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It is easy to verify that the set $(\mathbb{Z}/n\mathbb{Z})^\times$ is closed under multiplication in the sense that $a, b ∈ (\mathbb{Z}/n\mathbb{Z})^\times$ implies $ab ∈ (\mathbb{Z}/n\mathbb{Z})^\times$, and is closed under inverses in the sense that $a ∈ (\mathbb{Z}/n\mathbb{Z})^\times$ implies $a^{-1} ∈ (\mathbb{Z}/n\mathbb{Z})^\times$.

The question is the following:

Firstly, are $a$ and $b$ referring to each equivalence class of integers modulo $n$?

Secondly, by $a^{-1}$, what is this referring to? If $a$ is the equivalence class, I cannot see (or I am not sure) how I can make inverse set.

mookid
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Absolutio Zaka
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6 Answers6

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To the first question, just quoting your intro, "$a,b\in(\mathbb{Z}/n\mathbb{Z})^{\times}$". So if $\mathbb{Z}/n\mathbb{Z}$ is a set of equivalence classes, then yes, $a$ and $b$ are equivalence classes.

To the second question, $a^{-1}$ is an equivalence class such that $aa^{-1}$ is the equivalence class of $1$. In general, there is no quick formula to find an element of $a^{-1}$ given $a$. Instead, since a representative $A$ of $a$ and $n$ are relatively prime, the Euclidean algorithm provides solutions to the equation $$AB + nt = 1$$ Then mod $n$, $AB\equiv 1$. So the Euclidean algorithm will lead you to a representative of $a^{-1}$.

Now, to back-peddle a little bit, actually there is a rather simple formula for a representative of $a^{-1}$, given a representative $A$ of $a$. Take $B=A^{\varphi(n)-1}$, where $\varphi$ is Euler's totient function. Then $AB=A^{\varphi(n)}\equiv1\mod{n}$. The problem with this "simple" formula is that it's not computationally efficient. For example, if $n=97$, the Euclidean algorithm quickly tells us that $2^{-1}=49$. But this formula would give us $2^{96}$, which is tiresome to compute even if we reduce at every multiplication.

2'5 9'2
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The notation $a \in (\mathbb{Z} / n \mathbb{Z})^\times$ means the variable $a$ is being used to denote an element of $(\mathbb{Z} / n \mathbb{Z})^\times$.

If you are representing elements of $(\mathbb{Z} / n \mathbb{Z})^\times$ as equivalence classes, then $a$, being an element of $(\mathbb{Z} / n \mathbb{Z})^\times$, has a representation as an equivalence class.

The structure $(\mathbb{Z} / n \mathbb{Z})^\times$ is an abelian group. It has a group operation, an inverse, and a neutral element. As we are writing this group with multiplicative notation, we adopt the default notation ${}^{-1}$ for the inverse operation of the group.

As an aside, I think focusing on an object "being" an equivalence class really obscures the simplicity of what's going on. Equivalence classes are frequently just a technical set-theoretic trick.

One rather good notation for elements of the ring $\mathbb{Z} / n \mathbb{Z}$ is that elements are represented by integers. Not just integers in the range $\{ 0, 1, \cdots, n-1\}$, but any integer. The same element has many notations; for example, '3' and '10' are two separate notations for the same element of $\mathbb{Z} / 7 \mathbb{Z}$.

It may help to add a decoration to help keep track of when you are using '3' to denote an element of $\mathbb{Z}$ and when you are using it to denote an element of $\mathbb{Z} / 7 \mathbb{Z}$: common choices are $\overline{3}$ and $[3]$, or sometimes $[3]_7$.

If you are representing the elements of $\mathbb{Z} / n \mathbb{Z}$ as equivalence classes, then the element $[3]$ is notation for the element represented by the equivalence class of 3. Despite that fact, you usually shouldn't be thinking "$[3]$ is an equivalence class": you should be thinking "$[3]$ is the element of $\mathbb{Z} / n \mathbb{Z}$ that comes from the integer 3".

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Well $2\cdot 3\equiv 1\; \text{mod}\ 5$, so $2$ and $3$ are multiplicative inverses $\text{ mod } 5$.

How to find the inverse of a number modulo a prime number was the topic of one of my previous answers. Modulo a composite number, inverses don't always exist.

See Calculating the Modular Multiplicative Inverse without all those strange looking symbols for the way to find the inverse of $322$ mod $701$. It turns out to be $455$

Community
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Michael Hardy
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It is also standard to identify $(\mathbf Z/n \mathbf Z)$, which is typically defined algebraically as a set of equivalence classes, with the integers $\{0, \dots, n-1 \}$. One thing to note is that $(\mathbf Z/n \mathbf Z)$ is essentially finitary, so the representation as integers is sometimes useful (but less convenient algebraically)

David Harris
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First: yes, $a$ and $b$ represent the equivalence classes modulo $n$.

Second: Well, $a^{-1}$ is not the inverse as a rational number. Rather, it is the unique equivalence class modulo $n$ (if it exists) such that $aa^{-1}$ is congruent to 1 modulo $n$. And for existence, this inverse exists if and only if $a$ is coprime to $n$.

M Turgeon
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Firstly, yes.

Secondly, by $a^{-1}$ we mean the unique equivalence class such that multiplication of (the equivalence class) $a$ with $a^{-1}$ gives (the equivalence class of) $1.$

For example in $(\mathbb Z/4\mathbb Z)^\times=\{1,3\}$ we have $1\cdot 1\equiv 1$ so $1^{-1}=1$ and $3\cdot 3=9\equiv 1$ so $3^{-1}=3.$

Andrew
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