# Introductory Abstract Algebra – Group Theory, Part I

Hi all! In the next few posts, I’ll introduce everything you might want to know about abstract algebra. After this, I’ll try introducing either commutative algebra or basic algebraic number theory. And then I’ll post notes on algebraic geometry as I learn it.

Anyhow, the first thing one learns in algebra is the notion of a group. Let us think of the simplest polygon that you can draw; for me, it’s a square. What are the symmetries of a square? Well, there’s obviously clockwise rotation about $90^\circ$ – let’s call this operation $\mathrm{Rot}$. Then we can compose $\mathrm{Rot}$ with itself twice, and that’s still a symmetry of the square. In other words, $\mathrm{Rot}\circ\mathrm{Rot}$, which we will denote by $\mathrm{Rot}^2$, is a symmetry of the square. Similarly, $\mathrm{Rot}\circ\mathrm{Rot}\circ\mathrm{Rot}$, which we denote by $\mathrm{Rot}^3$, is also a symmetry of the square.

Let’s now consider anticlockwise rotation about $90^\circ$; we will denote this operation by $\mathrm{Rot}^{-1}$. This notation makes sense because $\mathrm{Rot}^{-1}\circ\mathrm{Rot}$ is the “identity” symmetry – it does absolutely nothing to the square. What you should note is that $\mathrm{Rot}^{-1}$ is the same as $\mathrm{Rot}^3$!

We can also flip about one of the corners. Let us label the vertices of the square $1,2,3,4$. Then the flips can be denoted $F_1,F_2,F_3,F_4$, respectively. The interesting thing is that $F_i\circ \mathrm{Rot}^n$ is still always a symmetry of the square! (Can you find the inverse to each of these operations? It’s quite easy!) Let us stop for a moment and take a look at the data.

We have a collection of symmetries of the square, namely the powers of $\mathrm{Rot}$ and the flips, $F_i$, and an operation on them – composition – that when you compose any two of these, you can get something another symmetry, there’s an inverse to each symmetry, and there’s an “identity” which does not do anything to the square. That’s very interesting – are there other sets that exhibit such behavior? You don’t even have to look too far to find something like this!

Consider the integers, $\mathbf{Z}$, with the operation of addition, $+$. Then if you add two integers, you still get an integer; given an integer $n$ you can always find it’s additive inverse, $-n$; the number $0$ does not do anything at all to a number on addition. So $\mathbb{Z}$ also exhibits similar behavior. Similarly with $\mathbb{Q}$, the rationals, under addition, and $\mathbb{R}$, the rationals, under addition. (Why not multiplication? The reason for $\mathbb{Q}$ and $\mathbb{R}$ is the same, and for $\mathbb{Z}$, there’s an additional reason.) How about $\mathbb{C}$ under addition? Under multiplication?

This suggests that these properties are something special, and indeed, they are! They motivate the definition of a group:

Definition 1: A group is a set $G$ with a binary operation $\circ: G \times G \to G$ and a map $\circ^{-1}:G\to G$, called the multiplication (note this is used in a general sense, as the “multiplication” could be, say, addition) and the inverse, respectively, such that the following conditions are satisfied:

• The multiplication is associative, i.e., $x\times (y\times z)=(x\times y)\times z$. Since I’m a category theorist, this corresponds to a commutative diagram which I sadly cannot draw, but here’s a picture:

Just replace $m$ with the $\circ$ above and $id$ is the identity map. Check to see how these are equivalent.

• There’s an element $1$ that is the multiplicative identity for this multiplication, i.e., such that $1\times x=x\times 1=x$. (Although denoted “$1$“, this is simply notation for any element which “does nothing” under the binary operation)
• For any element $g$, there’s another element, called the inverse, that is the image of $g$ under $\circ^{-1}$, such that multiplication with that element gives the multiplicative identity, i.e., there is a $g^{-1}$ such that $g\times g^{-1}=g^{-1}\times g=1$.

In fact, one can show that if we replaced the square with any regular $n$-gon in our discussion above, and consider the symmetric rotations, then we get a group, called the dihedral group $D_n$ of order $n$. You can show that the operation that we defined as composition is not commutative; that is, if I rotate clockwise and then flip the $n$-gon, it will not be the same as flipping it and then rotating it. However, “multiplication” (in this case addition) is commutative for the integers, the reals, and the rationals. This motivates:

Definition 2: A group is abelian if the multiplication is commutative.

One can study the size of a group as well:

Definition 3: Let $(G,\circ)$ be a group.The order of the group $G$ is denoted $|G|$, and is the number of elements in $G$. It’s always a whole number.

You can figure out what a finite group is. Here’s an example of a finite group: let $S_n$ denote the collection of all permutations of the set $\{1,\cdots,n\}$. Under composition, this is a finite group of order $n!$ (prove it!).

Alright, let’s move on. You know that there are inclusions $\mathbb{Z}\subset\mathbb{Q}\subset\mathbb{R}\subset\mathbb{C}$, and the operation on all these groups are the same (here we are using the multiplication of addition on the set $\mathbb{C}$ – I just gave away the answer to two questions above!). We can think of this as a “chain” of subgroups, motivating us to state:

Definition 4: Let $G$ be a group and $H$ a subset of $G$. It is a subgroup if, when we restrict the operations of $G$ to $H$, then $H$ is itself a group.

Obviously, we’d like some easy-to-check conditions under which a subset of a group is a subgroup. Here’s a statement which we won’t prove here:

Proposition 5: Let $H$ be a subgroup of a group $G$. Then the identity of $H$ is the identity of $G$, and the inverse of an element in $H$ is the same as its inverse in $G$.

We’re actually more interested in the following statement:

Proposition 6: A subset $H$ of a group $G$ is a subgroup if and only if it is closed under multiplication and inverses.

To prove this, we first note that the forward direction is obvious (it is the definition of a group). The reader can check that the multiplication coming from $G$ is well-defined. We only need to show that $H$ has an identity. Let $x\in H$. Then $H$ is closed under inverses, so $x^{-1}\in H$. But it’s also closed under multiplication, so $x\times x^{-1} = 1_x$ is also in $H$. Thus, $H$ satisfies the axioms of a group.

Here are some examples. The first is the example we showed before: $\mathbb{Z}\subset\mathbb{Q}\subset\mathbb{R}\subset\mathbb{C}$. You can check that the conditions of Proposition 6 are satisfied. Another one: We can interpret any symmetry as a matrix. So let $\mathrm{GL}_2(\mathbb{R})$ denote the group of $2\times 2$ matrices with entries in $\mathbb{R}$. (This is called the “General Linear Group of $2 \times 2$ matrices”) Then $D_n$ is a subgroup of $\mathrm{GL}_2(\mathbb{R})$. Yet another one: Let $\mathrm{SL}_2(\mathbb{R})$ denote the group of $2\times 2$ matrices with entries in $\mathbb{R}$ whose determinant is 1. Then it’s a subgroup of $\mathrm{GL}_2(\mathbb{R})$.

A very important thing in abstract algebra is the manipulation of finite groups. Here’s an example that illustrates this:

Proposition 7: Let $G$ be a finite group and $H\subseteq G$ that is closed under multiplication. Then it’s a subgroup.

To prove this (following here), we will use the property that $G$ is finite. To save space, let’s suppress the multiplication, and write $xy$ for $x\times y$, and $x^n$ for $x\times x\times \cdots\times x$. By Proposition 6, it suffices to show that $H$ is closed under inverses. Suppose $g\in H$. If this is the identity, $1$, then its inverse is … itself, so it’s in $H$. So it’s safe to assume that $g\neq 1$. Recall that $H$ is closed under multiplication; so the infinite sequence $g,g^2,\cdots$ has all entries in $H$. But $G$, and therefore $H$, is finite – so there must be some $i,j$ for which $g^i=g^j$. Without loss of generality, we can assume $i. So, $g^{i-j}=1$; but $g\neq 1$, so $i-j\neq 1$. Let $k=i-j-1$. Then let $h=g^k$. We observe that $g\times h=g^{1+i-j-1}=g^{i-j}=1$; so $gh=1$. We’ve constructed an inverse of $g$, so we’re done.

We can now turn to an important class of groups – the cyclic groups. I think that this is already a lot of information for one post, so I guess I’ll write this up in the next post.

S.D. (J.T.)