Basic Theorems Regarding Abelian Groups

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:

• 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 $G$).

• 4) For all $a, b \in G$ we have that $a \cdot b = b \cdot a$ (Commutativity of $\cdot$).

We will now look at some basic results regarding abelian groups.

Proposition 1: Let $(G, \cdot)$ be a group. If for all $a, b \in G$ we have that $(a \cdot b)^2 = a^2 \cdot b^2$ then $(G, \cdot)$ is an abelian group.

• Proof: Suppose that for all $a, b \in G$ we have that $(a \cdot b)^2 = a^2 \cdot b^2$. Then for all $a, b \in G$ we have that:

(1)

\begin{align} \quad (a \cdot b)^2 &= a^2 \cdot b^2 \\ \quad (a \cdot b) \cdot (a \cdot b) &= (a \cdot a) \cdot (b \cdot b) \\ \quad a^{-1} \cdot [(a \cdot b) \cdot (a \cdot b)] &= a^{-1} \cdot [(a \cdot a) \cdot (b \cdot b)] \\ \quad (a^{-1} \cdot a) \cdot [b \cdot (a \cdot b)] &= (a^{-1} \cdot a) \cdot [a \cdot (b \cdot b)] \\ \quad e \cdot [b \cdot (a \cdot b)] &= e \cdot [a \cdot (b \cdot b)] \\ \quad b \cdot (a \cdot b) &= a \cdot (b \cdot b) \\ \quad [b \cdot (a \cdot b)] \cdot b^{-1} &= [a \cdot (b \cdot b)] \cdot b^{-1} \\ \quad (b \cdot a) \cdot (b \cdot b^{-1}) &= (a \cdot b) \cdot (b \cdot b^{-1}) \\ \quad (b \cdot a) \cdot e &= (a \cdot b) \cdot e \\ \quad b \cdot a &= a \cdot b \end{align}

• Hence, $(G, \cdot)$ is an abelian group. $\blacksquare$

Proposition 2: Let $(G, \cdot)$ be a group. Then $G$ is an abelian group if and only if for all $a, b \in G$ we have that $(a \cdot b)^{-1} = a^{-1} \cdot b^{-1}$.

• Proof: $\Rightarrow$ Suppose that $G$ is an abelian group. Let $a, b \in G$. Then $(a \cdot b)^{-1} = b^{-1} \cdot a^{-1}$ and since $G$ is abelian, $b^{-1} \cdot a^{-1} = a^{-1} \cdot b^{-1}$. Thus $(a \cdot b)^{-1} = a^{-1} \cdot b^{-1}$.

• $\Leftarrow$ Suppose that for all $a, b \in G$ we have that $(a \cdot b)^{-1} =a^{-1} \cdot b^{-1}$. Then for all $a, b \in G$ we have that $b^{-1} \cdot a^{-1} = a^{-1} \cdot b^{-1}$. So:

(2)

\begin{align} \quad b^{-1} \cdot a^{-1} &= a^{-1} \cdot b^{-1} \\ \quad a \cdot b^{-1} \cdot a^{-1} &= b^{-1} \\ \quad b \cdot a \cdot b^{-1} \cdot a^{-1} &= e \\ \quad b \cdot a \cdot (a \cdot b)^{-1} &= e \\ \quad b \cdot a &= \cdot a \cdot b \end{align}

• Since this holds for all $a, b \in G$ we have that $G$ is an abelian group. $\blacksquare$

Proposition 3: Let $(G, \cdot)$ be a group. If for all $a \in G$ we have that $a = a^{-1}$ then $(G, \cdot)$ is an abelian group.

• Proof: Suppose that for all $a \in G$ that $a = a^{-1}$. Then $b \in G$ implies that $b = b^{-1}$. So for all $a, b \in G$ we have that:

(3)

\begin{align} \quad a \cdot b = a^{-1} \cdot b^{-1} = (b \cdot a)^{-1} \end{align}

• But $(b \cdot a) \in G$ since $a, b \in G$ and by definition, $G$ is closed under $\cdot$. Therefore $b \cdot a = (b \cdot a)^{-1}$ so:

(4)

\begin{align} \quad a \cdot b = b \cdot a \end{align}

• Hence, for all $a, b \in G$, $a \cdot b = b \cdot a$, so $(G, \cdot)$ is an abelian group. $\blacksquare$