The Additive Group of Integers Modulo n, Z/nZ

Recall from the Groups page that a group a set $G$ with a binary operation $\cdot : G \times G \to G$ where:

• 1) For all $a, b, c \in G$ we have that $(a \cdot b) \cdot c = a \cdot (b \cdot c)$ (Associativity of $\cdot$).

• 2) There exists an element $e \in G$ such that $a \cdot e = a$ and $e \cdot a = a$ (The existence of an identity for $\cdot$).

• 3) For all $a \in G$ there exists a $a^{-1} \in G$ such that $a \cdot a^{-1} = e$ and $a^{-1} \cdot a = e$ (The existence of inverses for each element in $G$).

We will now look at the additive group of integers modulo $n$ denoted $\mathbb{Z}/n\mathbb{Z} = \{ 0, 1, 2, ..., n - 1 \}$ for integers $n > 1$. For each of these groups, define the operation of $+$ for $x, y \in \mathbb{Z}/n\mathbb{Z}$ by $x + y$ to equal $x + y \pmod n$ (where the righthand $+$ denotes usual addition).

The operation $+$ above has a very nice interpretation. For any $n > 1$ we can interpret the operation of $+$ on $\mathbb{Z}/n\mathbb{Z}$ for $x + y$ by considering a circle numbered $0$ through $n - 1$ clockwise, starting at $x$, and then traveling clockwise around the circle a distance of $y$ to get $x + y$ as illustrated in the following diagram:

We will now show that $(\mathbb{Z}/n\mathbb{Z}, +)$ is a group. For $x, y \in \mathbb{Z}/n\mathbb{Z}$ we have the $x + y$ is the remainder of the sum $x + y$ on division by $n$. The only possible remainders are $0, 1, ..., n - 1$ and so $(x + y) \in \mathbb{Z}/n\mathbb{Z}$, so $\mathbb{Z}/n\mathbb{Z}$ is closed under $+$.

It's not hard to intuitively see that the remainder of $x + (y + z)$ upon division by $n$ is equal to the remainder of $(x + y) + z$ for all $x, y, z \in \mathbb{Z}/n\mathbb{Z}$, so indeed, $x + (y + z) = (x + y) + z$, so $+$ is associative.

For each $x \in \mathbb{Z}/n\mathbb{Z}$ we note that the remainder of $x + 0$ is equal to the remainder of $x$ upon division by $n$ and the remainder of $0 + x$ is equal to the remainder of $x$ upon division by $n$. In other words, $x + 0 = x$ and $0 + x = x$. Furthermore, $0 \in \mathbb{Z}/n\mathbb{Z}$, so $0$ is the identity element.

Lastly, if $x \in \mathbb{Z}/n\mathbb{Z}$ then $-x \in \mathbb{Z}/n\mathbb{Z}$ is such that $x + (-x) = 0$ and $(-x) + x = 0$.

Thus we conclude that $(\mathbb{Z}/n\mathbb{Z}, +)$ is a group.