Proof: Order of a Cyclic Group is Equal to the Order of its Generator

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Problem

Prove that the order of a cyclic group is equal to the order of its generator.

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Proof

Let $G = \langle a \rangle$ be a cyclic group generated by the element $a$. We will consider two cases based on whether the order of the generator is finite or infinite.

Case 1: The order of the generator $a$ is finite.

Let the order of $a$ be a finite positive integer $n$. This means $n$ is the least positive integer such that $a^n = e$, where $e$ is the identity element.

Consider the set of $n$ elements: $A = \{a^0, a^1, a^2, \dots, a^{n-1}\}$.

First, we show that these $n$ elements are all distinct.

Suppose they are not distinct, meaning two powers of $a$ in the set are equal. Let’s say $a^i = a^j$ for some integers $i$ and $j$ where $0 \le j < i < n$. By multiplying by the inverse of $a^j$, we get:

$$ a^i (a^j)^{-1} = a^j (a^j)^{-1} $$ $$ a^{i-j} = e $$

From our assumption $0 \le j < i < n$, it follows that $0 < i-j < n$. This shows that there is a positive integer $i-j$, which is smaller than $n$, that results in the identity element. This is a contradiction, as $n$ was defined as the least positive integer with this property. Therefore, our initial assumption must be false, and all the elements $\{e, a, a^2, \dots, a^{n-1}\}$ are distinct. This implies that the group $G$ contains at least $n$ elements.

Next, we show that $G$ does not contain any other elements.

Let $x$ be any element in $G$. Since $G$ is cyclic and generated by $a$, $x$ must be some power of $a$, so $x = a^m$ for some integer $m$. By the division algorithm, we can write $m = nq + r$, where $q$ and $r$ are integers and $0 \le r < n$.

Now we can substitute this into the expression for $x$:

$$ x = a^m = a^{nq+r} = (a^n)^q \cdot a^r $$

Since $a^n=e$, this becomes:

$$ x = (e)^q \cdot a^r = e \cdot a^r = a^r $$

Because $0 \le r < n$, this means $x = a^r$ is one of the elements in the set $\{e, a, a^2, \dots, a^{n-1}\}$. We have shown that any arbitrary element of $G$ is one of these $n$ elements.

Therefore, $G$ contains precisely $n$ elements, and the order of the group, $o(G)$, is $n$. This is the same as the order of the generator, $o(a)$.

Case 2: The order of the generator $a$ is infinite.

In this case, we must show that the group $G$ also has an infinite number of elements.

If no two powers of $a$ are equal, then elements like $a, a^2, a^3, \dots$ are all distinct, and the group would contain an infinite number of elements. –

Let’s assume for contradiction that two powers of $a$ are equal. Say $a^m = a^n$ for integers $m > n$. This would imply $a^{m-n} = e$ with $m-n > 0$. This means that $a$ has a finite order, which contradicts our assumption for this case.

Therefore, no two powers of $a$ can be equal, and the group $G$ contains an infinite number of distinct elements. In this scenario, the order of the group is infinite, which is equal to the order of its generator.