Assume that $G$ allows a direct decomposition $$G = E_1 × ··· × E_n$$ that is invariant under A, i.e., $E_i^a \in {E_1,...,E_n}$ for all $a ∈ A$ and $i \in {1,...,n}.$
Under the additional hypothesis that $A$ acts transitively on ${E_1,...,E_n}$, we compare the fixed-point groups $C_G(A)$ and $C_{E_i} (N_A(E_i))$.
Let $E ∈ {E_1,...,E_n}$ and $B := N_A(E)$, and let $S$ be a transversal for the cosets of $B$ in $A$. Then $$(+): G = \left\langle {{E^A}} \right\rangle = \mathop \times \limits_{s \in S} {E^s}$$
Under the above hypotheses the following hold:
[8.1.6] (a) $$C_G(A) = \left\{ {\prod\limits_{s \in S} {{e^s}|e \in {C_E}(B){\rm{\} }}} } \right\}$$
The proof:
Article: http://gen.lib.rus.ec/book/index.php?md5=C1F95CF5F64359D6DFD3C6E5BD7EB12F
I don't understand why
For every $(s, a) ∈ S × A$ there exists a unique $(b(s, a), s_a) ∈ B × S$ such that $sa = b(s, a)s_a.$ And $s_a=?$.
$\prod\limits_{s \in S} {{e^{b(s,a){s_a}}} = \prod\limits_{s \in S} {{e^{{s_a}}} = } } \prod\limits_{s \in S} {{e^s}} $
Thank you very much
Write $A$ as a disjoint union: $A = \bigcup_{s \in S} Bs$. As $sa \in A = \bigcup_{s \in S} Bs$ and the union is a disjoint union, we see that there exists unique $s_a \in S$ such that $sa \in B s_a$. So, there exists unique $b(s,a) \in B$ such that $sa = b(s,a)s_a$.
By definition (I found it on page 176 of the book http://web.math.ku.dk/~olsson/manus/GruFus/Kurzweil-Stellmacher_Theory%20of%20finite%20groups.pdf ), $C_G(B)$ is the subgroup of fixed points of $B$ in $G$. So, $e^b = e$ for any $b \in B$. In particular, $e^{b(s_a)} = e$. That's why $\prod_{s \in S}e^{b(s,a)s_a} = \prod_{s \in S}e^{s_a}$.
The last equality follows from the fact that if $s_1,s_2 \in S$ with $s_1 \ne s_2$, then $s_1a$ and $s_2a$ lie in different (right) cosets of $B$. Indeed, if $s_1a, s_2a \in Bs$, then $s_1a = b_1s$ and $s_2 = b_2s$ for some $b_1,b_2 \in B$. Then $s_1 = b_1b_2^{-1}s_2 \in Bs_2$, i.e. $s_1$ and s_2$ lie in the same coset.