On Lie rings of torsion groups

Abstract

We prove that the Lie ring associated to the lower central series of a finitely generated residually-p torsion group is graded nil.

1 Introduction

Let G be a group. A descending sequence of normal subgroups $G = G_1> G_2 > \cdots$ is called a central series if $[G_i, G_j] \subseteq G_{i+j}$ for all $i,j \ge 1$. The direct sum of abelian groups $L(G) = \oplus _{i \ge 1} G_i/G_{i+1}$ is a graded Lie ring with Lie bracket $[a_iG_{i+1}, b_jG_{j+1}] = [a_i, b_j]G_{i+j+1}$; $a_i \in G_i$, $b_j \in G_j$.

Of particular interest are the lower central series: $G_1 =G$, $G_{i+1} = [G_i, G]$, $i \ge 1$, and, for a fixed prime number p, the Zassenhaus series (see [7, 8]).

Let p be a prime number. We say that a group G is residually-p if the intersection of all normal subgroups of indices $p^i$, $i \ge 1$, is trivial.

A graded Lie ring $L = L_1 + L_2 + \cdots$ is called graded nil if for an arbitrary homogeneous element $a \in L_i$ the adjoint operator $ad(a): x \rightarrow [a,x]$ is nilpotent.

The main instrument in the study of the Burnside problem in the class of residually p-groups is the connection between torsion in the group G and graded nilness in the Lie algebra L(G) of the Zassenhaus series (see [6–9]).

In this paper we prove this connection for the lower central series and for an arbitrary central series of G.

Theorem 1

Let $\Gamma$ be a finitely generated residually-p torsion group. Let $\Gamma = \Gamma _1> \Gamma _2 > \cdots$ be the lower central series of $\Gamma$. Then the Lie ring $L(\Gamma ) = \oplus _{ i\ge 1} \Gamma _i / \Gamma _{i+1}$ is graded nil.

Note that the known important classes of torsion groups: Golod–Shafarevich groups (see [2, 3]), Grigorchuck groups [4] and Gupta–Sidki groups [5] are residually—p.

We say that a (possibly infinite) group $\Gamma$ is a p-group if for an arbitrary element $g \in \Gamma$ there exists $k \ge 1$ such that $g^{p^k} = 1$. Clearly, for a residually p-group being torsion and being a p-group are equivalent.

Theorem 2

Let $\Gamma$ be a p-group. Let $\Gamma = \Gamma _1> \Gamma _2 > \cdots$ be a central series. Then the Lie ring $L(\Gamma ) = \oplus _{i \ge 1} \Gamma _i / \Gamma _{i+1}$ is locally graded nil.

In other words, we claim that an arbitrary finitely generated graded subalgebra of $L(\Gamma )$ is graded nil.

2 Definitions and results

Let $\Gamma$ be a residually-p group and let G be its pro-p completion (see [1]). For an element $y \in G$ let $\langle y^G \rangle$ denote the closed normal subgroup of G generated by y. Let $[y^G, y^G]$ denote the closed commutator subgroup of $\langle y^G \rangle$.

For elements $g_1, g_2, \ldots , g_n \in G$ let $[g_1, g_2, \ldots , g_n] = [g_1, [g_2,[ \ldots , g_n]] \ldots ]$ be their left-normed commutator.

We will need the following equalities which can be found in [1]:

• (1) For an arbitrary integer $k \ge 1$ we have $[y,x]^k = [y^k,x]$ mod $[y^G, y^G]$;

• (2) $[y,x^k] = [y,x]^{{k \atopwithdelims ()1}} [y,x,x]^{{k \atopwithdelims ()2}} \ldots [y,\underbrace{x, \ldots x}_k]^{{k \atopwithdelims ()k}}$ mod $[y^G, y^G]$.

Let p be a prime number. The equalities (1) and (2) imply that

• (3) $c = [y,x^p][y,\underbrace{x, \ldots , x}_p]^{-1} = [y^{{p \atopwithdelims ()1}}, x][y^{{p \atopwithdelims ()2}},x,x] \ldots [y^{{p \atopwithdelims ()p-1}}, \underbrace{x, \ldots x}_{p-1}]$ mod $[y^G, y^G]$.

Hence $[y^p,x] = c [[y^p,x],x]^{- {p \atopwithdelims ()2}/p} \ldots [[y^p,x],\underbrace{x, \ldots x}_{p-2}]^{- {p \atopwithdelims ()p-1}/p}$ mod $[y^G, y^G]$.

Iterating this process and the use of equalities (1) and (2) we conclude that there exists an infinite sequence of nonnegative integers $k_i \ge 0$ such that

• (4) $[y^p,x ] = c [c,x]^{k_1} [c,x,x]^{k_2} \ldots$ mod $[y^G, y^G]$.

Let $\rho$ be a left-normed group commutator, $\rho = [g_1, \ldots g_m]$, where each element $g_i$ is either equal to y or to $x^{p^k}$ for some $k \ge 1$. Let $d_x(\rho )$ denote the sum of powers of x involved in $\rho$ and $d_y(\rho )$ the number of times the element y occurs in $\rho$.

Lemma 1

Let $\rho = [y,x^{p^{i_1}},x^{p^{i_2}}, \ldots , x^{p^{i_k}}]$, $d_y(\rho ) = 1$, $0 \le i_1,\ldots ,i_k \le l-1$. Then $\rho$ is a converging product of commutators $\sigma$ of types:

• (i) $\sigma = [y,x^{p^{j_1}},x^{p^{j_2}}, \ldots , x^{p^{j_s}}]$, where $d_y(\sigma ) = 1$, no more than one integer of $j_1, \ldots , j_s$ is different from 0 and $d_x(\sigma ) \ge d_x(\rho )$;

• (ii) $\sigma = [y, x^{p^{j_1}}, \ldots , y, \ldots ]$, where $d_y(\sigma ) \ge 2$ and $(d_y(\sigma )-1)p^l + d_x(\sigma ) \ge d_x(\rho )$.

Proof

Suppose that $i_{\alpha }, i_{\beta } \ge 1$, $1 \le \alpha \ne \beta \le k$. We will represent $\rho$ as a product of commutators of types (i) and (ii) and of commutators of type

• (iii) $\rho ' = [y,x^{p^{j_1}},x^{p^{j_2}}, \ldots , x^{p^{j_s}}]$, where $0 \le j_1, \ldots , j_s \le l - 1$ and $0 \le d_x(\rho ') > d_x(\rho )$

and type

• (iv) $\rho '' = [y,x^{p^{j_1}},x^{p^{j_2}}, \ldots , x^{p^{j_s}}]$, where $0 \le j_1, \ldots , j_s \le l - 1$ and $d_x(\rho '') = d_x(\rho )$, $s >k$.

Iterating we will get rid of commutators (iii) and (iv).

Without loss of generality we will assume that $\alpha = 1$, $\beta = 2$ and $1 \le i_1 \le i_2$.

By (3) we have

$[y,x^{p^{i_1}}] = [y^{{p \atopwithdelims ()1}}, x^{p^{i_1-1}}] \ldots [y^{{p \atopwithdelims ()p-1}}, \underbrace{x^{p^{i_1-1}}, \ldots , x^{p^{i_1-1}}}_{p-1}] [y,\underbrace{x^{p^{i_1-1}}, \ldots , x^{p^{i_1-1}}}_{p}]$

mod $[y^{G},y^{G}]$.

Now $\rho$ is a product of commutators of the form

$[y^{{p \atopwithdelims ()t}}, \underbrace{x^{p^{i_1-1}}, \ldots , x^{p^{i_1-1}}}_t, x^{p^{i_2}}, \ldots , x^{p^{i_k}}]$, $1 \le t \le p-1$; of the commutator $[y, \underbrace{x^{p^{i_1-1}}, \ldots , x^{p^{i_1-1}}}_p, x^{p^{i_2}}, \ldots , x^{p^{i_k}}]$ and of commutators that involve at least two occurrences of y and the powers $x^{p^{i_1-1}},x^{p^{i_2}}, \ldots , x^{p^{i_k}}$ (see [7]).

The latter commutators are commutators of type (ii). The commutator $[y, \underbrace{x^{p^{i_1-1}}, \ldots , x^{p^{i_1-1}}}_p, x^{p^{i_2}}, \ldots , x^{p^{i_k}}]$ satisfies condition (iv). Hence it remains to consider commutators of the form $[y^{{p \atopwithdelims ()t}}, \underbrace{x^{p^{i_1-1}}, \ldots , x^{p^{i_1-1}}}_t, x^{p^{i_2}}, \ldots , x^{p^{i_k}}]$.

Since $p | {p \atopwithdelims ()t}$, $1 \le t \le p-1$, we will consider the commutator

$$\begin{aligned}{}[y^p, \underbrace{x^{p^{i_1-1}}, \ldots , x^{p^{i_1-1}}}_t, x^{p^{i_2}}, \ldots , x^{p^{i_k}}]. \end{aligned}$$

Modulo longer commutators we can move the power $x^{p^{i_2}}$ to the left. By (4) we get

$$\begin{aligned}{}[y^p, x^{p^{i_2}}] = \sigma [\sigma ,x]^{s_1} [\sigma , x,x]^{s_2} \ldots \end{aligned}$$

where $\sigma = [y, x^{p^{i_2+1}}][y, \underbrace{x^{p^{i_2}}, \ldots , x^{p^{i_2}}}_p]^{-1}$ mod $[y^{G},y^{G}]$.

The commutators $[y, x^{p^{i_2+1}}, \underbrace{x^{p^{i_1-1}}, \ldots , x^{p^{i_1-1}}}_t, \ldots x^{p^{i_k}}]$ and

$[y,\underbrace{ x^{p^{i_2}}, \ldots , x^{p^{i_2}}}_p, \underbrace{x^{p^{i_1-1}}, \ldots , x^{p^{i_1-1}}}_t, \ldots x^{p^{i_k}}]$ are of type (iii) since $i_2 \ge i_1$ and therefore $p^{i_2+1} + p^{i_1 -1} > p^{i_2} + p^{i_1}.$

This finishes the proof of the lemma.

Consider again the commutator $\rho = [y, x^{p^{i_1}}, \ldots , x^{p^{i_k}}]$. Suppose that $x^{p^l} = 1.$ Consider the l-tuple $ind(\rho ) =(k_{l-1}, \ldots , k_0)$, $k_i \in { Z}_{\ge 0}$, where $k_i$ is the number of times i occurs among $i_1, \ldots , i_k$. Clearly, $k_0 + k_1 + \cdots k_{l-1} = k$.

Consider the length-lex order in ${Z^l}_{\ge 0}$: $(\alpha _1, \ldots , \alpha _l) > (\beta _1, \ldots , \beta _l)$ if either $\sum \alpha _i > \sum \beta _i$ or $\sum \alpha _i = \sum \beta _i$ and $(\alpha _1, \ldots , \alpha _l) > (\beta _1, \ldots , \beta _l)$ lexicographically.

Lemma 2

Let $x,y \in G$, $x^{p^l} = 1$, $y^{p^s} = 1$. A commutator $\rho = [y, x^{p^{i_1}}, \ldots , x^{p^{i_k}}]$ such that $d_x(\rho ) \ge (s+1)p^l$ can be represented as a product of commutators $\sigma = [y, x^{p^{j_1}}, \ldots , y, \ldots , x^{p^{j_q}}]$, where $d_y(\sigma ) \ge 2$ and $(d_y(\sigma )-1)p^l + d_x(\sigma ) \ge d_x(\rho )$.

Proof

We will show that $\rho$ is a (converging) product of commutators of the form $\sigma _1$ and $\sigma _2$, where $d_y(\sigma _1) \ge 2$, $(d_y(\sigma _1)-1)p^l + d_x(\sigma _1) \ge d_x(\rho )$ for commutators of the form $\sigma _1$, whereas commutators of the form $\sigma _2$ look as $\sigma _2 =[y,x^{p^{j_1}}, \ldots , x^{p^{j_t}}]$ with $d_x(\sigma _2) > d_x(\rho )$ or $d_x(\sigma _2) = d_x(\rho )$ and $ind(\sigma _2) > ind(\rho )$. Then, applying this assertion to commutators of the form $\sigma _2$ and iterating we will get rid of commutators $\sigma _2$.

We claim that at least one i, $0 \le i \le l-1$, occurs in $i_1, \ldots , i_k$ not less than p times. Indeed, otherwise $d_x(\rho ) \le (p-1)(1+p+ \cdots + p^{l-1})$, which contradicts our assumption that $d_x(\rho ) \ge (s+1)p^l$.

Suppose that i occurs in $i_1, \ldots , i_k$ not less than p times and i is the smallest in $\{i_1, \ldots , i_k\}$ with this property. Moving the occurrences of i to the left, modulo longer commutators, we assume $i_1 = \cdots = i_p = i$.

By (2) we have

$$\begin{aligned}{}[y,\underbrace{ x^{p^i}, \ldots , x^{p^i}}_p ] = [y,x^{p^{i+1}}][y^{{p \atopwithdelims ()1}}, x^{p^i}]^{-1} \ldots [y^{{p \atopwithdelims ()p-1}},\underbrace{x^{p^i}, \ldots , x^{p^i}}_{p-1}]^{-1} \tau _1 \ldots \tau _q, \end{aligned}$$

where $\tau _j$ are commutators that involve y at least twice.

The commutator $\sigma ' = [y, x^{p^{i+1}}, x^{p^{i_{p+1}}}, \ldots , x^{p^{i_k}}]$ has greater index than $\rho$. Indeed, $d_x(\sigma ') = d_x(\rho )$, but $ind(\sigma ')$ is lexicographically greater than $ind(\rho )$.

For a commutator $\tau '_j = [\tau _j, x^{p^{i_{p+1}}}, \ldots , x^{p^{i_k}}]$, we have

$$\begin{aligned} d_x(\tau '_j) \ge d_x(\rho ) - (p-1)p^i. \end{aligned}$$

Hence, $p^l(d_y(\tau '_j)-1) + d_x(\tau '_j) \ge p^l + d_x(\rho ) - (p-1)p^i > d_x(\rho )$.

Consider now the commutator $\rho ' =[y^p, x^{p^i}, x^{p^{i_p+1}}, \ldots , x^{p^{i_k}}]$.

We claim that there exists $j \in \{i_{p+1}, \ldots , i_k\}$ such that $j \ge i$. Indeed, otherwise all integers in $\{i_{p+1}, \ldots , i_k\}$ are smaller than i and therefore occur $\le (p-1)$ times. Hence,

$$\begin{aligned} d_x(\rho ) \le p p^i + (p-1)(1+p+ \cdots +p^{i-1}) = p^{i+1} + p^i - 1 < 2p^l \le (s+1)p^l, \end{aligned}$$

which contradicts the assumption of the lemma.

Moving $x^{p^j}$ to the right end in $\rho '$ modulo longer commutators we will assume that $i_k = j \ge i$.

Consider the commutator $\rho '' = [y^p, x^{p^i}, x^{p^{i_{p+1}}}, \ldots , x^{p^{i_{k+1}}}]$. We have $d_x(\rho '') = d_x(\rho ) - (p-1)p^i - p^j \ge sp^l$. By the induction assumption on s the commutator $\rho ''$ is a product of commutators w in $y^p$ and x, each commutator involves $\mu = \mu (w) \ge 2$ elements $y^p$ and $(\mu - 1)p^l + d_x(w) \ge d_x(\rho '')$. We will assume that $w = [w_1, \ldots ,w_{\mu }]$, $w_j = [y^p, \ldots ]$, $1 \le j \le \mu$.

Remark

Any commutator that has degree $\ge \mu +1$ in y and degree $\ge d_x(w)$ in x fits the requirements of the lemma since $\mu p^l + d_x(w) \ge d_x(\rho '') + p^l \ge d_x(\rho )$.

The commutator $[w_1, \ldots , w_{\mu },x^{p^j}]$ is equal to a product

$$\begin{aligned}{}[[w_1, x^{p^j}],w_2, \ldots , w_{\mu }][w_1,[w_2,x^{p^j}],\ldots ] \ldots [w_1, \ldots ,[w_{\mu },x^{p^j}]] \end{aligned}$$

modulo longer commutators (see the Remark above).

Consider $[w_1, \ldots ,[w_{\nu },x^{p^j}],\ldots , w_{\mu }]$.

In $[w_{\nu }, x^{p^j}]$ move $x^{p^j}$ to the left position next to $y^p$ modulo longer commutators (see the Remark above).

By (4), $[y^p, x^{p^j}] = c[c,x]^{k_1}[c,x,x]^{k_2} \ldots \tau _1 \ldots \tau _q$, where $c = [y,x^{p^{j+1}}][y, \underbrace{x^{p^j}, \ldots ,x^{p^j}}_p]^{-1}$; $\tau _1, \ldots \tau _q \in [y^{G},y^{G}]$; $d_x(\tau _1), \ldots , d_x(\tau _q) \ge p^j.$

If the commutator $[y^p,x^{p^j}]$ is replaced by one of $\tau _1, \ldots , \tau _q$ then see the Remark.

If $[y^p,x^{p^j}]$ is replaced by c then

$$\begin{aligned}&d_x([w_1, \ldots , w_{\nu - 1},c,w_{\nu +1},\ldots , w_{\mu }]) \ge \nonumber \\&\quad d_x([w_1, \ldots ,w_{\mu }, x^{p^j}]) + (p-1)p^j d_x(w) + p{j+1}. \end{aligned}$$

Hence, $(\mu -1)p^l + d_x([w_1, \ldots , w_{\mu }, x^{p^j}]) \ge (\mu - 1)p^l + d_x(\mu ) + p{j+1} \ge$ $d_x(\rho '') + p^{j+1} = d_x(\rho ) - (p-1)p^i - p^j + p^{j+1} = d_x(\rho ) + (p-1)(p^j - p^i)\ge d_x(\rho )$.

Since $d_y([w_1, \ldots ,w_{\mu }, x^{p^j}]) \ge 2$, this commutator satisfies the requirements of the lemma. If the commutators $[y^p,x^{p^j}]$ is replaced by $[c, \underbrace{x, \ldots , x}_t]^{k_t}$, then $(\mu - 1)p^l + d_x([w_1, \ldots , w_{\nu -1}, [c, \underbrace{x, \ldots , x}_t]^{k_1}, w_{\nu +1}, \ldots , w_{\mu }]) > d_x(\rho )$.

This finishes the proof of the lemma.

Lemma 3

Let $x \in G_i$, $x^{p^l} = 1$, $y \in G_j$, $y^{p^s} = 1$. Suppose that $j \ge 2ip^l$. Then $(yG_{j+1})ad(x G_{i+1})^{(s+1)p^l} = 0$ in the Lie algebra $L = \sum _{k=1}^{\infty } G_k/G_{k+1}$.

Proof

By Lemma 2 the group commutator $\rho = [y, \underbrace{x, \ldots , x}_{(s+1)p^l}]$ can be represented as a product of commutators $w =[w_1, \ldots , w_{\mu }],$ $\mu \ge 2$, where each $w_k$ is a commutator of the type $w_k = [y,x^{p^{j_1}}, \ldots , x^{p^{j_r}}]$, $(\mu - 1)p^l + d_x(w) \ge d_x(\rho ) = (s+1)p^l$.

By Lemma 1 each $w_k$ is a product of commutators of type (i) or (ii). A commutator of type (ii) just increases the degree in y. Let $[y, x^{p^{j_1}}, \ldots , x^{p^{j_r}}]$ be a commutator of type (i). So all $j_1, \ldots , j_r$, except possibly one, are equal to 0. This implies that

$$\begin{aligned}{}[y, x^{p^{j_1}}, \ldots , x^{p^{j_r}}] \in G_{j+i(p^{j_1} + \cdots + p^{j_r} - (p^{l-1} - 1))}. \end{aligned}$$

Hence, $w \in G_d$, where $d = \mu j + id_x(w) - \mu i(p^{l-1} -1) \ge j + (\mu -1)ip^l + (\mu -1)ip^l + id_x(w) - \mu i(p^{l-1}-1) \ge j + id_x(\rho ) + i[(\mu -1)p^l - \mu (p^{l-1}-1)]$.

Now it remains to notice that $(\mu - 1)p^l - \mu (p^{l-1} -1) >0$. We showed that $d > j + id_x(\rho )$, which implies the lemma.

Lemma 4

The Lie ring $L(\Gamma )$ is weakly graded nil, i.e., for arbitrary homogeneous elements $a,b \in L(\Gamma )$ there exists $n(a,b) \ge 1$ such that $b ad(a)^{n(a,b)} = 0$.

Proof

Let $a \in \Gamma _i$, $a^{p^l} = 1$. Let $n(a) = 2ip^l$. By Lemma 3, for an arbitrary element $b \in \Gamma _j$, $j \ge n(a)$, there exists an integer $n(a,b) \ge 1$ such that $[b,\underbrace{a,a, \ldots ,a}_{n(a,b)}] \in G_{j + in(a,b)+1}$.

Since $\Gamma$ is a torsion group it follows that for an arbitrary $k \ge 1$ the subgroup $\Gamma _k$ has finite index in $\Gamma$, hence $\Gamma _k$ is open in $\Gamma$. The subgroup $G_k$ is the completion of $\Gamma _k$. Hence $\Gamma \cap G_k = \Gamma _k$.

We proved that $b ad(a)^{n(a,b)} = 0$ in $L(\Gamma )$. Now let b be an arbitrary homogeneous element from $L(\Gamma )$. Then the degree of the element $b' = b ad(a)^{n(a)}$ is greater than n(a). Hence, $b ad(a)^{n(a) + n(a,b')} = b'ad(a)^{n(a,b')} = 0$, which finishes the proof of the lemma.

Lemma 5

Let L be a Lie algebra over a field $\mathrm { Z}/ p \mathrm {Z}$ generated by elements $x_1, \ldots , x_m$. Let $a \in L$ be an element such that $x_i ad(a)^{p^k} = 0$, $1 \le i \le m$. Then $L ad(a)^{p^k} = (0)$.

Proof

The algebra L is embeddable in its universal associative enveloping algebra U(L). Let $a^{p^k}$ be the power of the element a in U(L). For an arbitrary element $b \in L$ we have $b ad(a)^{p^k} = [b,a^{p^k}]$. If the element $a^{p^k}$ commutes with all generators $x_1, \ldots , x_m$ then $[L,a^{p^k}] = L ad(a)^{p^k} = (0)$, which finishes the proof of the lemma.

Lemma 6

Let L be a Lie ring generated by elements $x_1, \ldots , x_m$. Suppose that $p^l L = (0)$. Let $a \in L$ be an element such that $x_i ad(a)^{p^k} = 0$, $1 \le i \le m$. Then $L ad(a)^{p^k l} = (0)$.

Proof

By Lemma 5 we have $L ad(a)^{p^k} \subseteq pL$. Hence $L(ad(a)^{p^k})^l \subseteq p^l L= (0)$, which proves the lemma.

Proof of Theorem 1

Let $x_1, \ldots , x_m$ be generators of the group $\Gamma$. Then the elements $x_i\Gamma _2$, $1 \le i \le m$, generate the Lie ring $L(\Gamma )$. Let $p^l$ be the maximum of orders of the elements $x_1, \ldots , x_m$, so $x_i^{p^l} = 1$, $1 \le i \le m$. Then $p^l(x_i \Gamma _2) = 0$ in the Lie ring $L(\Gamma )$. Hence $p^l L(\Gamma ) = (0)$.

Let a be a homogeneous element of $L(\Gamma )$. By Lemma 4 there exists $k \ge 1$ such that $(x_i \Gamma _2)ad(a)^{p^k} = 0$ for $i = 1, \ldots , m$. Now Lemma 6 implies that $L(\Gamma )ad(a)^{p^k \cdot l} = (0)$, which finishes the proof of Theorem 1.

Proof of Theorem 2

Without loss of generality we assume that $\cap _i \Gamma _i = (1)$. We view the subgroups $\{ \Gamma _i | i \ge 1\}$ as a basis of neighborhoods of 1 thus making $\Gamma$ a topological group. Let G be a completion of $\Gamma$ in this topology. Let $G_i$ be the closure of $\Gamma _i$ in G. Then $G_i \cap \Gamma = \Gamma _i$ and $G = G_1> G_2 > \cdots$ is a central series of the group G. Arguing as in Lemmas 3, 4 we conclude that the Lie ring $L(\Gamma ) = \oplus _{i \ge 1} \Gamma _i/\Gamma _{i+1}$ is weakly graded nil. Choose homogeneous elements $a_1, \ldots , a_m \in L(\Gamma )$. Since $\Gamma$ is a p-group it follows that there exists $l \ge 1$ such that ${p^l} a_i = 0$, $1 \le i \le m$. Consider the subring $L'$ of $L(\Gamma )$ generated by $a_1, \ldots , a_m$, $p^l L' = (0)$. If a is a homogeneous element from $L'$ and $a_i ad(a)^{p^k} = 0$, $1 \le i \le m$, then by Lemma 6 we have $L' ad(a)^{p^k \cdot l} = (0)$, which finishes the proof of Theorem 2.