An unconventional intro to group theory

June 26, 2020.

I mean, we could start with the definition and axioms of a group (Wikipedia), but this is tiny theorems. The goal is to achieve intuition through tiny theorems rather than through big definitions.

I’m just going to repeat my short preface on group theory here:

In my short experience: when learning group theory for the first time, it is not about learning what the abstract notion of group is. It’s more about studying specific groups which will show up in all the other theories (field theory, Galois theory, representation theory, etc).

To understand this, we first need to talk about the parallel history of group theory.

In this section, we detail a parallel history of group theory.

Historically, the modern notion of group came about in several seemingly-unrelated lines of thought:

(1800s) Gauss’ study of modular arithmetic
- …which led to his notion of groups (modern-day finite abelian groups)
(1820s) Galois’ study of algebraic solutions to polynomial equations
- …which led to his notion of groups (modern-day symmetric groups).
(1840s) Cauchy’s study of permutation theory
- …which led to his notion of groups (modern-day permutation groups).
(1870s) Klein’s study of non-euclidean geometric transformations
- …which led to his notion of groups (modern-day symmetry groups)

It just happens that all these notions of groups have commonalities. For example, you can talk about

Gauss’ primitive roots modulo n, or
Galois’ proper decompositions, or
permutation theory’s even permutations, or
geometric translations of the plane.

Modern group theory characterizes all of these as normal subgroups of their respective groups. This means if we want to talk about an operation that works on all four of these groups, then we don’t need to have a special version of the operation for each group – we could just talk about that operation on, for example, normal subgroups. Similarly, the modern-day group abstracts away the similarities of these various groups, a process that started with Cayley. To this tiny timeline, we might add:

(1850s-70s) Cayley’s study of abstract groups, via permutation groups and matrix groups

…which led to our modern notion of groups today.

Sources: U of St Andrews, C.K. Fong via Carleton University (PDF)

In this section, we approach group theory without looking at big definitions.

I find a lot of group theory texts start from the abstract definition without introducing this history. For me, this historical view of “group theory saving you from having to define four different operations” does really well when it comes to thinking about groups.

One of my bigger mistakes was being way too caught up in abstract laws, like the group axioms I never mentioned at the top of this page. That was something I had to unlearn. It’s not abstract groups and laws that are important at first – it’s the relationships between specific groups and specific things.

Let’s start by looking at permutations, and use those to arrive at the definition of a group.

In this section, we define and explore permutations.

A permutation of a set is a reordering of elements and is often written like this: $(1~4~2~8~5~7)$ . This permutation maps $1\mapsto 4$ , $4\mapsto 2$ , $2\mapsto 8$ , $8\mapsto 5$ , $5\mapsto 7$ , and finally $7\mapsto 1$ . The set we’re permuting here is some set of integers that includes $\{1,2,4,5,7,8\}$ .

This notation is called disjoint cycle notation; $(1~4~2~8~5~7)$ is a cycle. You can think of them as functions taking an integer to another integer. More complex permutations are composed of cycles via right-to-left function composition: the permutation $(1~4)(2~3)$ swaps $2$ and $3$ , and then swaps $1$ and $4$ . The identity permutation (which does nothing) can be written as a length $1$ cycle consisting of any element, like $(1)$ .

In this section, we compose and invert permutations.

Consider the permutation $(2~3)(3~1)$ . This first swaps $1$ and $3$ , and then swaps $2$ and $3$ . Since the net effect is the same as the cycle $(1~2~3)$ , we consider them to be the same permutation, in the sense that the set $\{(2~3)(3~1),(1~2~3)\}$ is a set consisting of one element.

In group theory, the important thing to focus on is not the set of integers that these permutations are operating on, but rather manipulating the permutations themselves. In this case, we can see that the composition of two permutations $(2~3)$ and $(3~1)$ is equal to another permutation $(1~2~3)$ . If you think of permutations as objects instead of functions, we have combined two objects into one (via function composition). This idea is very important.

The second important idea is that every permutation can be “undone” by an inverse permutation. For instance, the inverse permutation of $(1~2~3~4~5)$ is $(1~5~4~3~2)$ . You can tell that it is the inverse because their composition is always the identity permutation $(1)$ .

So far we’ve shown three facts:

Composing two permutations gives you another permutation.
There is an identity permutation that does nothing when composed.
For every permutation, there is an inverse.

When these three facts are true for a set of permutations on $n$ elements, we call it a permutation group. In particular, the set of all $n!$ possible permutations on a set of $n$ elements is known as the symmetric group $S_n$ . I consider the permutation group foundational, but before we see why, let’s see more examples of groups.

In this section, we decompose permutations.

Every permutation can be described using cycle notation, a composition of cycles in which every element of the set we’re permuting appears exactly once. For example, the permutation $(1~2~3)$ on the set $\{1,2,3,4,5\}$ can be represented as $(1~2~3)(4)(5)$ , where $(4)$ and $(5)$ are both identity permutations that show that they are unaffected (“fixed”) by the permutation. This form is also known as a product of disjoint cycles.

You can always decompose any cycle of length $3$ or more, like $(a~b~c~d~e)$ , into a product of $2$ -cycles, also known as transpositions. In general, this “decomposition into transpositions” is not unique. For instance: $(1~2~3~4~5)=(1~2)(2~3)(3~4)(4~5)$ $(1~2~3~4~5)=(1~5)(1~4)(1~3)(1~2)$ $(1~2~3~4~5)=(5~4)(5~2)(2~1)(2~5)(2~3)(1~3)$ As you can see, this $5$ -cycle can be decomposed into any even number of transpositions. The first two forms are standard ways to decompose a cycle, while the third is just some random product of transpositions that happen to compose into $(1~2~3~4~5)$ .

It is a theorem that odd-length cycles decompose into an even number of transpositions, and even-length cycles decompose into an odd number of transpositions. We’ll prove this for all permutations:

Lemma: The identity permutation

(1)

cannot be decomposed as a product of an odd number of transpositions.

Let’s express the identity as a product of transpositions $(1)=\tau_1\tau_2\ldots\tau_n$ , and let $a$ be an arbitrary element that is moved by one of the transpositions. This means that $a$ appears in $\tau_i$ for some $i$ . Let $\tau_i$ be the transposition containing the rightmost occurrence of $a$ . Note that $\tau_i$ cannot ever be the leftmost $\tau_1$ , since that would mean there is only one occurrence of $a$ in the product, meaning the product doesn’t fix $a$ and is therefore not the identity $(1)$ .

So let $\sigma$ be the transposition to the left of $\tau_i$ . If $\sigma$ doesn’t include $a$ , then we can freely flip the ordering $\sigma\tau_i$ with $\tau_i\sigma$ and try again with $\sigma$ being the new transposition to the left of $\tau_i$ . Eventually we’ll find some $\sigma$ that contains $a$ , since otherwise there would again be only one occurrence of $a$ in the decomposition meaning that $a$ is not fixed by the identity $(1)$ .

Then, using the fact that $\sigma\tau_i=\sigma\tau_i(\sigma^{-1}\sigma)=(\sigma\tau_i\sigma^{-1})\sigma$ , we can continue moving $\tau_i$ (in the form $\sigma\tau_i\sigma^{-1}$ ) to the left of $\sigma$ . There are two cases:

$\sigma\tau_i\sigma^{-1}$ contains $a$ . In this case, use that as the new $\tau_i$ and repeat with $\sigma$ being the transposition to the left of the new $\tau_i$ .
$\sigma\tau_i\sigma^{-1}$ doesn’t contain $a$ . This can only happen if $\tau_i=\sigma$ , so our original $\sigma\tau_i$ is the identity, leaving us with a product that is length $n-2$ . Then by induction on $n$ we eventually obtain either $n=0$ or $n=1$ . $n=1$ is not possible since the identity is not a transposition, therefore we end at $n=0$ by repeatedly subtracting $2$ from $n$ , showing that $n$ can only be even.

Theorem: A permutation that decomposes into an even number of transpositions cannot decompose into an odd number of transpositions, and vice versa.

Towards contradiction, assume that the permutation $\sigma$ can be decomposed into both an even and an odd number of transpositions. Let $\sigma=e_1e_2\ldots e_n$ be the even decomposition and $\sigma=o_1o_2\ldots o_m$ be the odd decomposition.
Then since $(1)=\sigma\sigma^{-1}$ , we can write $(1)=e_1e_2\ldots e_no_m^{-1}\ldots o_2^{-1}o_1^{-1}$ . Since this is an even number of transpositions composed with an odd number of transpositions, the total is an odd number of transpositions.
But by the lemma above, $(1)$ cannot be composed of an odd number of transpositions, contradiction.

Thus we can always refer to a permutation as either even or odd. If $\sigma$ can be decomposed into an even number of transpositions it is an even permutation, otherwise it is an odd permutation.

Recall that the symmetric group $S_n$ is the set of all permutations on $n$ elements where three facts hold:

Composing two permutations gives you another permutation.
There is an identity permutation that does nothing when composed.
For every permutation, there is an inverse.

Note that if you take the set of all even permutations on $n$ elements, these three facts still hold:

Composing two even permutations gives you another even permutation.
There is an (even) identity permutation that does nothing when composed.
For every even permutation, there is an inverse even permutation.

This set is known as the alternating group $A_n$ .

In this section, we look at characterizations of the cyclic group $C_n$ .

Let’s look at permutations again.

The following set of permutations, $\{(1),(1~2~3~4),(1~3)(2~4),(1~4~3~2)\}$ , is “cyclic” in the sense that “multiplying” any element by $(1~2~3~4)$ will give you the next item in the set. For instance, $(1~2~3~4)(1~3)(2~4)=(1~4~3~2)$ , and $(1~2~3~4)(1~4~3~2)=(1)$ . You can also multiply by $(1~4~3~2)$ to get the previous element instead.

Now let’s look at some different structures, which are similar in many ways:

The set of $4$ th roots of unity, $\{1,i,-1,-i\}$ , is “cyclic” in the sense that multiplying any element by $i$ will give you the next item in the set. For instance, $i(-1)=-i$ , and $i(-i)=1$ . You can also multiply by $-i$ to get the previous element instead.

This set of integers mod $5$ , $\{1,2,4,3\}$ , is “cyclic” in the sense that multiplying any element by $2$ will give you the next item in the set. For instance, $2\cdot 4\equiv 3\mod 5$ , and $2\cdot 3\equiv 1\mod 5$ . You can also multiply by $3$ to get the previous element instead.

This set of rotation matrices, $\{\left[\begin{matrix}1&0\\0&1\end{matrix}\right],\left[\begin{matrix}0&-1\\1&0\end{matrix}\right],\left[\begin{matrix}-1&0\\0&-1\end{matrix}\right],\left[\begin{matrix}0&1\\-1&0\end{matrix}\right]\}$ , is “cyclic” in the sense that multiplying any element by $\left[\begin{matrix}0&-1\\1&0\end{matrix}\right]$ will give you the next item in the set. For instance, $\left[\begin{matrix}0&-1\\1&0\end{matrix}\right]\left[\begin{matrix}-1&0\\0&-1\end{matrix}\right]=\left[\begin{matrix}0&1\\-1&0\end{matrix}\right]$ , and $\left[\begin{matrix}0&-1\\1&0\end{matrix}\right]\left[\begin{matrix}0&1\\-1&0\end{matrix}\right]=\left[\begin{matrix}1&0\\0&1\end{matrix}\right]$ . You can also multiply by $\left[\begin{matrix}0&1\\-1&0\end{matrix}\right]$ to get the previous element instead.

In general, these all correspond to the same set $\{1,g,g^2,g^3\}$ where multiplication by $g$ gives you the next item in the set, and $g^4=1$ so multiplying by $g^3$ gives you the previous item. Later we’ll recognize this as the cyclic group of order $4$ . A cyclic group is a group defined by a single generator $g$ which, by taking all products and inverses, generates all the elements of the group. Order $4$ means that there are $4$ elements of the group.

Why $4$ elements? Normally, taking all products and inverses of an arbitrary symbol $g$ gives you the infinite set $\{\ldots,g^{-3},g^{-2},g^{-1},e,g,g^2,g^3,\ldots\}$ . This cyclic group is known as the infinite cyclic group.

The cyclic group of order $4$ , however, is a finite cyclic group. All finite cyclic groups are essentially the infinite cyclic group together with a relation $g^n=e$ , in this case $g^4=e$ . Here, $n$ is the order of the group.

In this section, we finally define groups as a generalization of the above.

In general, groups are sets $G$ together with some product $\cdot$ defined that follows three rules:

(identity) There is an identity element $e\in G$ whose product with any element $g\in G$ gives the same element $g$ .
(closure under product) Taking the product of two elements $g,h\in G$ gives you another element in the set, denoted $gh\in G$ .
(closure under inverse) Every element $g\in G$ has an inverse in the set, denoted $g^{-1}\in G$ , such that the product of any element with their inverse is the identity.

When these three are true, $(G,\cdot)$ is known as a group. For example, the integers under addition $(\ZZ,+)$ form a group. The following articles will dive into all aspects of groups, which I hope you’ll find interesting.

In this section, we outline some properties of groups.

Every group has a product $\cdot$ but we usually write $g\cdot h$ as just $gh$ .

There are a couple properties we know just from the axioms of a group:

The product of any $g$ with the identity $e$ is equal to $g$ .
The product of any $g$ with its inverse $g^{-1}$ is equal to $e$
The inverse of a product $(gh)^{-1}$ is equal to $h^{-1}g^{-1}$ .

Since $(h^{-1}g^{-1})(gh)=h^{-1}g^{-1}gh=h^{-1}h=e$ , it follows that $h^{-1}g^{-1}$ and $gh$ are inverses.

We will talk about groups in the abstract in the following explorations, where we will use these properties extensively.

lemmas we need

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Exploration 1: Commutativity