Is $\ker\varphi^2=(\ker\varphi)(\mathop{\rm im}\varphi\cap\ker\varphi^2)$ always true for group endomorphism $\varphi$?

Question

Is $\ker\varphi^2=(\ker\varphi)(\mathop{\rm im}\varphi\cap\ker\varphi^2)$ always true for group endomorphism $\varphi$?

It is trivial that $\ker \varphi^2 \supseteq (\ker \varphi) ( \mathop{\rm im}\varphi \cap \ker\varphi^2 )$, since $\ker\varphi \subseteq \ker \varphi^2$ and $ \mathop{\rm im}\varphi \cap \ker\varphi^2 \subseteq \ker\varphi^2$; and also $ \ker \varphi \cap ( \mathop{\rm im}\varphi \cap \ker\varphi^2) = \{\,e\,\} $, $ (\mathop{\rm im}\varphi \cap \ker\varphi^2) \lhd \ker \varphi^2 $.

But how about the opposite direction?

I thought this was wrong, but I'm not sure: I still haven't found any counterexamples yet.

PS: $\mathop{\rm im} \varphi = \{ \varphi(g) :g\in G \}$ and $\varphi^2$ means $\varphi \circ \varphi$.

Answer 1

I assume that $\text{Im} G$ means $\mathop{\rm im}\varphi$ and that $(\ker\varphi)*(\mathop{\rm im}\varphi \cap \ker\varphi^2) = (\ker\varphi)(\mathop{\rm im}\varphi \cap \ker\varphi^2) = \{gh\in G \mid g\in\ker\varphi,\: h\in \mathop{\rm im}\varphi \cap \ker\varphi^2\}$.

Then, since $\ker\varphi\subset\ker\varphi^2$ and $\ker\varphi^2$ is a subgroup, we get $(\ker\varphi)(\mathop{\rm im}\varphi \cap \ker\varphi^2)\subset (\ker\varphi^2)(\ker\varphi^2) \subset \ker\varphi^2$, so the inclusion "$\supset$" indeed holds.

On the other hand, "$\subset$" is not true in general:
Consider $G = (\mathbb{R}^2,+)$ and $\varphi:\mathbb{R}^2\to\mathbb{R}^2$ given by $e_1\mapsto 0$, $e_2\mapsto e_1$. Then $\ker\varphi^2 = \mathbb{R}^2$ and $\ker\varphi = \mathop{\rm im}\varphi = \mathbb{R}e_1$, and thus $$(\ker\varphi) + (\mathop{\rm im}\varphi \cap \ker\varphi^2) = \mathbb{R}e_1 + \mathbb{R}e_1 = \mathbb{R}e_1 \ne \mathbb{R}^2=\ker\varphi^2. $$

So I wonder if I have misunderstood your question.