The Abelian Group of Even Integers

# The Abelian Group of Even Integers

Recall from the Abelian Groups page that an Abelian group is a group with the extra property of commutativity on its operation $\cdot$. That is, an Abelian group is a set $G$ paired with a binary operation $\cdot : G \times G \to G$ where:

• For all $a, b \in G$ we have that $(a \cdot b) \in G$ (Closure under $\cdot$).
• For all $a, b, c \in G$ we have that $(a \cdot b) \cdot c = a \cdot (b \cdot c)$ (Associativity of $\cdot$).
• There exists an element $e \in G$ such that $a \cdot e = a$ and $e \cdot a = a$ (The existence of an identity for $\cdot$).
• 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 $G$).
• For all $a, b \in G$ we have that $a \cdot b = b \cdot a$ (Commutativity of $\cdot$).

We will now look at the Abelian group of even integers denoted $2 \mathbb{Z}$.

Let $2\mathbb{Z}$ be the set defined as:

(1)
\begin{align} \quad 2\mathbb{Z} = \{ z \in \mathbb{Z} : z = 2n \: \mathrm{where} \: n \in \mathbb{Z} \} = \{0, \pm 2, \pm 4, ... \} \end{align}

Define the operation $\cdot$ as standard addition so that for all $a, b \in 2\mathbb{Z}$ we have:

(2)
\begin{align} \quad a \cdot b = a + b \end{align}

We will now check all of the group axioms to verify that $2 \mathbb{Z}$ is a group under $\cdot$. Let $a, b, c \in \mathbb{Z}$. Then $a = 2n_1$, $b = 2n_2$, and $c = 2n_3$ for $n_1, n_2, n_3 \in \mathbb{Z}$. We have that $2 \mathbb{Z}$ is closed under $\cdot$ since:

(3)
\begin{align} \quad a \cdot b = 2n_1 + 2n_2 = 2(n_1 + n_2) \in 2 \mathbb{Z} \end{align}

We also have that $\cdot$ is commutative since:

(4)
\begin{align} \quad a \cdot (b \cdot c) = 2n_1 \cdot (2n_2 + 2n_3) = (2n_1 + 2n_2) + 2n_3 = (a \cdot b) \cdot c \end{align}

The identity element for $\cdot$ is the even integer $0 = 0n$ since:

(5)
\begin{align} \quad a \cdot 0 = 2n_1 + 0 = 2n_1 = a \end{align}

And similarly:

(6)
\begin{align} \quad 0 \cdot a = 0 + 2n_1 = 2n_1 = a \end{align}

For each $a \in 2\mathbb{Z}$ we have that the inverse of $a$ is $-a = -2n_1 = 2(-n_1) \in 2 \mathbb{Z}$ since:

(7)
\begin{align} \quad a \cdot (-a) = 2n_1 + (2(-n_1)) = 2(n_1 - n_1) = 2(0) = 0 \end{align}

And similarly:

(8)
\begin{align} \quad (-a) \cdot a = (2(-n_1)) + 2n_1 = 2(-n_1 + n_1) = 2(0) = 0 \end{align}

Therefore $(2\mathbb{Z}, \cdot)$ forms a group. We now only need to show that commutativity for $\cdot$ holds to show that $2 \mathbb{Z}$ is an Abelian group. We have that:

(9)
\begin{align} \quad a \cdot b = 2n_1 + 2n_2 = 2n_2 + 2n_1 = b \cdot a \end{align}

Therefore $(2\mathbb{Z}, \cdot)$ is an Abelian group.