Witt's identity

This fact is related to: commutator calculus
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Statement

In terms of right-action convention

Let $a,b,c$ be elements of an arbitrary group $G$. Then:

$[[a,b^{-1}],c{]}^b\cdot [[b,c^{-1}],a{]}^c\cdot [[c,a^{-1}],b{]}^a=e$

where $[x,y]=x^{-1}{y}^{-1}xy$ and $x^y=y^{-1}xy$, and $e$ is the identity element of the group.

Related results

• Jacobi identity
• Three subgroup lemma

Proof

In terms of right-action convention

Given: A group $G$, elements $a,b,c\in G$. $e$ is the identity element.

To prove: $[[a,b^{-1}],c{]}^b\cdot [[b,c^{-1}],a{]}^c\cdot [[c,a^{-1}],b{]}^a=e$ where $[x,y]:=x^{-1}{y}^{-1}xy$ and $x^y=y^{-1}xy$.

Proof: We start out with the first term on the left side:

$[[a,b^{-1}],c{]}^b=[a^{-1}ba{b}^{-1},c{]}^b=b^{-1}[a^{-1}ba{b}^{-1},c]b=b^{-1}b{a}^{-1}{b}^{-1}a{c}^{-1}{a}^{-1}ba{b}^{-1}cb=a^{-1}{b}^{-1}a{c}^{-1}{a}^{-1}ba{b}^{-1}cb$

Similarly, we have:

$[[b,c^{-1}],a{]}^c=b^{-1}{c}^{-1}b{a}^{-1}{b}^{-1}cb{c}^{-1}ac$

and:

$[[c,a^{-1}],b{]}^a=c^{-1}{a}^{-1}c{b}^{-1}{c}^{-1}ac{a}^{-1}ba$

Multiplying these, all terms cancel and we obtain the identity element, as desired.