Component of centralizer of involution normalizes component of group

Question

A component of the finite group $G$ is a subgroup $H$ satisfying the following conditions:

(1) $H$ is quasisimple (i.e. $H/Z(H)$ is simple and $[H,H]=H$)

(2) $H$ is subnormal in $G$

Exercise 6.5.2 in Kurzweil & Stellmacher, The Theory of Finite Groups is the following:

Let $t$ be an involution of $G$ and $E$ a component of $C_G(t)$. Then $E$ normalizes every component of $G$.

I'm quite stumped on this and don't how to make use of the fact that $t$ is an involution, or even where to begin.

For context, this question is in the section on the generalized Fitting subgroup $F^*(G)$. A few theorems proved in this and earlier sections that seem like they might be relevant:

(1) If $K$ is a component of $G$ and $L$ is subnormal in $G$, then either $K \le L$ or $[K,L]=1$.

(2) If $L$ is subnormal in $G$ then $F^*(L) = F^*(G) \cap L$

(3) $C_G(F^*(G)) \le F^*(G)$

There's more, but all the propositions and theorems of these sections deal with subnormal subgroups of $G$, whereas $C_G(t)$ need not be subnormal in $G$. So again, not sure where to use them.

So far, all I've been able to come with is the following: it suffices to assume that

$G = C \langle K^C \rangle$

where $C = C_G(t)$, $K$ is a component of $G$, and $\langle K^C \rangle$ denotes the subgroup generated by the conjugates of $K$ by elements of $C$, which is a central product of conjugates of $K$.

If it helps, we may as well assume $\langle K^C \rangle$ is a direct product of simple groups, on which $C$ acts. I attempted to use the strategy of "try to come up with a counterexample and see why it doesn't work". My (failed) counterexample was:

$G = (\langle t \rangle \times A_5) \rtimes (M_1 \times \ldots \times M_5)$

where $M_1, \ldots, M_5$ are isomorphic simple groups (let's say copies of the Monster group, or whatever) and $A_5$ acts by permuting the coordinates of this direct product, and $\langle t \rangle$ acts by permuting $M_1$ and $M_2$. Then $A_5$ doesn't normalize any of $M_1, \ldots, M_5$, which are components of $G$. The reason the counterexample fails, is because $A_5$ fails to be subnormal in $C_G(t)$. The actual components of $C_G(t)$ are $M_3, M_4, M_5$, and the diagonal set $\{(m,m)\}\subset M_1 \times M_2$, and sure enough each of these normalizes the components of $G$ which are $M_1, \ldots, M_5$.

Maybe this is a clue that $E(C_G(t)) \le E(G)$, perhaps? ($E(G)$ denotes the product of the components of $G$.) But that might just be a coincidence. At any rate, thinking about this example doesn't seem to have gotten me anywhere.

Answer 1

I think I've finally solved this. Let $C := C_G(t)$ as above. The key idea is to note that $C \cap E(G)$ is a (sub)normal subgroup of $C$, therefore by Theorem (1) we have $E \le E(G)$ or $[E,C \cap E(G)]=1$. The first case immediately implies $E \le E(G) \le N_G(K)$, so assume $[E,C \cap E(G)]=1$ henceforth. We now look for elements in $C \cap E(G)$ in order to make use of this.

Case 1: Suppose $K^t \ne K$. Then $[K, K^t]=1$ and it follows that the diagonal set $\{xx^t: x\in K\}$ is a subset of $C$ (and indeed a component), as in the above "failed" counterexample. Therefore $(xx^t)^\epsilon = xx^t$ for all $\epsilon \in E$. Take $x$ to be any non-central element of $K$, and recall that $E(G)/Z(E(G))$ is a direct product of simple groups which are the images of the components of $G$. This implies that $K^\epsilon = K$ or $K^\epsilon=K^t$. Hence $E$ acts by conjugation on the set $\{K, K^t\}$, and $E \cap N_G(K)$ is a subgroup of index at most 2. But $E$ is a quasisimple group, therefore it does not have a subgroup of index 2. Thus $E \le N_G(K)$ and we're done.

Case 2: $K^t=K$. In this case $t$ acts as an automorphism of $K$, and we need a lemma about fixed-point-free involutions. Let's state that first:

Lemma: Suppose $K$ is a finite group and $\tau$ is an involution in $\mathrm{Aut}(K)$ such that 1 is the only fixed point of $\tau$. Then $\tau(x) = x^{-1}$ for all $x\in K$, and in particular $K$ is an abelian group.

Proof: This is an exercise appearing in numerous group theory texts. We consider the map $f(x)=x^{-1}\tau(x)$ and show that $f$ is injective, hence surjective. Hence every element of $K$ has the form $x=y^{-1}\tau(y)$, which implies $\tau(x)=\tau(y^{-1})y = x^{-1}$. $\blacksquare$

For Case 2, according to this Lemma, we have $K \cap C \ne 1$ since $K$ is not abelian. Suppose that $K$ is simple (i.e. $Z(K)=1$). Then there exists $\alpha \in (K \cap C) \setminus Z(K)$. Since $[E, C \cap E(G)]=1$, this evidently implies $K^\epsilon = K$ for all $\epsilon \in E$, because $K$ is the only component of $G$ that contains $\alpha$. Thus $E \le N_G(K)$ as desired.

The last step, therefore, is to reduce to the case $Z(K)=1$. Let $Z := Z(E(G))$, which is the product of all the component centers, and let overline denote the natural projection onto $G/Z$. Then $\overline{t}$ acts as an involution of $\overline K = KZ/Z$, which is a non-abelian simple group, and a component of $\overline{G}$. Similarly, $\overline{E}$ is a component of $\overline{C}$, because $\overline E \cong E/(E\cap Z)$ which is the quotient of a quasisimple group by an abelian subgroup, hence quasisimple.

Therefore we may apply Case 2 to conclude that $\overline{E}$ normalizes $\overline{K}$. Translating back to $G$, this means that $K^\epsilon \le KZ$ for all $\epsilon \in E$. But evidently, $K$ is the only component of $KZ$, therefore $K^\epsilon = K$. $\blacksquare$