Today, to wrap things up, we will use Lagrange’s Theorem to prove that if is an element of the group , the order of evenly divides the order of .

So we have a group and an element . In order to apply Lagrange’s Theorem we need a subgroup. Where are we going to get one of those? We will have to make one using .

Suppose , that is, is the smallest natural number such that . Then consider the set

We will denote this set by . Clearly is a subset of . But I claim it is actually a subgroup!

To prove this, let’s first introduce the notation . This fits nicely: for example, notice that for positive and , we have ; this property continues to hold even when or are if we define . (As an aside, notice this also fits nicely with the notation for the inverse of —and we can extend it to .) With this notation, our set is just . Now, let’s show that is a subgroup of :

- It is obviously nonempty.
- It is closed under the group operation. Generally speaking, . If then all is well. On the other hand, if ends up greater than , we can always subtract until we are left with something less than , since . For example, if , then . In general, we can say that . You might notice that this is very much like the group we have been using as an example—in fact, when the group is
*isomorphic*to (that is, in some sense they are really “the same” group); more generally, if then is isommorphic to , since powers of are added . - It has inverses: the inverse of is , since . (This makes sense even when .)

This subgroup is called the *cyclic subgroup generated by* . In any group, we can generate a subgroup from each element in this way (though the same cyclic subgroup may be generated by multiple distinct elements). For example, in the group , the element generates the trivial one-element subgroup; generates the subgroup ; and both generate the subgroup ; and , , , and all generate the entire group , as you can check.

One more important thing to note is that every element of is distinct: remember, if then —but is the smallest power of that is equal to the identity. So this group really does have distinct elements.

Now we have a subgroup of , and Lagrange’s Theorem tells us that its order must be a divisor of the size of . But the order of is the same as the order of , that is, . So the order of any element must divide the order of the whole group.

And that concludes MaBloWriMo! This month of posting every day has been a lot of fun—I’m very glad I took up the challenge! I have these grand visions of stuff I want to write about, but often get bogged down writing long, monolithic posts, or sometimes I am just intimidated by the thought of it and never even start. This month, it was freeing to realize that the world would not end if I just post things in smaller chunks, perhaps without quite so much advance planning or editing. Posting frequently also meant I could maintain a lot more momentum—writing the individual posts went pretty quickly, since everything was still fresh in my head. Paradoxically, it seems that upping the posting frequency actually made things *easier*.

So I certainly can’t keep writing a post per day. But going forward I have decided to commit to **two posts per week**—I think that will be manageable while still taking advantage of the momentum generated by more frequent posting. I’m excited—I’ve got a bunch of ideas for things I want to write about, some new, and some that have been on the list for years. A small sampling: I want to explain the million-dollar P vs NP problem; write about the proof of the four-color theorem; finish my long-languishing series of posts on a certain combinatorial proof; and write some addenda to my series on hyperbinary numbers. I hope you enjoy—and please do continue to send ideas of things you might like me to write about!

I really enjoyed following you this month. Thank you!

Very glad to hear it! Thanks for leaving a comment.

Following this series has made a very stressful month a little bit more enjoyable. Thank you.

I’d love to see a continuation of the What I Do series.

You’re very welcome! I’m glad I could bring you some enjoyment.

Ah, yes, the What I Do series. That one is a bit more intimidating since it will require some deeper reflection to figure out what exactly I want to say. But I’ll try.

I’ve enjoyed this series of posts too. Not least because I’m reading a book on abstract algebra at the moment. I like your lighthearted, yet sufficiently rigorous presentation style 🙂 Thank you

Thanks, I’m glad you enjoyed it!