Uniqueness of Subgroups in a Finite Cyclic Group

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Problem

Prove that if $G$ is a finite cyclic group of order $n$, then for any positive integer $d$ that divides $n$, there exists exactly one subgroup of $G$ of order $d$.

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Definitions Used

1. Finite Cyclic Group: A group $G = \langle a \rangle$ of order $n$, where $|G| = \text{ord}(a) = n$.
2. Order of a Cyclic Subgroup: The order of a subgroup generated by an element $a^m$ is given by the formula: $$ \text{ord}(\langle a^m \rangle) = \text{ord}(a^m) = \frac{\text{ord}(a)}{\text{gcd}(m, \text{ord}(a))} $$
3. Subset Property of Finite Sets: If $A$ and $B$ are finite sets with the same number of elements ($|A|=|B|$), and $A$ is a subset of $B$ ($A \subseteq B$), then the sets must be equal ($A=B$).

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Resolution

The proof consists of two parts: first, proving the existence of such a subgroup, and second, proving its uniqueness.

Part 1: Existence

Let $G = \langle a \rangle$ be a finite cyclic group of order $n$. Let $d$ be a positive divisor of $n$. We can write $n = dk$ for some integer $k$.

Consider the subgroup $H = \langle a^k \rangle = \langle a^{n/d} \rangle$. The order of this subgroup is the order of its generator:

$$ |H| = \text{ord}(a^{n/d}) = \frac{n}{\text{gcd}(n/d, n)} $$

Since $n/d$ is a divisor of $n$, their greatest common divisor is $n/d$.

$$ |H| = \frac{n}{n/d} = d $$

Thus, a subgroup of order $d$ exists.

Part 2: Uniqueness

Now, we must prove that this subgroup $H = \langle a^{n/d} \rangle$ is the only one of order $d$.

Let $K$ be any arbitrary subgroup of $G$ such that $|K|=d$. Since every subgroup of a cyclic group is cyclic, $K$ must have a generator. Let’s say $K = \langle a^m \rangle$ for some integer $m$.

The order of $K$ is the order of its generator, $a^m$:

$$ |K| = \text{ord}(a^m) = \frac{n}{\text{gcd}(m,n)} $$

We are given that $|K|=d$, which leads to the equation:

$$ d = \frac{n}{\text{gcd}(m,n)} \quad \implies \quad \text{gcd}(m,n) = \frac{n}{d} $$

By the definition of the greatest common divisor, $\text{gcd}(m,n)$ must divide $m$. So, $n/d$ must be a divisor of $m$. This means we can write $m$ as a multiple of $n/d$:

$$ m = j \cdot \left(\frac{n}{d}\right) \quad \text{for some integer } j. $$

Now let’s look at the generator of $K$, which is $a^m$:

$$ a^m = a^{j(n/d)} = \left(a^{n/d}\right)^j $$

This expression shows that the generator of $K$, $a^m$, is a power of the element $a^{n/d}$. By definition, this means that $a^m$ is an element of the subgroup generated by $a^{n/d}$, which is our subgroup $H$.

$$ a^m \in \langle a^{n/d} \rangle = H $$

If the generator of a cyclic group $K$ is in another group $H$, then all elements of $K$ must also be in $H$. Therefore, $K$ is a subgroup of $H$ ($K \subseteq H$).

We now have two facts:

1. $K \subseteq H$
2. $|K| = |H| = d$

Since both are finite subgroups of the same order and one is a subset of the other, they must be the same subgroup.

$$ K = H $$

Because any arbitrary subgroup $K$ of order $d$ is identical to the specific subgroup $H$ we constructed, we have proven that there is exactly one subgroup of order $d$.