Hi all! I’m going to start off from where I left off – cyclic groups. Let be a group and be a subset of . Then the smallest subgroup of containing is called the subgroup of generated by . This definition makes sense because you can choose to be the intersection of all subgroups of containing (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 to be the smallest weakly saturated class of morphisms containing . Note that this too has the gist that it’s the smallest *something* containing *something*. But I’m digressing!

If we let and there is a subset of generating then we say that generates . That makes sense, right? Now if simply contained one element , then is called cyclic and we write . 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 is generated by , and is an infinite cyclic group. In other words, , where by we mean . 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 is a cyclic group generated by an element then . In fact any cyclic group arises this way.

Let . Consider the subgroup generated by . This is a cyclic group, and its order is called the order of . 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 is finite then the order of any element of divides .

Let us consider the group of order 3: . This is simply the integers mod 3. Under mod 3 addition, 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 such that if is the identity of then . Now does not have order since it isn’t the identity. So by Prop. 2 it divides . But wait! is prime! So the order of is either or . It’s not 1, so it’s . In other words the order of the subgroup generated by is . But this is itself, so … 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.)

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