Formula for commutator of element and product of two elements

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Statement

With the left-action convention

Suppose G is a group and x,y,z are elements of G. Then:

\! [x,yz] = [x,y] c_y([x,z]).

and:

\! [xy,z] = c_x([y,z])[x,z]

Here [a,b] = aba^{-1}b^{-1} and c_g(h) = ghg^{-1}.

With the right-action convention

Suppose G is a group and x,y,z are elements of G. Then:

\! [x,yz] = [x,z][x,y]^z

and:

\! [xy,z] = [x,z]^y[y,z]

here [a,b] = a^{-1}b^{-1}ab and g^h = h^{-1}gh.

Related facts

Applications

Proof

With the left-action convention

Given: A group G, elements x,y,z \in G

To prove: \! [x,yz] = [x,y]c_y([x,z]) and [xy,z] = c_x([y,z])[x,z] where \! [a,b] := aba^{-1}b^{-1} and c_a(b) = aba^{-1}.

Proof: We have:

\! [x,y]c_y([x,z]) = [xyx^{-1}y^{-1}]y[xzx^{-1}z^{-1}]y^{-1} = xyzx^{-1}z^{-1}y^{-1} = xyzx^{-1}(yz)^{-1} = [x,yz]

and:

\! c_x([y,z])[x,z] = x[yzy^{-1}z^{-1}]x^{-1}[xzx^{-1}z^{-1}] = xyzy^{-1}x^{-1}z^{-1} = (xy)z(xy)^{-1}z^{-1} = [xy,z]

With the right-action convention

Given: A group G, elements x,y,z \in G

To prove: \! [x,yz] = [x,z][x,y]^z and [xy,z] = [x,z]^y[y,z] where [a,b] := a^{-1}b^{-1}ab and a^b = b^{-1}ab.

Proof: We have:

\! [x,z][x,y]^z = [x^{-1}z^{-1}xz]z^{-1}[x^{-1}y^{-1}xy]z = x^{-1}z^{-1}y^{-1}xyz = x^{-1}(yz)^{-1}x(yz) = [x,yz]

and:

\! [x,z]^y[y,z] = y^{-1}[x^{-1}z^{-1}xz]y[y^{-1}z^{-1}yz] = y^{-1}x^{-1}z^{-1}xyz = (xy)^{-1}z^{-1}(xy)z = [xy,z]