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If $G$ is a finite group, embedded as a transitive subgroup of $S_n$ for some $n$, will every automorphism of $G$ extend to an inner automorphism of $S_n$?

I'm trying to connect the language that's used in many different sources on Hurwitz spaces and Nielsen classes. On Michael Fried's website, he notes that "absolute equivalence" on a Nielsen class $(g_1,\ldots,g_r)$ is an equivalence given by componentwise conjugation by an element $h\in N_{S_n}(G)$.

On the other hand, in some other sources, eg Volklein's chapter on "Moduli Spaces of Covers of the Riemann Sphere" in his book Groups as Galois groups, and in Fried's paper "The Inverse Galois Problem and Rational Points on Moduli Spaces", their notion of absolute equivalence on Nielsen classes seems to allow for componentwise application of any automorphism of $G$.

Also, does anyone have a pdf of Fried's 1977 paper "Fields of Definition of Function Fields and Hurwitz Families - Groups as Galois Groups". It's in the 1977 Volume of Communications in Algebra, but my university doesn't seem to have access to issues from that journal prior to 2000.


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Its definitely true that if you choose the regular representation of $G$ then every automorphism extends to an inner automorphism in $S_{|G|}$ (so $G$ has at least one transitive representation with the desired property), although in your case it seems you need the result to hold for a far wider class of transitive representations. – ARupinski Jan 20 '13 at 2:01
A pdf of the Fried paper you asked for is on his website: – Michael Zieve Aug 11 '13 at 15:48
up vote 6 down vote accepted

While the answers by Eric and ARupinski give negative examples for your question, here is the precise characterization for when the answer is yes: Let $\alpha$ be an automorphism of the transitive subgroup $G\le S_n$, and $G_1$ be the stabilizer of $1$ in $G$. Then $\alpha$ extends to an inner automorphism of $S_n$ if and only if $G_1$ and $\alpha(G_1)$ are conjugate in $S_n$.

The necessity of the condition is clear, and the sufficiency is a nice exercise. I believe the result is also in the permutation groups book by Dixon-Mortimer.

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In the last sentence of your first paragraph, you meant "Then $\alpha$ extends to an INNER automorphism ..." right? – oxeimon Mar 8 '14 at 23:38
@oxeimon: Yes, I fixed it. – Peter Mueller Mar 9 '14 at 10:00

No, this isn't even true if $G$ is $S_n$ itself: the symmetric group $S_6$ has an outer automorphism.

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Such automorphisms need not extend to inner automorphisms of $S_n$. Take the dihedral group $D_8$ sitting transitively inside $S_4$, say as the following set of permutations: $\{(1)(2)(3)(4), (13)(2)(4), (1)(3)(24), (12)(34), (14)(23), (1234), (13)(24), (1432)\}$. Now the sets of reflections $\{(13)(2)(4), (1)(3)(24)\}$ and $\{(12)(34),(14)(23)\}$ are equivalent to one another via an outer automorphism of $D_8$, but clearly this automorphism is not inner in $S_4$.

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