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Consider the following facts:

  • A Frobenius group is a transitive permutation group on a finite set, such that no non-trivial element fixes more than one point and some non-trivial element fixes a point.

  • The subgroup $H$ of a Frobenius group $G$ fixing a point of the set $X$ is called the Frobenius complement.

(The following is a theorem due to Frobenius.)

  • The identity element together with all elements not in any conjugate of H form a normal subgroup called the Frobenius kernel K.

  • The Frobenius group G is the semidirect product of K and H.


Do we can determine structure Automorphims of a frobenius group?

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So you are asking about the automorphism group of a Frobenius group, right? – user641 Feb 5 '12 at 16:18
As with your previous post on this topic, you need to be more specific, and to show that you given the question some thought. You can say various things about the automorphism groups of Frobenius groups, but what exactly do you mean by "determine structure"? Here are a few thoughts. A Frobenius group has trivial centre, so embeds in its automorphism group. The kernel is characteristic, so is fixed by all automorphisms. The complements form a conjugacy class of subgroups, so are permuted by automorphisms. That's enough for now! – Derek Holt Feb 5 '12 at 22:17

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