algebra | Math Counterexamples

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 →

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 →

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 →

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 →

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 →

Math Counterexamples

Tag Archives: algebra

A field that can be ordered in two distinct ways

An infinite field that cannot be ordered

Introduction to ordered fields

Mathematical exceptions to the rules or intuition