1. ## Solvable Group

Show $B=\left\{ \left. \left(\begin{array}{cc}x&z\\0&y\end{array}\right) \;\; \right| \;\; x,y\in \mathbb{F}_p^\times,\; z\in\mathbb{F}_p \right\}$ is a solvable group.

The group operation is matrix multiplication.

2. Originally Posted by mathman88
Show $B=\left\{ \left. \left(\begin{array}{cc}x&z\\0&y\end{array}\right) \;\; \right| \;\; x,y\in \mathbb{F}_p^\times,\; z\in\mathbb{F}_p \right\}$ is a solvable group.

The group operation is matrix multiplication.

Check the commutator group: it then should be "almost" obvious that it is abelian so...

Tonio

3. Originally Posted by tonio
Check the commutator group: it then should be "almost" obvious that it is abelian so...

Tonio
$[B,B] = \left\{ \left. \left(\begin{array}{cc}1&z\\0&1\end{array}\right) \;\; \right| \;\; z\in\mathbb{F}_p \right\}$

And it easy to see $[B,B]$ is abelian.

We then have $\{1\} \triangleleft [B,B] \triangleleft B$ and $[B,B]/\{1\} = [B,B]$ is cyclic, but how is $B/[B,B]$ cyclic?

4. Originally Posted by mathman88
$[B,B] = \left\{ \left. \left(\begin{array}{cc}1&z\\0&1\end{array}\right) \;\; \right| \;\; z\in\mathbb{F}_p \right\}$

And it easy to see $[B,B]$ is abelian.

How does this show $B$ is solvable?

(!!) Because then the derived series is finite, of course: $G\supseteq G'\supseteq G''=1$ and this an abelian series for the group...what definition of "solvable" do you have?

Tonio

5. Originally Posted by tonio
(!!) Because then the derived series is finite, of course: $G\supseteq G'\supseteq G''=1$ and this an abelian series for the group...what definition of "solvable" do you have?

Tonio
$G$ is solvable $\iff \{1\} \triangleleft G_i \triangleleft \ldots G_1 \triangleleft G_0=G$ and $\forall \; j, \; G_j/G_{j+1}$ is cyclic.

6. Originally Posted by mathman88
$G$ is solvable $\iff \{1\} \triangleleft G_i \triangleleft \ldots G_1 \triangleleft G_0=G$ and $\forall \; j, \; G_j/G_{j+1}$ is cyclic.

This is not the usual definition, and it's a rather misleading and even incorrect one (imo, of course. See following note), but it never minds: since the group $B$ is finitely generated then $B\slash [B,B]$ is finitely generated and thus can be decomposed in a direct product of cyclic groups...

Note: for your definition to work it MUST be that G is fin. generated, or at least that all its abelian factor groups are, otherwise it fails: an infinitely generated abelian group wouldn't be solvable according to your definition which, of course, is absurd.

Tonio