Notations

We use the notation \([x_1,\ldots,x_n]\) defined inductively by

\[ [x_1,\ldots,x_{n+1}] := [[x_1,\ldots,x_n],x_{n+1}]. \]

We also use the notation

\[ [x,y^n] := [\ldots [x,\underbrace{y],y],\ldots,y}_{n \text{ times}}] \]

 

For a group \(G\), the lower central series is given by \(\gamma_1(G) = G\) and \(\gamma_{n+1}(G) = [\gamma_{n}(G),G]\).

Each quotient \(\gamma_i(G) / \gamma_{i+1}(G)\) is abelian, so we may consider the direct sum of abelian groups \((L(G),+)\) given by:

\[ L(G) := \bigoplus_{i<\omega} \gamma_i(G)/\gamma_{i+1}(G) \]

We may extend the commutator \([x,y]\) to all elements of \(L(G)\) so that \((L(G),+,[.,.])\) is a Lie ring.

(Bilinearity: \([x, zy] = [x,y][x,z]^*\) ; Jacobie: \([x,y,z^*]\cdot [z,x,y^*]\cdot [y,z,x^*] = 1\)).

\(L(G)\) nilpotent \(\overset{?}{\implies}\) \(G\) nilpotent

(not clear in general: Tarski's monster)

Consider the Baker-Campbell-Hausdorff formula:

\[ H(x,y) = x + y + \frac{1}{2} [x,y] + \frac{1}{12}[x,x,y] - \frac{1}{12} [y,x,y] + \ldots \]

and its inverse:

\[ h_{1}(x,y) = xy[x,y]^{-\frac{1}{2}}[x,x,y]^{-\frac{1}{12}}[y,x,y]^{\frac{1}{12}}\ldots \] \[ h_{2}(x,y) = [x,y][x,x,y]^{\frac{1}{2}}[y,x,y]^{\frac{1}{2}}\ldots \]

If \(L\) is a Lie algebra over \(\mathbb F _p\) and of nilpotency class \(c<p\), then:

• \(H(x,y)\) is a finite sum
• the coefficients in \(H(x,y)\) make sense in \(\mathbb F _p\)

\(\leadsto\) "\(a\cdot b := H(a,b)\)" turns \(L\) into a group \(G = G_L\) of exponent \(p\) with the same domain as \(L\), same nilpotency class, etc.

Conversely, if \(G\) has exponent \(p\), nilpotency class \(c<p\) then "\(g+h := h_1(g,h)\)" and "\([g,h] := h_2(g,h)\)" give an \(\mathbb F_p\)-Lie algebra structure with the same domain \(L = L_G\).

The operations \(L \leadsto G_L\) and \(G\leadsto L_G\) are inverse of each other.

\(L_G\) nilpotent \(\iff\) \(G\) nilpotent

 

Local Lazard correspondence

In fact, the assumption \(c<p\) in the above Lazard correspondence is overkill.

\[ h_{1}(x,y) = xy[x,y]^{-\frac{1}{2}}[x,x,y]^{-\frac{1}{12}}[y,x,y]^{\frac{1}{12}}\ldots \] \[ h_{2}(x,y) = [x,y][x,x,y]^{\frac{1}{2}}[y,x,y]^{\frac{1}{2}}\ldots \]

If \(G\) is a group of exponent \(p\) in which every \(3\)-generated subgroup is nilpotent of class \(c<p\), then we may define "\(+ = h_1\)" and "\([.,.] = h_2\)" to get a Lie algebra.

Conversely for Lie algebras in which every \(3\)-generated subalgebra is nilpotent of class \(<p\).

This comes from the fact that the axioms of a Lie algebra are to be checked using \(3\) elements.

 

We use that:

• \(2\)-generated subgroups are nilpotent of class \(<p\) (in order to define \(+\) and \([.,.]\))
• \(3\)-generated subgroups are nilpotent of class \(<p\) (in order to check the bilinearity of \([.,.]\) over \(+\), and the Jacobie identity \([x,y,z]+[y,z,x]+[z,x,y] = 0\).)

Envelopping algebra

For each \(a\in L\), the map \(\mathrm{ad_a}: L\to L\) defined by \(\mathrm{ad}_a(x) = [x,a]\) is an element of \(\mathrm{End}(L)\).

Define

\[ A(L) \subseteq \mathrm{End}(L) \]

to be the associative subalgebra of \(\mathrm{End}(L)\) generated by \(\mathrm{ad}(L)\).

We call \(A(L)\) the envelopping associative algebra of \(L\).

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Global nilpotency

Globale nilpotency is known asymptotically:

For each \(n\), there exists \(N\) such that locally nilpotent \(n\)-Engel \(p\)-groups with \(p>N\) are (globally) nilpotent.

but false in general... ("bad primes" always exist!):

For all \(p\geq 5\) and \(n = p-2\) there exists a non-solvable \(n\)-Engel group of exponent \(p\).

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For small \(n\)'s

Here is what is known for \(n\leq 5\):

\(p=\)	\(2\)	\(3\)	\(5\)	\(7\)	\(>7\)
\(2\)-Engel	\(2\)	\(3\)	\(2\)	\(2\)	\(2\)
\(3\)-Engel	\(\infty\) \(\checkmark\)	\(4\)	\(\infty\) \(\checkmark\)	\(4\)	\(4\)
\(4\)-Engel	\(\infty\)	\(\infty\) \(\checkmark\)	\(\infty\) \(\checkmark\)	\(7\)	\(7\)
\(5\)-Engel	\(\infty\)	\(\infty\)	\(\infty\) \(\checkmark\)	\(\infty\)	\(10\)

Burnside 1902, Hopkins 1929, Heineken 1961, Traustason 1995, Havas & Vaughan-Lee 2005, Vaughan-Lee 2024 (!).

\(\omega\)-categorical \(5\)-Engel \(5\)-groups are nilpotent.

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Global nilpotency

Globale nilpotency is known asymptotically:

For each \(n\), there exists \(N\) such that \(n\)-Engel Lie algebras over \(\mathbb{F}_p\) with \(p>N\) are (globally) nilpotent.

but false in general... ("bad primes" always exist!):

For all \(p\geq 5\) and \(n = p-2\) there exists a non-solvable \(n\)-Engel Lie algebra over \(\mathbb F_p\).

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For small \(n\)'s

Here is what is known for \(n\leq 5\):

\(p=\)	\(2\)	\(3\)	\(5\)	\(7\)	\(>7\)
\(2\)-Engel	\(2\)	\(3\)	\(2\)	\(2\)	\(2\)
\(3\)-Engel	\(\infty\)	\(4\)	\(\infty\) \(\checkmark\)	\(4\)	\(4\)
\(4\)-Engel	\(\infty\)	\(\infty\) \(\checkmark\)	\(\infty\)	\(7\)	\(7\)
\(5\)-Engel	\(\infty\)	\(\infty\)	\(\infty\)	\(\infty\)	\(11\)

Higgins 1954, Kostrikin, Putcha 1971, Traustason 1993, Vaughan-Lee 2024 (!).

 

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Global nilpotency

Globale nilpotency is well-known "asymptotically":

For each \(n\) and \(p>n\) associative algebra of nilexponent \(n\) over \(\mathbb{F}_p\) are nilpotent.

(Optimal... there are non-nilpotent associative algebras of nilexponent \(p\) over \(\mathbb F _p\)).

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Exceptional Lazard

There are "exceptional" cases of groups of exponent \(p\) where

• \(2\)-generated subgroups are nilpotent of class \(<p\)
• \(3\)-generated subgroups are nilpotent of class \(\geq p\)

and still, the operations "\(+=h_1\)" and "\([.,.] = h_2\)" do define a Lie algebra structure, for which, defining "\(\cdot = H\)" gives back the group operation we started with. I call those exceptional Lazard.

Using the GAP package anupq (Greg Gamble, Werner Nickel, Eamonn O’Brien) I could identify two cases:

There is an example of a group of exponent \(3\) which is nilpotent of class \(3\) where all \(2\)-generated subgroups are nilpotent of class \(2\) where the Lazard correspondence does not apply.

\(\omega\)-categorical \(3\)-Engel \(5\)-groups are nilpotent.

We use:

Let \(G\) be a locally finite \(n\)-Engel \(p\)-group. with \(p\) odd. If \(r\) is such that \(p^{r-1}<n\leq p^r\), then \(G^{p^r}\) is nilpotent.

Let \(G\) be such group. Using Wilson's theorem, we may assume that \(G\) is characteristically simple. Using Abdollahi-Traustason (\(r = 1\)), we may assume that \(G\) has exponent \(5\). However, we know that \(3\)-Engel groups of exponent \(5\) and \(3\)-Engel Lie algebras over \(\mathbb{F}_5\) are in (exceptional) Lazard correspondence, so \(G\) is nilpotent.

Let \(G\) be a \(4\)-Engel group of exponent \(5\). Then \(G\) is a subdirect product of center-by-\(3\)-Engel groups and groups of nilpotency class at most \(10\).

\(\omega\)-categorical \(4\)-Engel \(5\)-groups are nilpotent.

Let \(G\) be such group. Again by Wilson and Abdollahi & Traustason (with \(r = 1\)), we may assume that \(G\) is a characteristically simple \(4\)-Engel group of exponent \(5\). Because \(G\) is characteristically simple, \(\gamma_{11}(G) = 1\) or \(\gamma_{11}(G) = G\). In the former case, we are done hence we may assume that \(G\) is center-by-\(3\)-Engel, i.e. \(G/Z(G)\) is \(3\)-Engel. If \(Z(G) = G\) we are done hence \(Z(G) = 1\) and \(G\) is \(3\)-Engel so we conclude using the previous result.

Using an argument similar as above, we conclude:

\(\omega\)-categorical \(5\)-Engel \(5\)-groups are nilpotent.

For the proof, we use an identity

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Let \(L\) be an \(\omega\)-categorical \(3\)-Engel Lie algebra over \(\mathbb F _5\).

1. By \([x,y,z,z] = [x,z,z,y]\), in particular, as elements of \(A(L)\), \(\mathrm{ad}_a^2\) and \(\mathrm{ad}_b^2\) commute for all \(a,b\in L\).
2. Fix \(a\in L\) and consider the (commutative) associative subalgebra \(B\) of \(A(L)\) generated by all \(\mathrm{ad}_b^2\), for \(b\in L\).
3. Define the quotient \(B_a\) of \(B\) by the relation \(f\sim g\) if and only if \(f(a) = g(a)\) for \(f,g\in B\). As \(B\) is commutative, \(B_a\) is a well-defined commutative associative algebra.

4. \(B_a\) is interpretable in \(L\), hence nilpotent by Cherlin, which implies that there is some \(k\) such that

\[ [a,b_1^2,\ldots,b_k^2] = 0. \]

(by \(\omega\)-categoricity, we may assume that \(k\) uniform.)

5. Let \(I = \mathrm{Span}([a,b^2] \mid a,b\in L)\). As \([a,b^2,c] = [a,c,b^2]\), \(I\) is an ideal of \(L\).
6. Because \(L/I\) is \(2\)-Engel it is nilpotent of class \(2\) hence the product of every \(3\) elements in \(L\) is of the form: \(\sum_i [a_i,b_i^2]\).

7. Let \(c = 3+2(k-1)\), and \(d_1,\ldots, d_c\in L\). We have \([d_1, d_2 ,d_3] = \sum_i [a_i,b_i^2]\). Then

\[ [d_1,\ldots ,d_{c}] = \sum_i[ [a_i,b_i^2],d_4,\ldots ,d_{c}] = \sum_i[ a_i,d_4,\ldots ,d_{c},b_i^2] \]

8. By immediate iteration, the right hand side is a sum of element of the form \([a,b_1^2,\ldots ,b_k^2]\) hence \([d_1,\ldots ,d_{c}] = 0\).

A more refined result

(Same for \(\omega\)-categorical \(3\)-Engel groups of exponent \(5\), using exceptional Lazard).

The proof is harder, it is a direct encoding of the argument of Cherlin, using a further quite miraculous identity which holds in \(3\)-Engel Lie algebras of characteristic \(5\):

Compare this to Cherlin's original result.

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