Question on Proof that $O_p(C/(C\cap F(G)) = 1$ for $C = C_G(F(G))$.

Question

I have a question on the proof of a lemma about the Fitting subgroup, I mention all used facts:

If $N \unlhd G$ and $A ~\mbox{char}~ G$ be a characteristic subgroup of $G$. Then i) $A$ is normal in $G$, and ii) if $N ~\mbox{char}~ G$, then $A ~\mbox{char}~ G$. Denote by $F(G)$ the Fitting subgroup of a group $G$.

Lemma: Let $C := C_G(F(G))$. Then for each prime number $p$ we have $$ O_P(C/(C\cap F(G))) = 1. $$

Proof: Let $P$ be the inverse image of $O_p(C/(C\cap F(G))$ in $C$. Then $P$ is normal in $G$ by the above mentioned facts, and $P$ is nilpotent since $C \cap F(G) \le Z(C)$ (here another Theorem is referred, which says that $G$ is nilpotent if $G/Z$ is nilpotent for each $Z\le Z(G)$). Hence $P \le F(G) \cap C$ and $O_p(C/(C\cap F(G)) = 1$.

As $F(G)$ is normal, the center is normal too (see here). By the facts cited, that $P \unlhd G$ would follow if $P$ is characteristic in $C$, as $O_p(C/(C\cap F(G))$ is a characteristic subgroup of $C/(C\cap F(G))$, therefore normal, and normality is preserved by inverse images, so that $P \unlhd C$ is all I can get, but I need $P ~\mbox{char}~ C$.

So I asked myself if the property of being characteristic is preserved by inverse images, but I found the following counterexample, let $G = H\times H$ be the external direct product, then $H \times \{1\}$ is not characteristic as could be seen by applying the automorphism $(x,y) \mapsto (y,x)$, but $G / (\{1\}\times H) \cong H$ and the subgroup $H \times \{1\}$ corresponds to $$ ((H\times\{1\})(\{1\}\times H)) / (\{1\}\times H)= H\times H/(\{1\}\times H) = G/(\{1\}\times H)$$ i.e. the whole group in the quotient, so trivially is characteristic . So this reasoning seems to be wrong....

So how is the reasoning in the above proof? How could we conclude that $P \unlhd G$?

Answer 1

Edit:

$O_p(C/C\cap F)$ is char in $C/C\cap F$ and $C/C\cap F$ is normal in $G/C\cap F$. Hence, $O_p(C/C\cap F)$ is normal in $G/C\cap F$. Then we have $P$ is normal in $G$. (See the Below Lemma:)

Lemma If $H$ Char in $K$ and $K$ is normal in $G$ then $H$ is normal $G$.

Proof: Let $g\in G$ then set $f:K\to K$ by $f(k)=gkg^{-1}.$ Then $f\in Aut(K)$ so $f(H)=H$. Thus, $H$ is normal in $G$.

Above part is related the your question.

Some Facts About Charactersitic subgroups

• $H$ is in Char in $K$ and $K$ in Char in $G$ then $H$ is Char in $G$

Proof: If $\sigma\in Aut(G)$ then $\sigma|_K\in Aut(K)$. Thus, $\sigma|_K(H)=H\implies \sigma(H)=H$.

• $H$ is Char in $G$ then image of $H$ need Not to be characteristic.

Example: $G=D_8$ and $H$ be the cylic group of order $4$ and $N=Z(G)$. Then $G/Z(G)\equiv Z_2\times Z_2$ and $H/Z(G)\cong Z_2$.

• $H$ is Char in $G$ and $H\leq K$ such that $K/H$ is Char in $G/H$. Then $K$ is Char in $G$.

Proof: Let $\sigma\in Aut(G)$. Set $\bar \sigma :G/H \to G/H$ by $xH\mapsto \sigma(x)H$. Notice that $\sigma(K)/H =\bar \sigma (K/H)$ but $\bar \sigma \in Aut(G/H)$ so $\bar \sigma (K/H)=K/H$. As a result, $\sigma(K)/H=K/H\implies \sigma(K)H=KH\implies \sigma(K)=K$.

Answer 2

Let $F = F(G).$ Then $C \cap F \leq Z(C)$ as $C = C_{G}(F)$. On the other hand, $Z(C) {\rm char} C \lhd G$, and is certainly nilpotent, so $Z(C) \leq F(G) = F$. Hence $C \cap F = Z(C).$ Any nilpotent normal subgroup $Y$ of $C/Z(C)$ has a nilpotent pre-image in $C$, so its full pre-image $X$ in $C$ is a nilpotent normal subgroup of $C$. If $Y \neq 1$, then $X > C$, so $F(C) > Z(C)$. But $F(C) {\rm char C} \lhd G$, so $F(C) \leq F \cap C \leq Z(C)$, a contradiction. Hence $C/Z(C)$ has no non-identity nilpotent normal subgroup.

(It is also true that if $H {\rm char} G$ and $U/H {\rm char} G/H$, then $U {\rm char} G$).