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 matrices** of dimension \(3\) over \(\mathbb{Q}_2\). Let \(t\) be the **diagonal matrix** with diagonal entries: \(1, 2, 1\) and put \(H = \langle t, N \rangle\). We will prove that \(H\) is **finitely generated** and that one of its **quotient group** \(G\) is **isomorphic** to a proper quotient group of \(G\). Continue reading A finitely generated soluble group isomorphic to a proper quotient group

# Tag Archives: algebra

# A (not finitely generated) group isomorphic to a proper quotient group

The basic question that we raise here is the following one: *given a group \(G\) and a proper subgroup \(H\) (i.e. \(H \notin \{\{1\},G\}\), can \(G/H\) be isomorphic to \(G\)?* A group \(G\) is said to be

**hopfian**(after

**Heinz Hopf**) if it is not isomorphic with a

**proper quotient group**.

All **finite groups** are hopfian as \(|G/H| = |G| \div |H|\). Also, all **simple groups** are hopfian as a simple group doesn’t have proper subgroups.

So we need to turn ourselves to infinite groups to uncover non hopfian groups. Continue reading A (not finitely generated) group isomorphic to a proper quotient group

# Converse of Lagrange’s theorem does not hold

**Lagrange’s theorem**, states that for any finite **group** \(G\), the **order** (number of elements) of every **subgroup** \(H\) of \(G\) divides the order of \(G\) (denoted by \(\vert G \vert\)).

Lagrange’s theorem raises the converse question as to whether every divisor \(d\) of the order of a group is the order of some subgroup. According to **Cauchy’s theorem** this is true when \(d\) is a prime.

However, this does not hold in general: given a finite group \(G\) and a divisor \(d\) of \(\vert G \vert\), there does not necessarily exist a subgroup of \(G\) with order \(d\). The **alternating group** \(G = A_4\), which has \(12\) elements has no subgroup of order \(6\). We prove it below. Continue reading Converse of Lagrange’s theorem does not hold

# A field that can be ordered in two distinct ways

For a short reminder about **ordered fields** you can have a look to following post. We prove there that \(\mathbb{Q}\) can be ordered in only one way.

That is also the case of \(\mathbb{R}\) as \(\mathbb{R}\) is a **real-closed field**. And one can prove that the only possible **positive cone** of a real-closed field is the subset of squares.

However \(\mathbb{Q}(\sqrt{2})\) is a subfield of \(\mathbb{R}\) that can be ordered in two distinct ways. Continue reading A field that can be ordered in two distinct ways

# An infinite field that cannot be ordered

## Introduction to ordered fields

Let \(K\) be a field. An **ordering** of \(K\) is a subset \(P\) of \(K\) having the following properties:

- ORD 1
- Given \(x \in K\), we have either \(x \in P\), or \(x=0\), or \(-x \in P\), and these three possibilities are mutually exclusive. In other words, \(K\) is the disjoint union of \(P\), \(\{0\}\), and \(-P\).
- ORD 2
- If \(x, y \in P\), then \(x+y\) and \(xy \in P\).

We shall also say that \(K\) is **ordered by \(P\)**, and we call \(P\) the set of **positive elements**. Continue reading An infinite field that cannot be ordered