Witt's identity

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This fact is related to: commutator calculus
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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.

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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


\! [[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.