Posts Tagged ‘derived series’

Solvable groups (2)

Posted: January 10, 2022 in Basic Algebra, Groups
Tags: , , , , , ,
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See the first part of this post here.

We now look at the connection between solvability of a group and the so-called higher commutative subgroups of the group. Recall that the commutator (or derived) subgroup of a group G, denoted G' or [G,G], is the subgroup generated by the set of commutators of the group, i.e.

G':=\langle [x,y]: \ x,y \in G\rangle,

where [x,y]:=xyx^{-1}y^{-1}. It is easily seen that G' is a normal subgroup of G and if N is a normal subgroup of G, then G/N is abelian if and only if G' \subseteq N. We now define the higher commutator subgroups of G.

Definition 3. The derived series of a group G is a sequence \{G^{(n)}\}_{n \ge 0} of subgroups of a group G defined inductively as follows

G^{(0)}=G, \ \ \ \ G^{(n)}=[G^{(n-1)},G^{(n-1)}], \ \ \ \ n \ge 1.

So G^{(1)}=[G^{(0)},G^{(0)}]=[G,G]=G' and G^{(2)}=[G^{(1)},G^{(1)}]=[G',G'], etc. Just like derivatives of a function, we may also write G'' for G^{(2)}, etc. It is clear that

G=G^{(0)} \supseteq G^{(1)} \supseteq G^{(2)} \supseteq \cdots,

and G^{(n)} is normal in G^{(n-1)} for all n \ge 1, because, by definition, G^{(n)} is the commutator subgroup of G^{(n-1)}. We show in the following theorem that in fact each G^{(n)} is normal in G.

Definition 4. Let G be a group. A series (1)=G_0 \subseteq G_1 \subseteq \cdots \subseteq G_n=G is said to be normal if G_i is normal in G for all 0 \le i \le n.

Theorem 2. Let G be a group.

i) For any n \ge 0, the subgroup G^{(n)} is normal in G.

ii) If G is solvable, and (1)=G_0 \subseteq G_1 \subseteq \cdots \subseteq G_n=G is a solvable series in G, then G^{(i)} \subseteq G_{n-i} for all 0 \le i \le n.

iii) G is solvable if and only if G^{(n)}=(1) for some integer n.

iv) G is solvable if and only if G has a normal series (1)=G_0 \subseteq G_1 \subseteq \cdots \subseteq G_n=G such that G_{i+1}/G_i is abelian for all 0 \le i \le n-1.

Proof. i) By induction. There’s nothing to prove for n=0 because G^{(0)}=G. Now suppose that n \ge 1 and G^{(n-1)} is normal in G. Since, by definition, G^{(n)} is the commutator subgroup of G^{(n-1)}, we only need to show that g[x,y]g^{-1} \in G^{(n)} for all g \in G, \ x,y \in G^{(n-1)}, and that’s easy

g[x,y]g^{-1}=gxyx^{-1}y^{-1}g^{-1}=(gxg^{-1})(gyg^{-1})(gx^{-1}g^{-1})(gy^{-1}g^{-1})

\begin{aligned}=(gxg^{-1})(gyg^{-1})(gxg^{-1})^{-1}(gyg^{-1})^{-1}=[gxg^{-1},gyg^{-1}] \in [G^{(n-1)},G^{(n-1)}]=G^{(n)}.\end{aligned}

Note that, since we assumed that G^{(n-1)} is normal in G, both gxg^{-1} and gyg^{-1} are in G^{(n-1)}.

ii) By induction. It’s clear for i=0 since G=G^{(0)}=G_n. Suppose now that 0 \le i \le n-1 and the result holds for i. So G^{(i)} \subseteq G_{n-i}. We also have G_{n-i}' \subseteq G_{n-i-1} because G_{n-i}/G_{n-i-1} is abelian. Thus G^{(i+1)} \subseteq G_{n-i}' \subseteq G_{n-i-1}, which proves the result for i+1.

iii) Suppose first that G is solvable, and (1)=G_0 \subseteq G_1 \subseteq \cdots \subseteq G_n=G is a solvable series in G. By ii), G^{(n)} \subseteq G_0=(1). Conversely, suppose that G^{(n)}=(1) for some n, and let G_i:=G^{(n-i)}, \ 0 \le i \le n. Then

(1)=G^{(n)}=G_0 \subseteq G_1 \subseteq \cdots \subseteq G_n=G^{(0)}=G. \ \ \ \ \ \ \ \ \ \ (*)

By i), G_i is normal in G for all i, and hence in G_{i+1}. Also, G_{i+1}/G_i are all abelian because

G_i=G^{(n-i)}=(G^{(n-i-1)})'=G_{i+1}'.

Thus (*) is a solvable series in G and hence G is solvable.

iv) It is clear that if G has a normal series (1)=G_0 \subseteq G_1 \subseteq \cdots \subseteq G_n=G such that G_{i+1}/G_i is abelian for all i, then G is solvable because every normal series is obviously subnormal. Conversely, if G is solvable, then, by iii), G^{(n)}=(1) for some n and then, as explained in the proof of iii), the series (*) has the required properties. \Box