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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. $c(H)\geq c(G) - 1$.
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  • $\begingroup$ 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 $c(H)=1$. $\endgroup$ Sep 20, 2023 at 10:10
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    $\begingroup$ Ah, you're asking whether we have such $H$ for any $G$, got it. $\endgroup$ Sep 20, 2023 at 10:14
  • $\begingroup$ @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 $n$. $\endgroup$
    – YCor
    Sep 20, 2023 at 10:14

1 Answer 1

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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.

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  • $\begingroup$ 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? $\endgroup$
    – mme
    Sep 20, 2023 at 22:33
  • $\begingroup$ @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. $\endgroup$
    – YCor
    Sep 21, 2023 at 6:05
  • $\begingroup$ I see, I agree the desired identity holds then. Thanks for the nice answer! $\endgroup$
    – mme
    Sep 21, 2023 at 11:20

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