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Let $A$ be a free associative ${\mathbb Z}$-algebra on generators $I = \{ x_i ~:~ I \in {\mathbb N} \}$. Setting the usual bracket $[a,b] = ab-ba$, it forms a free Lie ring $A^{(-)}$. The Dynkin operator $\delta : A \rightarrow A^{(-)}$ on associative Lie monomials in the free generating set $I$ is defined as $\delta (x_{i_1} x_{i_2} \dotsc x_{i_m}) = [\dotsc [x_{i_1}, x_{i_2}], \dotsc, x_{i_m}]$ (left normed commutator). I am trying to understand a proof of the following lemma in book by E.I.Khukhro ($p$-automorphisms of finite $p$-groups) stating : $\delta(x_{k+1}[x_1, \dotsc, x_k]) = [x_{k+1}, [x_1, \dotsc, x_k]]$.

The step while we prove $\delta (x_{k+1} x_k [x_1, \dotsc, x_{k-1}]) = [[x_{k+1}, x_k], [x_1, \dotsc, x_{k-1}]]$, the argument says "we replace $x_{k+1}x_k$ by the commutator $[x_{k+1}, x_k]$ and then regard this commutator as a new variable, which implies this step.

The replacement seems to change the definition of Dynkin operator by changing the variables. It is unclear how to derive this step. On another thought, a direct approach seems to look like a modified version of the lemma, which require another proof.

Is there any general formula for $\delta(x_{k+l} \dotsc x_{k+1}[x_1, \dotsc, x_k])$ with a proof via induction on $k+l$?

Let $A$ be a free associative ${\mathbb Z}$-algebra on generators $I = \{ x_i ~:~ I \in {\mathbb N} \}$. Setting the usual bracket $[a,b] = ab-ba$, it forms a free Lie ring $A^{(-)}$. The Dynkin operator $\delta : A \rightarrow A^{(-)}$ on associative monomials in the free generating set $I$ is defined as $\delta (x_{i_1} x_{i_2} \dotsc x_{i_m}) = [\dotsc [x_{i_1}, x_{i_2}], \dotsc, x_{i_m}]$ (left normed commutator). I am trying to understand a proof of the following lemma in book by E.I.Khukhro ($p$-automorphisms of finite $p$-groups) stating : $\delta(x_{k+1}[x_1, \dotsc, x_k]) = [x_{k+1}, [x_1, \dotsc, x_k]]$.

The step while we prove $\delta (x_{k+1} x_k [x_1, \dotsc, x_{k-1}]) = [[x_{k+1}, x_k], [x_1, \dotsc, x_{k-1}]]$, the argument says "we replace $x_{k+1}x_k$ by the commutator $[x_{k+1}, x_k]$ and then regard this commutator as a new variable, which implies this step.

The replacement seems to change the definition of Dynkin operator by changing the variables. It is unclear how to derive this step. On another thought, a direct approach seems to look like a modified version of the lemma, which require another proof.

Is there any general formula for $\delta(x_{k+l} \dotsc x_{k+1}[x_1, \dotsc, x_k])$ with a proof via induction on $k+l$?

The Dynkin operator on associative Lie monomials in the free generators $\{ x_1, x_2, \dotsc \}$ generating a Lie ring $L$ is defined as $\delta (x_{i_1} x_{i_2} \dotsc x_{i_m}) = [\dotsc [x_{i_1}, x_{i_2}], \dotsc, x_{i_m}]$ (left normed commutator). I am trying to understand a proof of the following lemma in book by E.I.Khukhro ($p$-automorphisms of finite $p$-groups) stating : $\delta(x_{k+1}[x_1, \dotsc, x_k]) = [x_{k+1}, [x_1, \dotsc, x_k]]$.

The step while we prove $\delta (x_{k+1} x_k [x_1, \dotsc, x_{k-1}]) = [[x_{k+1}, x_k], [x_1, \dotsc, x_{k-1}]]$, the argument says "we replace $x_{k+1}x_k$ by the commutator $[x_{k+1}, x_k]$ and then regard this commutator as a new variable, which implies this step.

The replacement seems to change the definition of Dynkin operator by changing the variables. It is unclear how to derive this step. On another thought, a direct approach seems to look like a modified version of the lemma, which require another proof.

Is there any general formula for $\delta(x_{k+l} \dotsc x_{k+1}[x_1, \dotsc, x_k])$ with a proof via induction on $k+l$?

Let $A$ be a free associative ${\mathbb Z}$-algebra on generators $I = \{ x_i ~:~ I \in {\mathbb N} \}$. Setting the usual bracket $[a,b] = ab-ba$, it forms a free Lie ring $A^{(-)}$. The Dynkin operator $\delta : A \rightarrow A^{(-)}$ on associative Lie monomials in the free generating set $I$ is defined as $\delta (x_{i_1} x_{i_2} \dotsc x_{i_m}) = [\dotsc [x_{i_1}, x_{i_2}], \dotsc, x_{i_m}]$ (left normed commutator). I am trying to understand a proof of the following lemma in book by E.I.Khukhro ($p$-automorphisms of finite $p$-groups) stating : $\delta(x_{k+1}[x_1, \dotsc, x_k]) = [x_{k+1}, [x_1, \dotsc, x_k]]$.

The step while we prove $\delta (x_{k+1} x_k [x_1, \dotsc, x_{k-1}]) = [[x_{k+1}, x_k], [x_1, \dotsc, x_{k-1}]]$, the argument says "we replace $x_{k+1}x_k$ by the commutator $[x_{k+1}, x_k]$ and then regard this commutator as a new variable, which implies this step.

The replacement seems to change the definition of Dynkin operator by changing the variables. It is unclear how to derive this step. On another thought, a direct approach seems to look like a modified version of the lemma, which require another proof.

Is there any general formula for $\delta(x_{k+l} \dotsc x_{k+1}[x_1, \dotsc, x_k])$ with a proof via induction on $k+l$?