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Commutator (group)

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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 It then follows that 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:

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 $\begin{align*}(x^y,(y,z)) &= (x^{-1})^y \cdot (z,y) \cdot x^y \cdot (y,z) \\&= (x^{-1})^y \cdot z^{-1} (z)^y \cdot x^...$ By cyclic permutation of variables, we thus find $\begin{align*}(x^y, (y,z))(y^z, (z,x))(z^x, (x,y)) &= f(z,x,y)^{-1} f(x,y,z) f(x,y,z)^{-1} f(y,z,x) f(y,z,x)^{-1} f(z,x,y...$

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 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 $G$ .

The group $A$ normalizes the group $(A,B)$ .
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 $(A,(B,C))$ .
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 $x\in G$ , $a\in A$ , $b\in B$ , $c \in C$ , $\begin{align*}(a,(b,c)x) &= a^{-1}x^{-1}(c,b)a(b,c)x = (a,x)x^{-1}a^{-1}(c,b)a(b,c)x \\&= (a,x) (x,((b,c),a)) (a,(b,c...$ 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, $\begin{align*}(a,(b,c)) &= (a, (c,b) \cdot (b,c)^2) \\&= (a,(b,c)^2)\Bigl( (b,c)^2,\bigl((c,b),a \bigr) \Bigr) \cdot ...$ 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 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 $(a,(b,c)) = \bigl[ (b^c, (c,a')) \cdot (c^{a'}} , (a',b)) \bigr]^{-1} \in (B,(C,A)) \cdot (C, (B,A)) ,$ as desired. $\blacksquare$

See also

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Category: Group theory

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