Let $x$ be an involution in a finite group $G$ and $K$ be a component of $C_G(x)$. Show that $K\subseteq E(G)$ or $[C_{E(G)}(x),K]=1$.
Here, $E(G)$ is the layer of the group $G$.
My attempt: Suppose $E(G)\subseteq C_G(t)$. Then $C_{E(G)}(x)=E(G)$. Now, $K$ is a component of $C_G(x)$ and $E(G)$ is a subnormal subgroup of $C_G(x)$, so $K\subseteq E(G)$ or $[K,C_{E(G)}(x)]=1$, and we're done.
I don't know how to handle the general case, specially where to use the fact that $t$ is an involution?
Edit: I think the argument is as follows: Since $E(G)$ is a normal subgroup of $G$, so $C_{E(G)}(x)=E(G)\cap C_G(x)$ is a normal subgroup of $C_G(x)$. Combining with the fact that $K$ is a component of $C_G(x)$, we get that $K\subseteq C_{E(G)}(x)\subseteq E(G)$ or $[K,C_{E(G)}(x)]=1$.