Let $L$ be a (non-restricted) Lie algebra over a field of prime characteristic $p,$ $UL$ be its universal enveloping algebra and $a\_1,\dots, a\_p \in L$ (the number of elements is equal to the characteristic). I can prove that the element $\sum\_{\sigma \in S\_p} a\_{\sigma(1)}\cdot{}\dots{}\cdot a\_{\sigma(p)}$ of $UL$ lies in $L.$ But I want to have an evident formula for this element on the language of commutators. 


**Question:** Is it true that 
$$\sum\_{\sigma \in S\_p} a\_{\sigma(1)}\cdot{}\dots{}\cdot a\_{\sigma(p)} = \sum\_{\sigma\in S\_p\ :\ \sigma(1)=1} [a\_{1},a\_{\sigma(2)},\dots,a\_{\sigma(p)}]$$
?


Here $S\_p$ denotes the symmetric group and $[x\_1,\dots,x\_n]=[[x\_1,\dots,x\_{n-1}],x\_n].$ I can prove this formula for $p=2,3.$




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**Answer:** Yes, it is true. This formula can be found here: 
<http://msp.org/agt/2006/6-5/agt-v6-n5-p08-p.pdf>
on page 2252, line 6 (without prove). Jie Wu explained me a prove. 



One identity in Lie algebras
If you expand the commutator by the possible $2^{n-1}$ sign choices you get 
$$[a\_1,a\_2,\dots,a\_n]=\sum\_{r=1}^{n-1}(-1)^r\sum\_{k\_1> k\_2>\cdots> k\_r>1}a\_{k\_1}a\_{k\_2}\cdots a\_{k\_r}W(k\_1,k\_2,\dots,k\_r)$$
where the expression $W(k\_1,k\_2,\dots,k\_r)$ stands for the word $a\_1a\_2\cdots a\_n$ with the letters $a\_{k\_i}$ removed. Now this tells us that when you expand the sum
$$\sum\_{\sigma\in S\_p, \sigma(1)=1}[a\_{\sigma(1)},a\_{\sigma(2)},\dots,a\_{\sigma(p)}],$$
a word where $a\_1$ is in the $r$-th position appears with coefficient $(-1)^r\binom{p-1}{r}$. Wison's theorem tells you that $(-1)^r\binom{p-1}{r}\equiv 1\pmod{p}$ and your equality follows.


