# 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.

## 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.