Centralizer/Normalizer of abelian subgroup of a finite simple group

Question

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..

Answer 1

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!

Answer 2

Fusion 101: Suppose $G$ is a finite group whose Sylow 2-subgroup is $P \cong C_4 \times C_2$. Set $p=2$ for convenience. We can show that $G$ has a normal subgroup of index 8, so not only is $G$ not a simple group, it is a semi-direct product.

Look at $N_G(P)/ C_G(P) \leq \newcommand{\Aut}{\operatorname{Aut}}\Aut(C_4 \times C_2) \cong D_8$. Since $P$ is already a Sylow $2$-subgroup of $G$, $P$ is a Sylow 2-subgroup of $N_G(P)$ as well. Hence $N_G(P)/C_G(P)$ cannot contain any new elements of order two from $P/C_P(P) = P/P = 1$. Since $\Aut(P)$ is a 2-group, the only subgroup it has of odd order is also $1$, and $N_G(P)/C_G(P)=1$ and $N_G(P) = C_G(P)$. However, that means that $x^g$ is never in $P$ unless $x^g=x$ already:

Proposition: (Burnside fusion) Let $p$ be a prime, and let $G$ be a finite group with an abelian Sylow $p$-subgroup $P$. If $x, y \in P$ are conjugate in $G$, then they are already conjugate in $N_G(P)$.

Proof: Let $x, y \in P$ and suppose $y=x^g$ is conjugate to $x$ by some element $g \in G$. Let $Y=C_G(Y)$. Since $y \in Z(P)$, $P \leq Y$ is a Sylow $p$-subgroup. However, $x^g \in Z(P^g) = P^g$ so $P^g \leq Y$ is also a Sylow $p$-subgroup. By Sylow's theorem $P$ and $P^g$ are conjugate in $Y$, say $P = (P^g)^z$ for some $z \in Y = C_G(y)$. Now $x^{gz} = y^z = y$, with $gz \in N_G(P)$. Hence if $y=x^g$ for $g\in G$, then $y = x^h$ for some $h=gz \in N_G(P)$. $\square$

This is important for the transfer map: $T_P^G(x) = \prod (x^g)^{n_g}$ is a product of conjugates of $x$ raised to a power that puts them back inside $P$ with $\sum n_g = [G:P]$, by the transfer evaluation lemma (Isaacs's 5.5). However, if $x$ starts in $P$, then the only time $(x^g)^{n_g}$ is back in $P$ is when $(x^{n_g})^g = x^{n_g}$. That means $T(x) = x^{[G:P]}$ and $\ker(T) \cap P = 1$. However, the kernel of $T$ clearly contains every Sylow $q$-subgroup for $q \neq p$ since the image is a $p$-group, and also every $[g,h]$ since the image is abelian. In other words $\ker(T) \geq O^p(G) [G,G]$. If one is a bit more careful, one can actually show equality. We definitely have $T$ is surjective (it is an isomorphism when restricted to $P$), so we get that $\ker(T)$ is a normal subgroup of $G$ with $[G:\ker(T)] = |P| = 8$.

The Schur–Zassenhaus theorem gives that $G = P \ltimes \ker(T)$ is a semi-direct product.

The same argument works for $P\cong C_8$, but Derek gives a simple argument in this case.

Both of these cases are known as trivial fusion. When $P$ is abelian, you only need to check the $p'$-part of $\Aut(P)$. When $P$ is not abelian, then you get fancier versions (Frobenius's $p$-complement theorem says you only need to check certain subgroups of $\Aut(Q)$ for $Q \leq P$; Alperin–Goldschmidt gives exactly which $Q \leq P$ and which subgroups of $\Aut(Q)$ need to be checked).