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Commutator (group)

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In a group, the commutator of two elements $a$ and $b$, denoted $(a,b)$ or $[a,b]$, is the element $a^{-1}b^{-1}ab$. If $a$ and $b$ commute, then $(a,b)=e$. More generally, $(a,b) = (ba)^{-1}ab$, or \[ab = ba(a,b) .\] It then follows that \[(a,b)(b,a) = e .\] We also have \[x^y = y^{-1}xy = x(x,y) = (y,x^{-1})x ,\] where $x^y$ denote the image of $x$ under the inner automorphism $\text{Int}(y^{-1})$, as usual.

Relations with Commutators

Proposition. For all $x,y,z$ in a group, the following relations hold:

  • $(x,yz) = (x,z)(x,y)^z = (x,z)(z,(y,x))(x,y)$;
  • $(xy,z) = (x,z)^y (y,z) = (x,z)((x,z),y)(y,z)$;
  • $(x^y,(y,z))(y^z,(z,x))(z^x,(x,y)) = e$;
  • $(x,yz)(z,xy)(y,zx) = e$;
  • $(xy,z)(yz,x)(zx,y) = e$.

Proof. For the first equation, we note that \[(x,yz) = x^{-1}z^{-1}y^{-1}xyz = (x^{-1}z^{-1}xz)z^{-1}(x^{-1}y^{-1}xy)z = (x,z) (x,y)^z .\] From the earlier relations, \[(x,y)^z = (z,(x,y)^{-1})(x,y) = (z,(y,x))(x,y) ,\] hence the relation. The second equation follows from the first by passing to inverses.

For the third equation, we define $f(a,b,c) = ca (b)^c$. We then note that \begin{align*} (x^y,(y,z)) &= (x^{-1})^y \cdot (z,y) \cdot x^y \cdot (y,z) \\ &= (x^{-1})^y \cdot z^{-1} (z)^y \cdot x^y \cdot (z^{-1})^y \cdot z &= (x^{-1})^y \cdot z^{-1} \cdot y^{-1}(zxz^{-1})y \cdot z \\ &= \bigl[(x^{-1})^y \cdot z^{-1}y^{-1}\bigr] zx \cdot y^z \\ &= f(z,x,y)^{-1} f(x,y,z) . \end{align*} By cyclic permutation of variables, we thus find \begin{align*} (x^y, (y,z))(y^z, (z,x))(z^x, (x,y)) &= f(z,x,y)^{-1} f(x,y,z) f(x,y,z)^{-1} f(y,z,x) f(y,z,x)^{-1} f(z,x,y) \\ &= e. \end{align*}

For the fourth equation, we have \[(x,yz)(z,xy)(y,zx) = (yzx)^{-1}(xyz)(xyz)^{-1} (zxy) (zxy)^{-1}(yzx) = e .\] The fifth follows similarly. $\blacksquare$

Commutators and Subgroups

If $A$ and $B$ are subgroups of a group $G$, $(A,B)$ denotes the subgroup generated by the set of commutators of the form $(a,b)$, for $a\in A$ and $b\in B$.

The group $(A,B)$ is trivial if and only if $A$ centralizes $B$. Also, $(A,B) \subseteq A$ if and only if $B$ normalizes $A$. If $A$ and $B$ are both normal (or characteristic), then so is $(A,B)$, for if $f$ is an (inner) automorphism, then \[f((a,b)) = (f(a),f(b)).\]

See also

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