# Introductory Abstract Algebra – Group Theory, Part II

Hi all! I’m going to start off from where I left off – cyclic groups. Let $G$ be a group and $S$ be a subset of $G$. Then the smallest subgroup $H$ of $G$ containing $S$ is called the subgroup of $G$ generated by $S$. This definition makes sense because you can choose $H$ to be the intersection of all subgroups of $G$ containing $S$ (check that the intersection of subgroups is itself a subgroup). For the advanced reader: many such structures arise this way. For example, recall that in my post on chunks we had defined the weakly saturated class of morphisms generated by a set $S$ to be the smallest weakly saturated class of morphisms containing $S$. Note that this too has the gist that it’s the smallest something containing something. But I’m digressing!

If we let $H=G$ and there is a subset $S$ of $G$ generating $H$ then we say that $S$ generates $G$. That makes sense, right? Now if $S$ simply contained one element $a$, then $G$ is called cyclic and we write $G=\langle a\rangle$. This may seem quite complicated, but it isn’t. Really. Again examples are all over the place. Consider the integers (under addition), for example. We observe that $\mathbb{Z}$ is generated by $1$, and is an infinite cyclic group. In other words, $\mathbb{Z}=\{1^i|i\in\mathbb{Z}\}$, where by $1^2$ we mean $1+1$. In fact, any infinite cyclic group is isomorphic (think of this as saying that there is an invertible map to the integers such that the group structure is preserved by that map; this is called an invertible group homomorphism) to $\mathbb{Z}$! This is in fact a result from a more general case, which we leave to the reader to prove:

Proposition 1: If $G$ is a cyclic group generated by an element $a$ then $G=\{a^i|i\in\mathbb{Z}\}$. In fact any cyclic group arises this way.

Let $g\in G$. Consider the subgroup generated by $g$. This is a cyclic group, and its order is called the order of $g$. Take the following result as a black box for now; we’ll prove it after introducing the important result known as Lagrange’s Theorem.

Proposition 2: If $G$ is finite then the order of any element of $g$ divides $|G|$.

Let us consider the group of order 3: $\{1,2,3\}$. This is simply the integers mod 3. Under mod 3 addition, $1$ generates this group. Can this be generalized? I.e. is any group of prime order cyclic? Let’s try proving this.

So let us suppose $g\in G$ such that if $e$ is the identity of $G$ then $g\neq e$. Now $g$ does not have order $1$ since it isn’t the identity. So by Prop. 2 it divides $|G|$. But wait! $|G|$ is prime! So the order of $g$ is either $1$ or $|G|$. It’s not 1, so it’s $|G|$. In other words the order of the subgroup generated by $g$ is $|G|$. But this is $G$ itself, so … $G$ is cyclic!

Proposition 3: Any group of prime order is cyclic.

In fact, we can classify all cyclic groups! See Theorem 217 here for that. For those interested in reading more on these things I refer you here (I wanted to introduce the concepts and give motivation, not write up lecture notes! :-P). We’d skipped over the proof of Prop. 2, and I really want to prove it (actually a generalization of it) – this involves things known as cosets. We’ll introduce those next time!

S.D. (N.K.)