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I feel sort of silly asking this question. Unless I'm very much mistaken the paper I'm reading assumes the following statement:

Let $G$ be a finite group. We may embed it via the Cayley embedding into an ambient permutation group $G \leq S_{|G|}$. Then any automorphism of $G$ comes from conjugation by an element in $N_{S_{|G|}}(G)$.

Is this statement true?

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  • $\begingroup$ Note that left and right Cayley embeddings may be different (although this doesn't affect the answer to your question). $\endgroup$
    – S. Carnahan ♦
    Commented Jun 11, 2010 at 19:29
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    $\begingroup$ The normalizer you mention is called the holomorph. It is the semi-direct product of G and Aut(G) and is very often used as a containing group where automorphisms become group elements. I think you'll find this in most group theory texts, though the permutation description is clearest in Burnside. Remember an automorphism is just a permutation of the elements of a group. $\endgroup$
    – Jack Schmidt
    Commented Jun 12, 2010 at 15:16
  • $\begingroup$ This result reminds me of the Skolem-Noether theorem on automorphisms of simple subalgebras of central simple algebras, or at least the special case where the CSA is $M_n(K)$. Is there any connection here? $\endgroup$
    – Pete L. Clark
    Commented Jun 12, 2010 at 18:13

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The statement is true. Let $g \in G$ and $\pi \in Aut(G)$. Let $\lambda_{g}$ be the corresponding left translation by $g$. Regard $\pi$ and $\lambda_{g}$ as elements of $Sym(G)$. Then for all $x \in G$,

$(\pi \lambda_{g} \pi^{-1})(x) = (\pi \lambda_{g}) ( \pi^{-1}(x)) = \pi( g \pi^{-1}(x)) = \pi(g) x = \lambda_{\pi(g)}(x)$.

Thus $\pi \lambda_{g} \pi^{-1} = \lambda_{\pi(g)}$.

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    $\begingroup$ You would think this would be a more famous fact. I was always under the impression that outer automorphisms are much more mysterious than inner automorphisms. $\endgroup$
    – James D. Taylor
    Commented Jun 11, 2010 at 19:27
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    $\begingroup$ This observation is the basis of a famous construction of Hall. If you iteratively form the embeddings: $G_{0} \to G_{1} \to \cdots \to G_{n} \to \cdots$ where $G_{0} = Sym(5)$ and $G_{n} \to G_{n+1} = Sym(G_{n})$, then the union is a simple locally finite group $G$ such that: (a) every finite group embeds in $G$; and (b) every isomorphism between finite subgroups is induced by an inner automorphism of $G$. $\endgroup$
    – Simon Thomas
    Commented Jun 11, 2010 at 19:34
  • $\begingroup$ That sounds interesting. Do you have a reference? $\endgroup$
    – James D. Taylor
    Commented Jun 11, 2010 at 19:35
  • $\begingroup$ Outer automorphisms are much more mysterious than inner ones: look at the symmetric group $G=S_6$, for instance. You get no practical help in studying its outer automorphisms just by knowing the general fact you quoted. Here the left or right Cayley embedding involves an enormous symmetric group. $\endgroup$
    – Jim Humphreys
    Commented Jun 11, 2010 at 19:36
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    $\begingroup$ en.wikipedia.org/wiki/Hall%27s_universal_group $\endgroup$
    – Simon Thomas
    Commented Jun 11, 2010 at 19:37

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