Abelian Groups

# Abelian Groups

Recall from the Groups page that a group is a set $G$ paired 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 groups that contain an addition property - namely that the operation $\cdot$ is commutative.

 Definition: If $(G, \cdot)$ is a group where $\cdot$ is commutative for all $a, b \in S$, i.e., $a \cdot b = b \cdot a$ for all $a, b \in G$, then $(G, \cdot)$ is called an Abelian Group.

The term "Commutative Group" means the same thing as "Abelian Group".

The group $(\mathbb{R}, +)$ and $(\mathbb{C}, +)$ are clearly abelian groups as the reader should verify. One such example of a group that is not abelian is the group $(G, \cdot)$ where $G$ is the set of $2 \times 2$ matrices with real entries whose determinants are nonzero and $\cdot$ is defined to be matrix multiplication.

We first show that $G$ is not commutative. Consider the following matrices $A, B \in S$ where $A = \begin{bmatrix} 1 & 1\\ 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 1& 1\\ 1 & 0 \end{bmatrix}$. Then:

(1)
\begin{align} \quad A \cdot B = \begin{bmatrix} 1 & 1\\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1& 1\\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 2 & 1\\ 1 & 0 \end{bmatrix} \end{align}

We also have that

(2)
\begin{align} B \cdot A = \begin{bmatrix} 1& 1\\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 1\\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0& 2\\ 1 & 1 \end{bmatrix} \end{align}

Therefore $A \cdot B \neq B \cdot A$ and $\cdot$ is not commutative on $G$.

We now show that the three properties of a group hold. From linear algebra, recall that matrix multiplication is associative and so for all $A, B, C \in S$ we have that $A \cdot (B \cdot C) = (A \cdot B) \cdot C$.

The identity element in $G$ is the matrix $I = \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$ (where $\det I = 1 \neq 0$. Furthermore, since $\det A, \det B \neq 0$ we have that inverse matrices $A^{-1}, B^{-1}$ exist where $\det A^{-1}, \det B^{-1} \neq 0$ and $AA^{-1} = I$ and $A^{-1}A = I$ for all $A \in S$.

The product of any two $2 \times 2$ matrices is a $2 \times 2$ matrix, and since inverse elements exist, we have that $A*B \in S$ because the the product $B^{-1} \cdot A^{-1}$ has $\det (B^{-1} \cdot A^{-1}) \neq 0$ implying $\det (A \cdot B) \neq 0$, so $G$ is closed under the operation $\cdot$.

Therefore $(G, \cdot)$ is a group that is not abelian.