MaBloWriMo 12: Groups and Order

Continuing our discussion of groups (see here and here), today I want to discuss the concept of order, which is defined both for groups themselves and for the elements of a group.

The order of a group simply means the size of the set G. So instead of “a group with 8 elements” you will often hear “a group of order 8” instead.

Elements of a group also have an order. Recall the example \mathbb{Z}_8, the group with elements \{0, 1, \dots, 7\}, and with binary operation addition \pmod 8. Consider what happens if we start with 1 and keep combining it with itself according to the group operation. 1 +_8 1 = 2, then 1 +_8 1 +_8 1 = 3, and so on, until we get to adding eight copies of 1, giving 1 +_8 \dots +_8 1 = 0. If we add up copies of 2, on the other hand, it takes only 4 copies to reach 0. What about 3? Well, 3 +_8 3 = 6; three copies yield 1; four copies yield 4; and so on. As you can verify, we actually need eight copies of 3 before we get to 0.

In general, if G is a group and g \in G is some element of the group, the order of g is defined as the smallest number of copies of g which combine to yield the identity element. So in \mathbb{Z}_8, 1 has order eight, 2 has order four, 3 also has order eight, and so on. 0 itself has order 1, because one copy of 0 is already 0.

A few questions immediately suggest themselves:

  1. Does every group element have a well-defined order?
  2. How does the order of group elements relate to the order of the group?

To answer the first question, recall that the integers form a group under addition, and obviously if you start with some nonzero integer n, you can add n to itself as many times as you like but you will never get zero! In such cases we say that the order is infinite.

OK, but the group \mathbb{Z} itself has infinitely many elements. Let’s refine our question a bit:

  1. (revised) Does every element of a finite group have a well-defined, finite order?

Even in a finite group, you could imagine having some element g \in G such that no matter how much you combine it with itself, you will never get the identity element. Can this happen? (Also, as a tangential challenge, can you come up with an example of an infinite group where some elements other than the identity do have a finite order?)

For now, I’ll let you think about these questions! Tomorrow, we’ll answer question 1; the next day, we’ll talk about a simple answer to question 2 which will be sufficient for our purposes (there is also a more nuanced answer which is harder to prove).

As an aside, remember that we’re trying to prove the Lucas-Lehmer test, and currently we’re trying to understand the related number \omega = 2 + \sqrt{3}. The point of all this stuff about groups is that (1) we’re going to construct a group containing \omega as an element, and then (2) we’re going to figure out something about the order of \omega in that group. We’ll see that if the Lucas-Lehmer test didn’t work, then the order of \omega would be in contradiction to the answer to question 2.

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About Brent

Assistant Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.
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One Response to MaBloWriMo 12: Groups and Order

  1. Pingback: MaBloWriMo 13: Elements of finite groups have an order | The Math Less Traveled

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