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.)


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