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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.

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

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

[tex] \overline{0}, \overline{1}, ..., \overline{n - 1} [/tex]

This is defined on page 8.

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

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

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 [tex] \varphi(n) [/tex] which gives the number of integers less than or equal to n that are relatively prime to n.

Since [tex] \varphi(1) = 1 [/tex] 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 [tex] \overline{0} [/tex], but this equivalence class fails to meet the requirement for being in the multiplicative group! That is, [tex] \nexists

\ \overline{c} \in \mathbb{Z} / n\mathbb{Z} \ [/tex] with [tex] \ \overline{0} \cdot \overline{c} = \overline{1}[/tex]

So it would seem that the group is actually empty.

So which is it?

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

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

[tex] \overline{0}, \overline{1}, ..., \overline{n - 1} [/tex]

This is defined on page 8.

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

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

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 [tex] \varphi(n) [/tex] which gives the number of integers less than or equal to n that are relatively prime to n.

Since [tex] \varphi(1) = 1 [/tex] 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 [tex] \overline{0} [/tex], but this equivalence class fails to meet the requirement for being in the multiplicative group! That is, [tex] \nexists

\ \overline{c} \in \mathbb{Z} / n\mathbb{Z} \ [/tex] with [tex] \ \overline{0} \cdot \overline{c} = \overline{1}[/tex]

So it would seem that the group is actually empty.

So which is it?

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