Group Theory/Commutators, solvable and nilpotent groups

{{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,
 * $$\begin{equation*}

t \end{equation*}$$}}

{{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
 * $$(1, \ldots, 1, \overset{\overset{k\text{-th entry}}{\downarrow}}{g^{-1}}, 1, \ldots, 1)(h_1^{-1}, \ldots, h_n^{-1})(1, \ldots, 1, \overset{\overset{k\text{-th entry}}{\downarrow}}{g}, 1, \ldots, 1)(h_1, \ldots, h_n) = (1, \ldots, \overset{\overset{k\text{-th entry}}{\downarrow}}{g^{-1}h^{-1}gh}, 1 \ldots, 1)$$

is in $$N$$.}}

{{definition|solvable|}}

{{proposition|group is solvable iff maximal normal subgroup is solvable|Let $$G$$ be a group, and let $$H \triangleleft G$$ be a maximal normal subgroup. Then $$G$$ is solvable if and only if $$H$$ is solvable.}}