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:

$\![[a,b^{-1}],c]^{b}=[a^{-1}bab^{-1},c]^{b}=b^{-1}[a^{-1}bab^{-1},c]b=b^{-1}ba^{-1}b^{-1}ac^{-1}a^{-1}bab^{-1}cb=a^{-1}b^{-1}ac^{-1}a^{-1}bab^{-1}cb$

Similarly, we have:

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

and:

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

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