Finite Engel group is nilpotent.

Question

A group $G$ is said to be $n$ engel if

$$[x,[x, \dots ,[x,y]]\dots ]=1,$$

where $x$ appears $n$ times, and this holds for all $x,y\in G$.

We know there is infinite order engel group which is not nilpotent.

But what can we say about finite order engel groups, are they always nilpotent?

Answer 1

So that this question does not remain listed as unanswered, and for the benefit of any future readers, I will provide an answer. This follows the suggestions in the comments of Nicky Hekster, to look at the exposition in Derek J.S. Robinson - A Course in the Theory of Groups.

In Robinson this result is

Theorem 12.3.4 (Zorn) A finite Engel group is nilpotent.

To prove it, we need to recall the following: (For the benefit of the reader I shall include the same numbering as in Robinson so that the proofs for these may be looked up if you wish).

Notation and preliminaries: $L(G)$ is the set of left Engel elements of $G$. If $G$ is an Engel group then $G=L(G)$. The Hirsch-Plotkin radical of $G$ is the unique maximal normal locally nilpotent subgroup, I will denote this as $HP(G)$. The Fitting subgroup is the subgroup generated by all the normal nilpotent subgroups of a group $G$ and is written $Fit(G)$. Note that if $G$ is finite then $Fit(G)$ is nilpotent, and in particular in the case $G$ is finite, if $G=Fit(G)$ we can conclude $G$ is nilpotent.

We also need the following two Theorems:

Theorem 9.1.9 (O.J Schmidt) Assume that every maximal subgroup of a finite group $G$ is nilpotent but $G$ itself is not nilpotent. Then $G$ is soluble. _{^{(Note that the theorem says more than this but we do not need the rest)}}.

and

Theorem 12.3.3 (Gruenberg). Let $G$ be a soluble group. Then $L(G)$ coincides with the Hirsch-Plotkin radical. _{^{(Note that the theorem says more than this but we do not need the rest)}}

We can now prove the result.

Proof of Theorem 12.3.4

Suppose that this is false. Then let $G$ be a minimal counter example, that is, $G$ is a finite Engel group with smallest order subject to being not nilpotent. Thus every proper subgroup of $G$ must be nilpotent. Thus $G$ is soluble by the Theorem of Schmidt. Then by the Theorem of Gruenberg above, we know that if $G$ is soluble then $L(G)$ coincides with the Hirsch-Plotkin radical. Since $G$ is finite, the Hirsch-Plotkin radical of $G$ coincides with the Fitting subgroup of $G$, and since $G$ is Engel we know that $G=L(G)$. Thus we have $$G=L(G)=HP(G)=Fit(G).$$ Thus $G$ must be nilpotent. $\square$