5
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Actually, I was studying about the Engel groups and this question arises that weather finite $n$ Engel groups are nilpotent or not. $\endgroup$ – MANI Jan 3 at 10:00
  • 1
    $\begingroup$ The answer is YES (Zorn, 1936), see groupsstandrews.org/2009/Talks/Traustason.pdf $\endgroup$ – Nicky Hekster Jan 3 at 10:09
  • $\begingroup$ #Nicky Hekster thanks for the reply, $groupsstandrews.org/2009/Talks/Traustason.pdf$ here only statement is given please tell me that how to prove this result. $\endgroup$ – MANI Jan 3 at 10:14
  • $\begingroup$ You can find a proof in the book of Derek J.S. Robinson - A Course in the Theory of Groups. The proof runs as follows: take a minimal counterexample, then every proper subgroup is nilpotent, whence solvable by a famous Theorem of Schmidt (also to be found in the book of Robinson). Then the Hirsch-Plotkin radical (the unique maximal local nilpotent subgroup) equals $G$, which means that $G$ is nilpotent, since it is finite. Please look up the details in the book. $\endgroup$ – Nicky Hekster Jan 3 at 11:01
  • 3
    $\begingroup$ This looks like a reasonable request about the state of knowledge concerning a research topic rather than a "here is my problem solve it for me" question, so I would not vote for closing it for lack of context. $\endgroup$ – Derek Holt Jan 3 at 11:40
5
$\begingroup$

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. $\square$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.