# Commutator (group)

This article refers to commutators of groups, not to be confused with commutator groups.

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 1. 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 $$f((a,b)) = (f(a),f(b)).$$ Note also that since $(a,b) = (b,a)^{-1}$, $(A,B)=(B,A)$.

Lemma. Let $A,B$ be a closed subsets of $G$ (not necessarily a subgroups); denote by $(A,B)$ the subgroup of $G$ generated by elements of the form $(a,b)$, for $a\in A$, $b\in B$. Then $(A,B)^a \subseteq$.

Proof. Let $a,a'$ be elements of $A$ and $b$ be an element of $B$. Then $$(a,b)^{a'} = a'^{-1}a^{-1}b^{-1}aba' = (aa',b)b^{-1}a'^{-1}aba'$$

Proposition 2. Let $A,B,C$ be three subgroups of $G$.

1. The group $A$ normalizes the group $(A,B)$.
2. If the group $(B,C)$ normalizes $A$, then the set of elements $(a,(b,c))$, for $a\in A$, $b\in B$, $c \in C$, generates the group $(A,(B,C))$.
3. If $A$, $B$, and $C$ are normal subgroups of $G$, then $(A,(B,C))$ is a subgroup of the group $(C,(B,A)) \cdot (B, (C,A))$.

Proof. For the first, we note that for any $a,a' \in A$ and any $b \in B$, $$(a,b)^{a'} = (aa',b) (a',b)^{-1},$$ by Proposition 1.

For the second, we have for any $x\in G$, $a\in A$, $b\in B$, $c \in C$, \begin{align*} (a,(b,c)x) &= a^{-1}x^{-1}(c,b)a(b,c)x = (a,x)x^{-1}a^{-1}(c,b)a(b,c)x \\ &= (a,x) (x,((b,c),a)) (a,(b,c)) . \end{align*} Since $(B,C)$ normalizes $A$, the element $((b,c),a)$ lies in $A$. It then follows from induction on $n$ that for all $b_i \in B$, $c_i \in C$, the element $$\biggl( a, \prod_{i=1}^n (b_i,c_i) \biggr)$$ lies in the subgroup generated by elements of the form $(a,(b,c))$. Similarly, \begin{align*} (a,(b,c)) &= (a, (c,b) \cdot (b,c)^2) \\ &= (a,(b,c)^2)\Bigl( (b,c)^2,\bigl((c,b),a \bigr) \Bigr) \cdot (a,(c,b)) , \end{align*} lies in the subgroup generated by elements of the form $(a,(b,c))$; it then follows that $(a,(c,b))$ does. Then using the observation $$(a,(c,b)x) = (a,x)(x, ((c,b),a)) (a,(c,b)),$$ we prove by induction on $n$ that the element $$\biggl(a, \prod_{i=1}^n (b_i,c_i)^{\pm 1} \biggr)$$ lies in the subgroup generated by elements of the form $(a,(b,c))$. This proves the second result.

For the third part, we first note that since $A,B,C$ are normal subgroups, so are $(A,(B,C))$, $(C,(B,A))$, and $(B,(C,A))$; in particular, $(C,(B,A)) \cdot (B,(C,A)) = (B,(C,A)) \cdot (C,(B,A))$ is a group. From the previous result of this proposition, it suffices to show that $(a,(b,c))$ lies in this group, for all $a\in A$, $b\in B$, and $c\in C$. To that end, we note that since $A$ is normal, there exists $a' \in A$ such that $a = a'^{b}$. Then by the third result of Proposition 1, $$(a'^b , (b,c)) \cdot (b^c , (c,a')) \cdot (c^{a'} , (a',b)) = e,$$ or

$(a,(b,c)) = \bigl[ (b^c, (c,a')) \cdot (c^{a'}} , (a',b)) \bigr]^{-1} \in (B,(C,A)) \cdot (C, (B,A)) ,$ (Error compiling LaTeX. ! Extra }, or forgotten \$.)

as desired. $\blacksquare$