Group Theory/Commutators, solvable and nilpotent groups

Definition (commutator):

Let $G$ be a group and let $a,b\in G$ . Then the commutator of $a$ and $b$ is defined to be the element

[a,b]:=a^{-1}b^{-1}ab\in G

.

{{definition|commutator

Proposition (commutators form a subgroup):

Let $G$ be a group. Then the set $\{[a,b]|a,b\in G\}$ forms a subgroup of $G$ .

Proof: By the subgroup criterion, it is sufficient to show that for $a,b,c,d\in G$ , the element $[a,b][c,d]^{-1}$ is of the form $[e,f]$ for suitable $e,f\in G$ . Indeed,

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\Box

Definition (commutator subgroup):

Let $G$ be a group. Then the commutator subgroup of $G$ is defined to be $[G,G]$ .

Definition (perfect):

A group $G$ is called perfect if and only if $[G,G]=G$ .

Proposition (subdirect normal product of perfect groups is direct):

Let $H_{1},\ldots ,H_{n}$ be perfect groups, and let

N\trianglelefteq H_{1}\times \cdots \times H_{n}

be a subdirect product which is simultaneously a normal subgroup of their outer direct product. Then in fact $N=H_{1}\times \cdots \times H_{n}$ .

Proof: It suffices to show that whenever $k\in \{1,\ldots ,n\}$ and $g,h\in H_{k}$ , then

(1,\ldots ,{\overset {\overset {k{\text{-th entry}}}{\downarrow }}{g^{-1}h^{-1}gh}},1\ldots ,1)\in N

,

since $H_{k}$ is a perfect group. Thus, let $h\in H_{k}$ be arbitrary, and pick $(h_{1},\ldots ,h_{n})\in N$ , where $h_{j}\in H_{j}$ for all $j\in \{1,\ldots ,n\}$ , such that $h_{k}=h$ . Since $N$ is a subgroup and normal, the element