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. 
 A: 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 $89$.
