Proof: The Converse of Lagrange’s Theorem Holds for Cyclic Groups
Problem
Prove that the converse of Lagrange’s Theorem holds for finite cyclic groups. That is, if $G$ is a finite cyclic group of order $n$, and if $d$ is any positive integer that divides $n$, then there exists a subgroup of $G$ of order $d$.
Definitions Used
- Lagrange’s Theorem: If $H$ is a subgroup of a finite group $G$, then the order of $H$ must divide the order of $G$.
- Converse of Lagrange’s Theorem (General Statement): If $d$ is a divisor of the order of a finite group $G$, then there exists a subgroup of $G$ of order $d$. (Note: This is generally false, but we prove it true for cyclic groups).
- Finite Cyclic Group: A group $G$ is a finite cyclic group of order $n$ if there exists an element $a \in G$ (the generator) such that $G = \langle a \rangle = \{e, a, a^2, \dots, a^{n-1}\}$. The order of the group is $|G| = n$, which is also the order of the generator, $\text{ord}(a) = n$.
- Order of an Element: The order of an element $g$, denoted $\text{ord}(g)$, is the smallest positive integer $m$ such that $g^m = e$. The order of a cyclic subgroup $\langle g \rangle$ is equal to the order of the element $g$.
Resolution
Let $G$ be a finite cyclic group of order $n$. Let $a$ be a generator for $G$, so $G = \langle a \rangle$. By definition, this means $|G| = \text{ord}(a) = n$.
Let $d$ be a positive integer that divides $n$. Our goal is to construct a subgroup of $G$ that has order $d$.
Since $d$ is a divisor of $n$, we can write $n = dk$ for some integer $k$. This also means $k = n/d$.
Consider the element $b = a^k = a^{n/d}$ in $G$. Let’s form the subgroup $H$ generated by this element: $H = \langle b \rangle = \langle a^{n/d} \rangle$.
To complete the proof, we must show that the order of this subgroup $H$ is exactly $d$. The order of a cyclic subgroup is the order of its generator, so we need to find $\text{ord}(b) = \text{ord}(a^{n/d})$.
Let’s raise our element $b$ to the power of $d$:
$$ b^d = (a^{n/d})^d = a^{(n/d) \cdot d} = a^n $$Since the order of $a$ is $n$, we know that $a^n = e$ (the identity element). So, $b^d = e$. This tells us that the order of $b$ must be a divisor of $d$.
Now, let $m$ be the order of $b$. We know $m \le d$. By definition, $m$ is the smallest positive integer such that $b^m = e$.
$$ b^m = (a^{n/d})^m = a^{(nm)/d} = e $$Since the order of $a$ is $n$, for $a^x = e$, it must be true that $n$ divides $x$. Therefore, $n$ must divide $(nm)/d$.
$$ n \mid \frac{nm}{d} $$This implies that $\frac{nm/d}{n}$ must be an integer. Simplifying the fraction, we get that $m/d$ must be an integer. This means $d$ must divide $m$.
So we have two conditions:
- $m \le d$
- $d \mid m$
The only positive integer $m$ that satisfies both conditions is $m = d$.
Therefore, the order of the element $b = a^{n/d}$ is $d$. This means the subgroup $H = \langle a^{n/d} \rangle$ is a subgroup of order $d$.
Conclusion
We have shown that for any divisor $d$ of the order $n$ of a finite cyclic group $G$, we can always construct a subgroup of order $d$. This subgroup is generated by the element $a^{n/d}$, where $a$ is a generator of $G$.
Thus, the converse of Lagrange’s theorem holds true for all finite cyclic groups.