Sketch 1

Examples of groups and their properties

Huh...

it seems like groups are super interesting, and I still don't know them like the back of my hand. That's why this sketch is dedicated to them.

Date Started: December 29, 2023
Date Finished: Unfinished

A list of groups to consider...

I studied much of what I know about algebra from Dummit and Foote as well as Lang, so many of the examples I will think of were inspired by these sources. If anyone has other ideas of useful groups to consider, contact me, I would love to learn to sketch them.

Integers, Rational Numbers, Real Numbers

Residues modulo \(n\)

Locally Compact Abelian Groups

Infinite Groups

This section covers mainly integers, rational numbers, and real numbers. While not present in this section, I will also treat infinite matrices in the infinite matrices section as an example of an infinite noncommutative group.

Integers

The integers \(\mathbb{Z}\) form a group with respect to addition, with identity 0, and the inverse of \(x\) being denoted \(-x\). In a sense, I think that in some sense it is fair to think of \(\mathbb{Z}\) being the "completion" or "closure" of the natural numbers (including zero) with respect to inverses, noting that \(\mathbb{N}\) forms a monoid. It also turns out that \(\mathbb{Z}\) is the smallest ring containing \(\mathbb{N}\).

The only subgroups of \(\mathbb{Z}\) are sets of the form \((k) = \{kn : n \in \mathbb{Z}\}\), and it is clear that with \((1) = \mathbb{Z}\), \(a\) divides \(b\) is equivalent to \((b) \le (a)\) (\(\le\) denoting subgroup). In other words, the lattice of \(\mathbb{Z}\) is precisely the same as \(\mathbb{Z}\) considered as a poset with divisibility as the order. The centralizer and normalizer are trivially all of \(\mathbb{Z}\) since \(\mathbb{Z}\) is commutative.

Now I would like to categorize normal subgroups and group homomorphisms to and from \(\mathbb{Z}\). All subgroups are trivially normal since \(\mathbb{Z}\) is commutative. Additionally, it is interesting that every group homomorphism \(\varphi : \mathbb{Z} \rightarrow G\) is of the form \(1 \mapsto g\), so that \(n \mapsto g^n\). In this way the collection of homomorphisms from \(\mathbb{Z}\) to a group \(G\) appear to give us all cyclic subgroups of \(G\). Now we can define \(\text{ord}(g)\) to be the order of the cyclic subgroup generated by \(g\), and we see that now the kernel is precisely \((\text{ord}(g))\) and \(\text{im}(\varphi) \cong \mathbb{Z}/(\text{ord}(g))\).

Proposition 1.