My concern is to look for a classification of :
- Centralizers/Normalizer of elementary abelian subgroups of a simple group.
- Centralizers/Normalizer of abelian subgroups(Not necessarily elementary) of a simple group.
Suppose we have a copy of $\mathbb{Z}_8$ in a non abelian simple group $G$ then $C_G(\mathbb{Z}_8)$ would be of order at least $8$ as $\mathbb{Z}_8$ is abelian.
But does this say anything about possible order of $C_G(\mathbb{Z}_8)$ atleast under some conditions?
And does this hold if that abelian subgroup is not actually $\mathbb{Z}_8$ but something like $\mathbb{Z}_4\times \mathbb{Z}_2$ or $\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_2$
As $G$ is simple $\mathbb{Z}_8$ would not be a normal subgroup assuming $|G|>8$ so Normalizer of $\mathbb{Z}_8$ in $G $ i.e., $N_G(\mathbb{Z}_8)$ would not be whole $G$.
But does this say anything about possible order of $N_G(\mathbb{Z}_8)$ atleast under some conditions?
And does this hold if that abelian subgroup is not actually $\mathbb{Z}_8$ but something like $\mathbb{Z}_4\times \mathbb{Z}_2$ or $\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_2$ ?
This question got stuck in my mind after reading some other question so at present i do not have an considerable ideas to explain.
So, please give some hints to proceed further..
Best Answer
Firstly, any group with a nontrivial cyclic Sylow $2$-subgroup has a normal $2$-complement, so it cannot be simple. There is a very elementary proof of that. Look at the regular permutation representation. Then the image of a generator of a cyclic Sylow $2$-subgroup is an odd permutation, so intersecting with the alternating group gets you a subgroup of index $2$, and then you get the normal $2$-complement by induction.
Secondly, Burnside's Transfer Theorem states that if $P \in {\rm Syl}_p(G)$ and $P \le Z(N_G(P))$, then $G$ has a normal $p$-complement. So, in particular, (assuming $P$ nontrivial), $G$ cannot be simple. Note that this can only apply if $P$ is abelian.
In particular, if $G$ has an abelian Sylow $2$-subgroup with $N_G(P)=P$, then $G$ is not simple. This applies whenever ${\rm Aut}(P)$ is a $2$-group, which is the case when $P = C_4 \times C_2$. So if $G$ is simple and $|P| = 8$, then $P$ must be elementary abelian. In that case $|{\rm Aut}(P)|= 168$, so the possibilities for $|N_G(P)/C_G(P)|$ are $3$, $7$ and $21$.
In the ealier question Isaacs 5C10, you would still need to eliminate the case $|N_G(P)/C_G(P)|=3$. I am guessing that there are other results on transfer in Isaacs book that would enable you to do that (there is a result called the Focal Subgroup Theorem for example), but unfortunately I don't have Isaacs' book right now!