I asked the same question on stackexchange, but I didn't get an answer, so I am reposting it here in hope of one (or an appropriate reference to a textbook or otherwise). I am assuming all groups finite. Suppose $A$ is an elementary abelian $2$-group and $C$ is a cyclic group of odd order acting fixed-point-freely on $A$.

My question is: does this situation yield any structural information about their semi-direct product $\Gamma=A\rtimes C$?

For instance, $Z(\Gamma)=1$ obviously, since the action has no fixed points. Another immediate observation is that $\Gamma$ is solvable, but not nilpotent, since $C$ is self-normalizing. A more specific question: is it true in this situation that any two self-normalizing subgroups (of the same order) are conjugate? This last bit would follow at once from Carter's theorem if these subgroups were nilpotent, but that's not the case here.


1 Answer 1


I think that one reason that you got no responses on stackexchange was that your question was too vague, and you should ask more specific questions, like you have done here.

The answer to your question is no. Let $G$ be the subgroup of $S_8$ generated by the permutations

$$(1,2)(3,4), (5,6)(7,8), (1,2,3)(5,6,7).$$

This has order 48, and is a subdirect product of two copies of $A_4$. It is the semidirect product of $A = \langle (1,2)(3,4), (1,3)(2,4), (5,6)(7,8), (5,7)(6,8) \rangle$ and $C = \langle (1,2,3)(5,6,7) \rangle$. The subgroups $H1 = \langle (1,2)(3,4), (1,2,3)(5,6,7) \rangle$ and $H2 = \langle(5,6)(7,8), (1,2,3)(5,6,7) \rangle$ are self-normalizing of order 12, but not conjugate in $G$.

More generally, your groups are Frobenius groups with complement $C$, so all of the standard results that apply to Frobenius complements apply to $C$ (but since you are already assuming that $C$ is cyclic, that might not help much).

  • 1
    $\begingroup$ Thanks. Is there a similar counter-example in the case $|C|=|A|-1$? $\endgroup$
    – user13040
    May 29, 2012 at 14:26
  • $\begingroup$ I think the answer is no, but I don't have time or energy to try and write down a proof! If $|C|=|A|−1=2^n−1$, then $G$ is isomorphic to the affine group ${\rm AGL}(1,2^n)$, and the non-nilpotent self-normalizing subgroups are of the form ${\rm AGL}(1,2^m)$ for $m|n$. I believe that it can be shown that there is a unique class of such subgroups for each $m|n$. $\endgroup$
    – Derek Holt
    May 29, 2012 at 15:32
  • $\begingroup$ I will look for a proof. Thanks anyway. $\endgroup$
    – user13040
    May 29, 2012 at 16:24

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