2. (Facts used: facts (1), (2), (3)): G is automorphism-faithful in \operatorname{Aut}(G): By fact (1), G is centralizer-free in \operatorname{Aut}(G), and by fact (2), G is normal in \operatorname{Aut}(G). Combining these with fact (3), we obtain that G is automorphism-faithful in \operatorname{Aut}(G).
  3. (Given data used: G is characteristic in \operatorname{Aut}(G)): For any automorphism \sigma of \operatorname{Aut}(G), there exists an element \tau \in \operatorname{Aut}(G) such that conjugation by \tau has the same effect as \sigma on the subgroup G: First, observe that since G is characteristic in \operatorname{Aut}(G), the restriction of \sigma to G equals an automorphism of G, so there is an automorphism \tau of G having the same effect. Note that (by fact (4)) an automorphism \tau of G manifests itself as the inner automorphism by \tau on G = \operatorname{Inn}(G). Thus, c_\tau and \sigma have the same effect on G, viewed as a subgroup of \operatorname{Aut}(G).
  4. \sigma = c_\tau, and thus, \sigma is inner: By step (3), \sigma^{-1}c_\tau is an automorphism of \operatorname{Aut}(G) whose restriction to G = \operatorname{Inn}(G) is the identity map. Step (2) now forces \sigma^{-1}c_\tau to be the identity map on the whole group \operatorname{Aut}(G), yielding \sigma = c_\tau.
  5. Step (1) shows that \operatorname{Aut}(G) is centerless, and steps (3) and (4) show that every automorphism is inner. This completes the proof of completeness.

Proof using automorphism group action lemma

Given: A centerless group G, embedded naturally as G = \operatorname{Inn}(G) in the automorphism group \operatorname{Aut}(G). Further, G is a characteristic subgroup of \operatorname{Aut}(G) under this embedding.

To prove: \operatorname{Aut}(G) is complete: the center of \operatorname{Aut}(G) is trivial, and any automorphism of \operatorname{Aut}(G) is inner.

Proof: Set K = H = \operatorname{Aut}(G) and N = \operatorname{Inn}(G). By assumption, N is characteristic in K = H, so H \le N_G(N). Let \sigma be any automorphism of K. Observe that:

  • \sigma restricts to an automorphism of H, because H = K.
  • \sigma restricts to an automorphism of N, because by assumption N is characteristic in K.

Further, the natural map

\rho:H \to \operatorname{Aut}(N) = \operatorname{Aut}(G)

is in this case the identity map. Thus, the automorphism group action lemma tells us that for any automorphism \sigma of K = \operatorname{Aut}(G), we have:

\sigma = c_\alpha

where \alpha is the automorphism of N = G induced by \sigma.