Is the following claim correct in general?

Question

Modified Fitting Lemma:

Claim : If $G$ is a finite group and $f$ is an endomorphism of $G$, then for some $n \ge 1$, $G = \operatorname{ Ker } (f^{n}) \times \operatorname{ Im } (f^{n})$

Where $\text{ Ker } f^n$ means the kernel of $f^{n}$.

One Example which I have tried is identity map assume $n=1$ kernel will be trivial so $G = \operatorname { Ker } f \times \operatorname{ Im } f$.

Is the claim is true can anyone explain how? If not give an counter example.

Answer 1

$\renewcommand{\phi}{\varphi}$I have a version of Fitting's Lemma for non-commutative groups (which is contained in the proof of Theorem 4.1 of this 1985 paper, see also Theorem 4.2 of this 2013 paper), but you all you get in general is a semidirect product, the image of $f^{n}$ being not necessarily normal, as shown by Derek Holt's $S_{3}$ example.

The version from the 2013 paper sounds

Let $G$ be a group which satisfies the ascending chain condition on normal subgroups, and the descending chain condition on subgroups. (Of course a finite group satisfies both.)

Let $\phi$ be an endomorphism of $G$. Then there is a natural number $n$ such that $\ker(\phi^{n}) = \ker(\phi^{n+i})$ and $G^{\phi^{n}} = G^{\phi^{n+i}}$ for each $i \ge 0$. We have:

1. $G$ is the semidirect product of the normal subgroup $K = \ker(\phi^{n})$ by the subgroup $H = G^{\phi^{n}}$,
2. the restriction of $\phi$ to $K$ is nilpotent, and
3. the restriction of $\phi$ to $H$ is an automorphism.