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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.

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The converse is not true.

As a subnormal abelian subgroup belongs to the locally nilpotent radical it would follow that a bounded left Engel element always belongs to the locally nilpotent radical. It is however well known that the latter is not the case.

See  'Left 3-Engel elements in groups of exponent 5' by G. Traustason. There he gives an argument in the introduction that shows that bounded left Engel elements do not have to belong to the locally nilpotent radical.

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