# 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 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

# 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