# This group is “dual” to the Mathieu group $M_{23}$. Is it known?

Inspired by this question, in particular by the indeed elegant description of the Mathieu group $M_{23}$ it starts with, I am wondering about the following:

Instead of $C$, defined as the multiplicative subgroup of order $23$ in the field $F=\mathbb F_{2^{11}}$ with $2^{11}=23\cdot89+1$ elements, let us take the "complementary" subgroup $D$ of order $89$. Knowing that $M_{23}$ is the group of additive maps of $F$ to itself which permute the set $C$, what about the group $G$ of additive maps of $F$ to itself which permute the set $D$ instead? Naively, I would expect this group $G$ to be also a simple group by "duality", but it cannot be a sporadic one because of the divisor $89$.

What can be said about $G$?

Of course, any composite pernicious Mersenne number which is a product of two primes defines two "dual" groups in a similar way. Is there any interesting relationship between the groups of such a pair? Disclaimer: I am not a group theorist.

• Naïvely, I would expect $G$ to be the cyclic group of order $89$ given by $D$ itself (acting on $F$ by multiplication). This should be the "typical" situation (barring cases like $23$ in $2^{11}$ when something magical happens). – Gro-Tsen Dec 29 '17 at 13:16
• @Gro-Tsen If that is the "typical" situation (by the way, why do you think that?), that would make the exceptional cases even more magical, between order 89 and order 10200960, sure enough... Isn't there any software to check that quickly? Or maybe it is essentially sufficient to show that $G$ must be simple, which would rule out the sporadic groups? – Wolfgang Dec 29 '17 at 13:30
• My general intuition is that "typically" you don't have any more automorphisms than the ones you can see. (This is, of course, just a vague general idea, and I don't have any proof in this particular example. Nor do I know how to check this with GAP in a reasonable amount of time.) I agree that it makes the 23 case quite magical. And it would indeed be interesting to know if there's a general result that for any $F=\mathbb{F}_{2^n}$ and subgroup $D$ of $F^\times$ of prime order the group of additive maps of $F$ preserving $D$ is always simple. I have no idea. – Gro-Tsen Dec 29 '17 at 13:50
• @Gro-Tsen Presumably (for $n=2$) you count $S_3=\text{PSL}(2,2)$ as an honorary simple group? :) – Jeremy Rickard Dec 29 '17 at 15:23

This group is the semidirect product $H=C_{89}\rtimes C_{11}$. Note that $H\le G$, where $C_{89}$ is multiplication by elements of order $89$ (and $1$), and $C_{11}$ is generated by the Frobenius automorphism $x\mapsto x^2$ of $\mathbb F_{2^{11}}$.
As $89$ is prime, $G$ is a primitive group. Primitive groups of such small degrees (actually up to degree $4000$, if I remember right, possibly with the exception of some affine cases) have been classified. In this case $G$ is either $A_{89}$, $S_{89}$, or the Sylow $89$-subgroups are normal. The first two cases cannot hold, because $\lvert\text{GL}_{11}(\mathbb F_2)\rvert$ is way too small (or because $13$ doesn't divide the order $2^{55}(2-1)(2^2-1)\dots(2^{11}-1)$ of $\text{GL}_{11}(\mathbb F_2)$).
Thus $G$ normalizes $C_{89}$. Let $g$ be an additive bijection of $\mathbb F_{2^{11}}$ which fixes $1$. Write $g$ as an additive polynomial map, i.e. $g(x)=\sum_{i=0}^{10}a_ix^{2^i}$. Then there is an integer $m$ with $1\le m\le 88$ such that $\sum_{i=0}^{10}a_i\zeta^{2^i}=\zeta^m$ for each element $\zeta$ of order $89$ (or $1$). Let $X$ be a variable, and $r(X)$ be the remainder of the division of $\sum_{i=0}^{10}a_iX^{2^i}$ by $X^{89}-1$. So $r(X)-X^m$ has at least $89$ roots, but degree $\le 88$, so $r(X)=X^m$. However, the remainders of $2^i$ modulo $89$, $0\le i\le 10$, are pairwise distinct. This forces that precisely one of the $a_i$ is $1$, the others are $0$, and $m$ is a power of $2$ modulo $89$.
• That looks nice, but I don't understand the role of $H$. Also: Why "13 doesn't divide the order ..."? Do you mean 23? A – Wolfgang Dec 29 '17 at 20:57
• @Wolfgang: No, I mean $13$ (actually $23$ divides the order of the linear group). Note that $2^{12}\equiv1\pmod{13}$, while $2^6$ and $2^4$ are not congruent $1$ modulo $13$. – Peter Mueller Dec 29 '17 at 21:01
• OK now (only now!) I see for $H$. In fact, you define $H:=C_{89}\rtimes C_{11}$ and then you show $G=H$. :) – Wolfgang Dec 30 '17 at 16:32