# A finitely generated soluble group isomorphic to a proper quotient group

Let $\mathbb{Q}_2$ be the ring of rational numbers of the form $m2^n$ with $m, n \in \mathbb{Z}$ and $N = U(3, \mathbb{Q}_2)$ the group of unitriangular matrixes of dimension $3$ over $\mathbb{Q}_2$. Let $t$ be the diagonal matrix with diagonal entries: $1, 2, 1$, put $H = < t, N >$ and $w=\left( \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right)$. Then the group $G=H/< w^2 >$ is finitely generated soluble and isomorph to a proper quotient subgroup.

For more details you can see here.

What is a simple proof that $G$ is soluble?

• From quickly glancing at the link provided, it seems that this issue of solvability is not addressed and you are looking to fill in the details. But why are you convinced that a simple proof must exist? What does simple mean here? Are you aware of another proof that uses something like a sledgehammer? – RghtHndSd Dec 14 '14 at 18:33
• The group of all upper triangular matrices is solvable, and this is a quotient of a subgroup, and hence solvable. It's the finitely generated property that's interesting. – Derek Holt Dec 15 '14 at 7:59
• the given ring has a standard notation, namely $\mathbb{Z}[1/2]$; $\mathbb{Q}_2$ (borrowed from your link) is a very misleading notation since it denotes $2$-adic numbers and is often used in the same context. – YCor Mar 22 '16 at 0:13