Thursday 24 January 2013

Basic properties of nilpotent and solvable groups (unfinished)

Theorem If $G$ is nilpotent or solvable so is any subgroup.
proof: Let $G$ have the series $$1 = G_0 \lhd G_1 \lhd \cdots \lhd G_n = G.$$ Suppose $H \le G$ then $$1 = G_0 \cap H \lhd G_1 \cap H \lhd \cdots \lhd G_n \cap H = H$$ is a series for $H$ by ABC.
Now suppose $G$ was nilpotent, by the alternative characterization of a central factor we have that $[G,G_i] \le G_{i-1}$ and we deduce $H$ is nilpotent from $[G \cap H,G_i \cap H] \le G_{i-1} \cap H$.
Now suppose $G$ was solvable we deduce that $H$ is too from the fact that if $G_{i}/G_{i-1}$ is abelian so is $G_{i} \cap H/G_{i-1} \cap H \simeq G_{i-1}(G_i \cap H)/G_{i-1} \le G_i/G_{i-1}$ (by ABC) as it's the subgroup of an abelian group.

Theorem If $G$ is nilpotent or solvable so is any quotient.
proof: Suppose $G$ was solvable, $N \unlhd G$ and let $\nu : G \to G/N$ be the natural inclusion homomorphism which we know to be surjective. Take $G$s composition series as before. Let us call $\nu(G_i)$ by $\bar H_i$ so by Noether(-1) and the fact it's a composition series $\bar H_i \lhd \bar H_{i+1}$ (so we have a series for $G/N$) also it's an abelian series since $\bar H_{i+1}/\bar H_i$ is abelian: let $\bar x, \bar y \in \bar H_{i+1}$ say $\bar x = \nu x$, $\bar y = \nu y$ then since $H_{i+1}/ H_i$ is abelian $xy = yxd$ for some $d \in H_i$, so taking $\nu$ we find $\bar x \bar y = \bar y \bar x \bar d$ for some $\bar d \in \bar H_i$, thus: $$(\bar x \bar H_i)(\bar y \bar H_i) = \bar y \bar x \bar d \bar H_i = \bar y \bar x \bar H_i = \bar y \bar H_i \bar x \bar H_i.$$

Theorem If $H$ and $K$ are both nilpotent or both solvable so is $H\times K$
proof:

Proposition Nilpotent groups aren't closed under taking extensions. Example $S_3$ is an extension of $A_3 \simeq C_3$ by $S_3/A_3 \simeq C_2$ but isn't nilpotent.

Theorem Solvable groups are closed under taking extensions: Suppose $K \unlhd G$ then if $K$ and $G/K$ are both solvable so is $G$.
proof:

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