Basic Theorems Regarding Abelian Groups

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 $*$. That is, an Abelian group is a set $G$ paired with a binary operation $* : G \times G \to G$ where:

  • For all $a, b \in G$ we have that $(a * b) \in G$ (Closure under $*$).
  • For all $a, b, c \in G$ we have that $(a * b) * c = a * (b * c)$ (Associativity of $*$).
  • There exists an element $e \in G$ such that $a * e = a$ and $e * a = a$ (The existence of an identity for $*$).
  • For all $a \in G$ there exists a $a^{-1} \in G$ such that $a * a^{-1} = e$ and $a^{-1} * a = e$ (The existence of inverses for each element in $G$).
  • For all $a, b \in G$ we have that $a * b = b * a$ (Commutativity of $*$).

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

Theorem 1: Let $(G, *)$ be a group. If for all $a, b \in G$ we have that $(a * b)^2 = a^2 * b^2$ then $(G, *)$ is an abelian group.
  • Proof: Suppose that for all $a, b \in G$ we have that $(a * b)^2 = a^2 * b^2$. Then for all $a, b \in G$ we have that:
(1)
\begin{align} \quad (a * b)^2 = a^2 * b^2 \\ \quad (a * b) * (a * b) = (a * a) * (b * b) \\ \quad a^{-1} * [(a * b) * (a * b)] = a^{-1} * [(a * a) * (b * b)] \\ \quad (a^{-1} * a) * [b * (a * b)] = (a^{-1} * a) * [a * (b * b)] \\ \quad e * [b * (a * b)] = e * [a * (b * b)] \\ \quad b * (a * b) = a * (b * b) \\ \quad [b * (a * b)] * b^{-1} = [a * (b * b)] * b^{-1} \\ \quad (b * a) * (b * b^{-1}) = (a * b) * (b * b^{-1}) \\ \quad (b * a) * e = (a * b) * e \\ \quad b * a = a * b \end{align}
  • Hence, $(G, *)$ is an abelian group. $\blacksquare$
Theorem 2: Let $(G, *)$ be a group. If for all $a \in G$ we have that $a = a^{-1}$ then $(G, *)$ 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:
(2)
\begin{align} \quad a * b = a^{-1} * b^{-1} = (b * a)^{-1} \end{align}
  • But $(b * a) \in G$ since $a, b \in G$ and by definition, $G$ is closed under $*$. Therefore $b * a = (b * a)^{-1}$ so:
(3)
\begin{align} \quad a * b = b * a \end{align}
  • Hence, for all $a, b \in G$, $a * b = b * a$, so $(G, *)$ is an abelian group. $\blacksquare$
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