# Is there a subgroup of a non-abelian $p$-group $G$ with a large nilpotency class?

Let $$G$$ be a non-abelian $$p$$-group ($$p\ne2$$). Does there exist a group $$H\subset G$$ such that both 1, 2 are satisfied?

1. $$|H| = |G|/p$$.
2. $$c(H)\geq c(G) - 1$$.
• Maybe I'm misunderstanding the question, but can't you just take abelian $H$ and $G=H\rtimes C_p$ a nontrivial semidirect product? Then $c(G)=2$ and $c(H)=1$. Sep 20, 2023 at 10:10
• Ah, you're asking whether we have such $H$ for any $G$, got it. Sep 20, 2023 at 10:14
• @AchimKrause the question is whether $H$ exists for every $G$. By the way, your assertion is not true: if $H=C_p^n$, then according the action, the semidirect product $C_p^n\rtimes C_p$ can have any class between $1$ and $n$.
– YCor
Sep 20, 2023 at 10:14

Not always. Here is an answer in the realm of Lie algebras, and below I'll make it an answer in the realm of $$p$$-groups (a counterexample of order $$p^{10}$$ and class $$5$$ (and exponent $$p$$) in which every proper subgroup has class $$\le 3$$).

Fix an arbitrary ground field. Consider the vector space of basis $$(e_i)_{1\le i\le 10}$$. Make it a Lie algebra with the brackets (using the shortcut $$i.j|x$$ to mean $$[e_i,e_j]=x=-[e_j,e_i]$$, and other brackets being meant to be zero):

$$\begin{gather*} 1.2|e_3,\qquad 1.3|e_4,\;2.3|e_5,\qquad 1.4|e_6,\;1.5|e_7,\;2.4|e_7,\;2.5|e_8 \\ 1.7|e_9,\;2.6|-2e_9,\;1.8|e_{10},\;2.7|-2e_{10}. \end{gather*}$$

Then this satisfies Jacobi, and also satisfies that $$[x,[x,[x,[x,y]]]]=0$$ for all $$x$$, $$y$$ in the Lie algebra; the latter making use of the choice of $$-2$$ coefficients.

This is also a (Carnot) graded Lie algebra with $$e_1$$, $$e_2$$ of degree $$1$$, $$e_3$$ of degree $$2$$, $$e_4$$, $$e_5$$ of degree $$3$$, $$e_6$$, $$e_7$$, $$e_8$$ of degree $$4$$, and $$e_9$$, $$e_{10}$$ of degree $$5$$.

One then sees that the natural action of $$\mathrm{GL}_2$$ in the degree 1 space extends to the whole Lie algebra.

(Well, this Lie algebra was precisely produced to satisfy this, namely starting with the 2-generator free 5-step-nilpotent Lie algebra, observing that the degree 5 component splits as $$\mathrm{GL}_2$$-module as sum of a 4-dimensional and a 2-dimensional module, and killing the 4-dimensional module. The resulting Lie algebra is the free Lie algebra in the variety of 5-step-nilpotent Lie algebras satisfying the additional law $$[x,[x,[x,[x,y]]]]=0$$.)

Hence, all codimension-1 subalgebras of the above Lie algebra are isomorphic. Hence it is enough to see that a single one, say the one with basis $$(e_i)_{2\le i\le 10}$$ is 3-step nilpotent. Namely, its derived subalgebra has basis $$(e_5,e_7,e_8,e_9,e_{10})$$ and the next step in the central series has basis $$(e_8,e_{10})$$ and is central therein.

Now consider this Lie algebra over $$\mathbf{Z}/p\mathbf{Z}$$ with $$p>5$$. Then Lazard proved that the Baker–Campbell–Hausdorff formula produces a group law for which the lower central series coincides with the Lie-algebraic one, and subgroups also coincide with subalgebras. Thus the resulting group works.

• I assume the second 2.7 should read 2.8. Beyond this, are the defining brackets correct as written? Taking $x = e_1 + e_2$ and $y = e_2$ I seem to find $[x, [x, [x, [x, y]]]] = -2e_9 - e_{10}$. Have I repeatedly made some arithmetic error?
– mme
Sep 20, 2023 at 22:33
• @mme thanks for noticing. Indeed the correction is rather that the first 2.7 should have been 2.6 (now fixed) while 2.8 is zero.
– YCor
Sep 21, 2023 at 6:05
• I see, I agree the desired identity holds then. Thanks for the nice answer!
– mme
Sep 21, 2023 at 11:20