Proof: Every Cyclic Group is Abelian
Problem
Prove that if a group $G$ is cyclic, then it must be abelian.
Definitions Used
- Cyclic Group: A group $G$ is cyclic if there exists a generator element $g \in G$ such that every element in $G$ can be written as an integer power of $g$. That is, for any $x \in G$, $x = g^k$ for some integer $k \in \mathbb{Z}$.
- Abelian Group: A group $(G, *)$ is abelian if its binary operation is commutative. That is, for any two elements $a, b \in G$, it must be true that $a * b = b * a$.
- Laws of Exponents in Groups: For any element $g$ in a group and any integers $m, n$, we have $g^m * g^n = g^{m+n}$.
Resolution
Let $G$ be a cyclic group with a generator $g$. Our goal is to show that $G$ is abelian. To do this, we must show that for any two arbitrary elements in $G$, the commutative property holds.
Let $x$ and $y$ be any two elements of $G$.
Since $G$ is cyclic with generator $g$, every element in $G$ is an integer power of $g$. Therefore, we can write $x$ and $y$ as:
$$ x = g^m \quad \text{for some integer } m \in \mathbb{Z} $$ $$ y = g^n \quad \text{for some integer } n \in \mathbb{Z} $$Now, let’s compute the product $x * y$:
$$ x * y = g^m * g^n $$Using the laws of exponents, we can add the exponents:
$$ x * y = g^{m+n} $$Next, let’s compute the product in the reverse order, $y * x$:
$$ y * x = g^n * g^m $$Again, using the laws of exponents:
$$ y * x = g^{n+m} $$Since addition of integers is commutative, we know that $m+n = n+m$. Therefore:
$$ g^{m+n} = g^{n+m} $$This leads us to the conclusion:
$$ x * y = y * x $$Since we have shown that for any two arbitrary elements $x, y \in G$, the commutative property $x * y = y * x$ holds, the group $G$ is, by definition, an abelian group.