Witt's identity

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

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.