Commutators and solvable groups

We begin our discussion with the following basic notion.

Definition 8.1.1   Let $G$ be a group and let $a,b \in G$. The product $aba^{-1} b^{-1}$ is called the commutator of $a$ and $b$. We write $[a,b] = aba^{-1} b^{-1}$.

Clearly $[a,b] = e$ if and only if $a$ and $b$ commute.

Definition 8.1.2   Let $G'$ be the subgroup of $G$ which is generated by the set of all commutators of elements of $G$, i.e., $G' = gp(\{[x,y] \ \vert\ x,y \in G\})$. $G$ is called the commutator (or derived) subgroup of $G$.

If we recall our discussion on the subgroup generated by a subset of a group $G$, then we know that $G'$ consists of all finite products of commutators and inverses of commutators. (See Proposition 5.1.1.) However, the inverse of a commutator is once again a commutator (see exercise 1 for this section). It then follows that $G$ is precisely the set of all finite products of commutators, i.e., $G$ is the set of all elements of the form

$$h_1h_2 ... h_n$$

where each $h_i$ is a commutator of elements of $G$.

The following proposition shows that the commutator subgroup is always normal.

Proposition 8.1.3   If $G$ is a group, then $G'\lhd G$.

Proof: If $h = [a,b]$ for $a,b \in G$, and $x\in G$, $xhx^{-1} = [xax^{-1}, xbx^{-1}]$ is again a commutator of elements of $G$. Now from our previous comments, an arbitrary element of $G'$ has the form $h_1h_2 ... h_n$, where each $h_i$ is a commutator. Thus $x(h_1h_2 ... h_n)x^{-1} = (xh_1x^{-1}) (xh_2x^{-1}) ... (xh_nx^{-1})$ and since by the above each $xh_ix^{-1}$ is a commutator $x(h_1h_2 ... h_n)x^{-1} \in G'$. Using Proposition 6.1.4, we have proven $G \lhd G$. $\Box$

We next contend that the factor group $G/G'$ is an abelian group and that actually $G'$ is the smallest normal subgroup that enjoys this property. (Note that if $G$ is a finite group and if $\vert N\vert > \vert G'\vert$, where $N \lhd G$ such that $G/N$ is abelian, then $\vert G/G'\vert > \vert G/N\vert$, so we are actually discovering the ``largest'' abelian homomorphic image of $G$.)

Theorem 8.1.4   $G/G'$ is an abelian group. Moreover, if $N \lhd G$ such that $G/N$ is abelian, then $G'\subset N$.

Proof: In order to establish the first part of the theorem, let $aG'$ and $bG'$ be any two elements of $G/G'$. Then

$$\begin{array}{rr} [aG',bG']&=aG'\cdot bG' (aG')^{-1}(bG')^{-... ...1}G'\cdot b^{-1}G'\\ & =aba^{-1}b^{-1}G' \\ &=G', \end{array}$$

since $[a,b] \in G'$. In other words, any two elements of $G/G'$ commute (recall that in the factor group $G/G'$, $G'$ functions as the identity element). Hence $G/G'$ is abelian.

Next let $N \lhd G$. If $N$ does not contain $G'$, then $N$ certainly cannot contain all commutators of elements of $G$ (recall that the group generated by a set is the smallest subgroup containing that set - see §5.1). Thus let $a,b \in G$ be such that $[a,b] \notin N$. Then $[aN, bN] = aba^{-1}b^{-1}N = [a,b]N\not= N$. Hence $G/N$ is non-abelian. Taking the contrapositive completes the proof. $\Box$

In general, we know (see exercise 8 for Section 6.1) that a normal subgroup of a normal subgroup need not be normal in the whole group. However, the following theorem shows that in the special case of the commutator subgroup of a normal subgroup, we can state that this is normal in the entire group.

Theorem 8.1.5   Let $N \lhd G$, a group, and let $N'$ be the commutator subgroup of $N$. Then $N'\lhd G$.

Proof: Let $c = [a,b]$ where $a,b \in N$. Then for an arbitrary $x\in G$, we have

$$\begin{array}{c} xcx^{-1}=xaba^{-1}b^{-1}x^{-1}\\ =(xax^{-1... ...^{-1})^{-1}(xbx^{-1})^{-1}\\ =[xax^{-1},xbx^{-1}]. \end{array}$$

Since $N \lhd G$, $xax^{-1}$ and $xbx^{-1}\in N$, we have $xcx^{-1} \in N$, where $c$ was an arbitrary commutator of elements of $N$. It follows directly from this that $N \lhd G$ (WHY?). $\Box$

We consider next the following sequence of subgroups of an arbitrary group G:

$$...\subset G'''\subset G''\subset G'\subset G ,$$ (8.1)

where $G''$ is the commutator subgroup of $G'$, $G'''$ is the commutator subgroup of $G''$, etc.

Definition 8.1.6   If the above sequence of subgroups of a group G given in (8.1) contains the trivial subgroup, i.e., $\{e\}$, then the group $G$ is called solvable.

If $G$ is abelian, then $G = \{e\}$, and so an abelian group is solvable. The converse is false, e.g., $S_3$ can be shown to be solvable (see exercise 2 for this section), but of course $S_3$ is non-abelian. We also observe that $A_n$ for $n \geq 5$ is not solvable. We have $A_n'=A_n$, for $n \geq 5$, since by Theorem 6.3.2 we know that $A_n$ is simple (and non- abelian) for $n \geq 5$. We mention in passing that the Feit-Thompson Theorem alluded to earlier (see the beginning of Section 6.3) states: Any group of odd order is solvable.