A question from Isaac's Book

Question

Let $A$ act via automorphisms on $G$, and $(|A|,|G|)=1$ and $G=HK$ where $H,K$ are $A$ invariant subgroups of $G$. Show that $$C_G(A)=C_H(A)C_K(A)$$

I can solve the question when I assume $H\cap K =1$ but in general I could not solve. ($3.E.2$)

Edit: I think there is a problem in the solution of Derek at last step.

When we have $A^h =A^{t} \implies A^{ht^{-1}}=A \implies ht^{-1}\in N_H(A)$ so how can we conclude that $ ht^{-1}\in C_H(A)$ ? Can anyone explain this ?

Answer 1

Here is the solution from Kurzweil–Stellmacher via @j.p.

Lemma: Suppose the finite group $A$ acts on the finite group $G$ with $\gcd(|A|,|G|)=1$ and at least one of $A,G$ is solvable. Let $U$ be an $A$-invariant subgroup of $G$ and $g \in G$ such that $(Ug)^A = Ug$. Then there exists $c \in C_G(A)$ such that $Ug = Uc$.

Proof: This is Isaacs's Theorem 3.27 on page 101, and Kurzweil–Stellmacher's 8.2.1 page 184: If $Ug$ is an $A$-invariant coset of $U$ in $G$, then $Ug$ is a coset containing an element $c \in C_G(A)$. That just means $Ug = Uc$. $\square$

Proposition: Suppose the finite group $A$ acts on the finite group $G$ with $\gcd(|A|,|G|)=1$ and at least one of $A,G$ is solvable. Suppose further that $G=HK$ for $A$-invariant subgroups $H,K \leq G$. Then $C_G(A) = C_H(A) C_K(A)$.

Proof: This is Isaacs's Exercise 3E.2 page 107 and Kurzweil–Stellmacher's 8.2.11 page 188. Let $hk \in C_G(A) \leq HK$ with $h \in H, k \in K$. Then, for each $a\in A$, $hk = (hk)^a = h^ak^a$ so $[a,h] = (h^{a})^{-1} h = k^a k^{-1} = [ a, k^{-1} ] \in H \cap K$. Set $U= H \cap K$. Then $(hU)^a = h^a U = h^a [a,h] U = h U$, so the lemma applies to give $hU = c_hU$ for some $c_h \in C_H(A)$ (where $H$ takes the place of $G$ in the lemma). Similarly, $(Uk)^a = U k^a = U [a,k^{-1}]^{-1} k^a = Uk$, so there is some $c_k \in C_K(A)$ with $Uk = Uc_k$. In particular, there are $u_h,u_k \in H \cap K$ with $h=c_h u_h$ and $k = u_k c_k$. Then $hk = c_h u_h u_k c_k$ and $(hk)^a = c_h (u_h u_k)^a c_k$, so $(u_h u_k)^a = u_h u_k \in C_{H \cap K}(A)$. Hence $hk = c_h ( u_h u_k ) c_k \in C_H(A) C_{H\cap K}(A) C_K(A) = C_H(A) C_K(A)$ as was to be shown. $\square$

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Derek Holt's argument proves the lemma as well, but perhaps the proof's brevity makes it confusing. In his argument he has $ht^{-1} \in N_H(A)$. Hence $[ ht^{-1}, A ] \leq H \cap A = 1$, the first $H$ because $H$ is $A$-invariant and $ht^{-1} \in H(H \cap K)^{-1} = H$, and the $A$ because $ht^{-1} \in N_H(A)$. Hence $ht^{-1}$ centralizes $A$ and $ht^{-1} \in C_H(A)$.

Answer 2

Let's work in the semidirect product $G \rtimes A$. Suppose that $g = hk \in C_G(A)$. Then $A^h = A^{k^{-1}} \subset HA \cap KA = (H \cap K)A$.

Now, the solvability of $G$ or $A$ means we can use the conjugacy part of the Schur-Zassenhaus Theorem, and deduce that $A$ and $A^h$ are conjugate in $(H \cap K)A$. So there exists $t \in H \cap K$ with $A^h = A^{k^{-1}} = A^t$, and now $g = (ht^{-1})(tk) \in C_H(A)C_K(A)$.