Formula for commutator of element and product of two elements

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 $$g^h = h^{-1}gh$$.

Applications

 * Class two implies commutator map is endomorphism
 * Commutator map is homomorphism if commutator is in centralizer
 * Subgroup normalizes its commutator with any subset

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:

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