# Is it always true that $g$ is bounded left-Engel element iff it belongs to an abelian subnormal subgroup?

Suppose $$G$$ is a group. Let’s call $$g \in G$$ a bounded left-Engel element iff $$\exists n \in \mathbb{N} \forall h \in G [h, g]_n = e$$. Here $$[h, g]_n$$ is defined by recurrence:

$$[h, g]_n = \begin{cases} [[h, g]_{n-1}, g] & \quad g > 0 \\ h & \quad g = 0 \end{cases}$$

Is it always true that $$g$$ is bounded left-Engel element iff it belongs to an abelian subnormal subgroup?

Suppose $$g$$ belongs to an abelian subgroup $$H$$, which is subnormal in $$G$$ of length $$n$$ (We call $$H < G$$ subnormal of length $$n$$, iff $$\exists \{H_k\}_{k = 0}^n$$ such that $$H_0 = H$$, $$H_n = G$$ and $$\forall 0 < k < n-1 H_n \triangleleft H_{n+1}$$)

Now we will prove, that $$g$$ is a bounded left-Engel element by induction:

Base: If $$n = 0$$, then $$H = G$$ is abelian and the statement is trivially true for any element.

Step: Suppose, it is true for $$n-1$$. Suppose $$H$$ is an abelian subgroup, which is subnormal in $$G$$ of length $$n$$, and $$g \in H$$. Then there exists $$K \triangleleft G$$, such that $$H$$ is subnormal in $$K$$ of length $$n - 1$$. So, by the supposition of induction $$\exists k \in \mathbb{N} \forall h \in K [h, g]_k = e$$. Now it is sufficient to prove, that $$\forall h \in G [h, g] = (hgh^{-1})g^{-1} \in K$$, which is rather obvious.

And so we have proved, that every element of an abelian subnormal subgroup is bounded left-Engel.

However, the inverse statement seems to be more difficult, and I do not know how to prove it. I have tried to construct the corresponding subnormal series here: $C_G^n(g) \triangleleft C_G^{n + 1}$?, but, according to what was said in the comments, the series constructed that way are not always subnormal.