proof of the converse of Lagrange’s theorem for finite cyclic groups

The following is a proof that, if $G$ is a finite cyclic group and $n$ is a nonnegative integer that is a divisor of $|G|$, then $G$ has a subgroup of order $n$.

Proof.

Let $g$ be a generator of $G$. Then $|g|=|⟨g⟩|=|G|$. Let $z\in \mathbb{Z}$ such that $nz=|G|=|g|$. Consider $⟨g^z⟩$. Since $g\in G$, then $g^z\in G$. Thus, $⟨g^z⟩\le G$. Since $|⟨g^z⟩|=|g^z|={\displaystyle \frac{|g|}{\gcd (z,|g|)}}={\displaystyle \frac{nz}{\gcd (z,nz)}}={\displaystyle \frac{nz}{z}}=n$, it follows that $⟨g^z⟩$ is a subgroup of $G$ of order $n$. ∎

Title	proof of the converse of Lagrange’s theorem for finite cyclic groups
Canonical name	ProofOfTheConverseOfLagrangesTheoremForFiniteCyclicGroups
Date of creation	2013-03-22 13:30:27
Last modified on	2013-03-22 13:30:27
Owner	Wkbj79 (1863)
Last modified by	Wkbj79 (1863)
Numerical id	10
Author	Wkbj79 (1863)
Entry type	Proof
Classification	msc 20D99
Related topic	CyclicRing3
Related topic	ProofThatGInGImpliesThatLangleGRangleLeG