# The Integers Modulo n

1. Sep 2, 2006

### BSMSMSTMSPHD

We are currently using Dummit & Foote's Abstract Algebra book in a gradute course of the same name. Recently, I had an issue concerning the additive and multiplicative integer groups mod n, which I brought to the professor's attention. The issue deals specifically with the way these groups are defined in the text.

$$\mathbb{Z} / n\mathbb{Z} = \{ \overline{a} \ | \ a \in \mathbb{Z}, 0 \leq a < n \}$$

Thus, for any Natural Number n, there are exactly n equivalence classes in $$\mathbb{Z} / n\mathbb{Z} \ ,$$ namely:

$$\overline{0}, \overline{1}, ..., \overline{n - 1}$$

This is defined on page 8.

Later, on page 10, the group of integers modulo n under multiplication is defined as:

$$( \mathbb{Z} / n\mathbb{Z} ) ^x = \{ \overline{a} \in \mathbb{Z} / n\mathbb{Z} \ | \ \exists \ \overline{c} \in \mathbb{Z} / n\mathbb{Z} \$$ with $$\ \overline{a} \cdot \overline{c} = \overline{1} \}$$

So, here's my question. What are the elements of this group when n = 1?

I have no problems with the groups for larger n. In each case, you just get the set of equivalence classes in the additive group such that their representative values are relatively prime to n. This is given as a proposition on the same page.

It's easy to see that if n is prime, then this group contains n - 1 elements. The number of elements for any n can be calculated by Euler's totient function $$\varphi(n)$$ which gives the number of integers less than or equal to n that are relatively prime to n.

Since $$\varphi(1) = 1$$ it should be that the group of integers mod 1 under multiplication should contain 1 element. But, if we follow the definition from above, it seems to me that it should, in fact, be empty!

The only element in the additive group mod 1 is $$\overline{0}$$, but this equivalence class fails to meet the requirement for being in the multiplicative group! That is, $$\nexists \ \overline{c} \in \mathbb{Z} / n\mathbb{Z} \$$ with $$\ \overline{0} \cdot \overline{c} = \overline{1}$$

So it would seem that the group is actually empty.

So which is it?

Last edited: Sep 2, 2006
2. Sep 2, 2006

### AKG

Why should it be empty by that definition? When n = 1, then:

$$\overline{1} = \overline{0}$$.

In other words, 0 and 1 belong to the same equivalence class modulo 1. In fact, every integer belongs to the same equivalence class modulo 1. $\mathbb{Z}/1\mathbb{Z} = \{\mathbb{Z}\}$.

Last edited: Sep 2, 2006
3. Sep 2, 2006

### BSMSMSTMSPHD

I thought $$\mathbb{Z} / 1\mathbb{Z} = \{ \overline{0} \}$$

4. Sep 2, 2006

### AKG

Yes, and:

$$\overline{0} = \mathbb{Z}$$

5. Sep 2, 2006

### AKG

Modulo n, $\bar{c}$ denotes the equivalence class of c, which is {z in Z : n|(c-z)}. So modulo 1, $\bar{0}$ = {z in Z : 1|(0-z)} = {z in Z : 1|-z} = {z in Z} = Z, because 1 divides every integer. That is, if z is any integer, then there exists an integer k such that k x 1 = z, namely choose k = z.

6. Sep 2, 2006

### BSMSMSTMSPHD

Okay, you've shed some light on the subject for me, but I'm still iffy...

So, what is $$( \mathbb{Z} / 1\mathbb{Z} ) ^x$$ ?

Are you saying that it contains only one element since every integer is in the same equivalence class?

7. Sep 2, 2006

### AKG

$$(\mathbb{Z}/1\mathbb{Z})^{\times} = \mathbb{Z}/1\mathbb{Z}$$

These two things are identical as groups. That is, they both have the same underlying set (which, yes, has only one element) but also have the same operation. Although we look at the thing on the left and call its operation multiplication (mod 1) and we look at the thing on the right and call its operation addition (mod 1), they are technically the same operation in this special case. Just for additional information, and to be entirely clear, the underlying set for this group is:

$$\{\mathbb{Z}\} = \{\overline{0}\} = \{\overline{1}\} = \{\overline{125853}\} = \{\overline{-823}\}$$

8. Sep 2, 2006

### AKG

Also, why wouldn't you put this thread in the Linear and Abstract Algebra forum instead of in the Calculus and Analysis forum, which this thread has nothing to do with?

9. Sep 2, 2006

### BSMSMSTMSPHD

Oh whoops. Ha ha... I know why I did that. Because I've been taking analysis courses all year and I'm just used to coming to this link. Sorry.

I do understand it now, though. I think it finally hit me while I was at the store today. The key was realizing that these equivalence classes can have different names, and that, most importantly, when n = 1 only, 0 and 1 are in the same equivalence class (as you already stated).

So, for example, $$\mathbb{Z} / 2\mathbb{Z} = \{ \overline{0}, \overline{1} \}$$

But, I could also call it $$\{ \overline{6}, \overline{3} \}$$ or $$\{ \overline{-50}, \overline{2001} \}$$ or, in general $$\{ \overline{2k}, \overline{2m+1} \}$$ for integers k and m.

When n = 1, all of the integers are in the same equivalence class. So, like you said, it doesn't matter which integer I use to represent it.

Thanks for your help and patience. I diserve to be ridiculed for this given its simplicity and where I put the post, but you did a great job.

10. Sep 2, 2006

### AKG

You're very welcome, glad I could help!
Yup, you seem to get it now. Just note one thing: when you write Z/2Z, it's implied you're talking about a group, that is, a set WITH an operation. When you write {6~, 3~} (the tildes denote overlines) that denotes a set consisting of two equivalence classes, and it would be the underlying set of the group Z/2Z. Of course, if the context is right, then you can refer to a group by its underlying set, so you will get away with writing Z/2Z = {6~, 3~} 99.99% of the time. I would assume you know this already, but I'm covering all bases just to be safe.