Show $\displaystyle 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.

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- Apr 12th 2010, 11:27 AMmathman88Solvable Group
Show $\displaystyle 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. - Apr 12th 2010, 11:38 AMtonio
- Apr 12th 2010, 11:47 AMmathman88
$\displaystyle [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 $\displaystyle [B,B] $ is abelian.

We then have $\displaystyle \{1\} \triangleleft [B,B] \triangleleft B $ and $\displaystyle [B,B]/\{1\} = [B,B] $ is cyclic, but how is $\displaystyle B/[B,B] $ cyclic? - Apr 12th 2010, 11:58 AMtonio
- Apr 12th 2010, 12:02 PMmathman88
- Apr 12th 2010, 12:14 PMtonio

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 $\displaystyle B$ is finitely generated then $\displaystyle B\slash [B,B]$ 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