Is the Fitting subgroup compatible with surjective homomorphisms?

Question

If $G$ is a group, its Fitting subgroup $F_G$ is generated by all normal nilpotent subgroups of $G$. By the Fitting Lemma, it is equal to the union of the normal nilpotent subgroups. If $G$ is finite then $F_G$ is also nilpotent but that doesn't hold in general.

I have two related questions that I couldn't find an answer for in the literature (maybe because it is too easy), also not by myself since I have almost no knowledge about concrete computations of the Fitting subgroup.

First question: I have read here that we have $\varphi(F_G) = F_G$ for every surjective homomorphism $\varphi : G \to G$. What is a reference for this result? Does it also hold when $G$ is infinite? (The literature is mostly concerned with finite groups, for which the claim is trivial.)

Second question: If $\varphi : G \to H$ is any surjective homomorphism, I have checked $\varphi(F_G) \subseteq F_H$, but I assume $\varphi(F_G) = F_H$ may fail. What is an example? My feeling is that there must be an extreme example, so a group $G$ with $F_G = 1$ but which surjects onto a nilpotent group $H$, so that $F_H = H$. I suspect that it might work when $G$ is a free group of rank $ > 1$, but I am not sure how to compute $F_G$ in that case.

Answer 1

A couple of examples related to the second question.

$S_{3}$ has a Fitting subgroup $A_{3}$ of order $3$, but the quotient $S_{3}/A_{3}$, of order $2$, is its own Fitting subgroup.

For another example, $S_{5}$ has trivial Fitting subgroup, but $S_{5}/A_{5}$, of order $2$, is its own Fitting subgroup.

Answer 2

$\newcommand{\NN}{\mathbb{N}}$Let me try and address the first question. Consider the following (infinite!) group \begin{equation*} G = \prod_{i \in \NN} H_{i} \times \prod_{j \in \NN} C_{i}, \end{equation*} where each $H_{i}$ is isomorphic to $S_{3}$, and each $C_{i}$ is a cyclic group of order $2$. If $K_{i}$ is the subgroup of order $3$ of $H_{i}$, then the Fitting subgroup of $G$ is \begin{equation*} \prod_{i \in \NN} K_{i} \times \prod_{j \in \NN} C_{i}. \end{equation*} Now consider the quotient $G/K_{0}$. This is clearly isomorphic to $G$ itself, as the copy $H_{0}$ of $S_{3}$ is sent to a copy of the cyclic group of order $2$. Let us fix a specific isomorphism $\psi : G \to G/K_{0}$, which sends $H_{i}$ to $H_{i+1}$, $C_{i}$ to $C_{i-1}$, for $i > 0$, and $C_{0}$ to $H_{0}/K_{0}$.

So there is a surjective homomorphism $\varphi : G \to G$ which is the composition of the natural surjective homomorphism $\pi : G \to G/K_{0}$ and $\psi^{-1}$.

Now note that $\pi$ sends the Fitting subgroup of $G$ to \begin{equation*} \prod_{i > 0} K_{i} \times \prod_{j \in \NN} C_{i}, \end{equation*} which is properly contained in the Fitting subgroup of $G/K_{0}$, which is \begin{equation*} \prod_{i > 0} K_{i} \times \color{red}{H_{0}/K_{0}} \times \prod_{j \in \NN} C_{i}. \end{equation*} Since $\psi$ and $\psi^{-1}$ are isomorphisms, and thus preserve the Fitting subgroup, one sees that the image of the Fitting subgroup of $G$ under $\varphi$ is properly contained in the Fitting subgroup of $G$.