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 $n$

Dihedral group

Quaternions

Klein-4 Group

$S_n$

$A_n$

Matrices

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 $\phi :\mathbb{Z}\to G$ 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 $\text{im}(\phi )\cong \mathbb{Z}/(\text{ord}(g))$.

Proposition 1.