# 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 $|g|=|\langle g\rangle|=|G|$. Let $z\in{\mathbb{Z}}$ such that $nz=|G|=|g|$. Consider $\langle g^{z}\rangle$. Since $g\in G$, then $g^{z}\in G$. Thus, $\langle g^{z}\rangle\leq G$. Since $\displaystyle|\langle g^{z}\rangle|=|g^{z}|=\frac{|g|}{\gcd(z,|g|)}=\frac{nz}{% \gcd(z,nz)}=\frac{nz}{z}=n$, it follows that $\langle g^{z}\rangle$ is a subgroup of $G$ of order $n$. ∎

Title proof of the converse of Lagrange’s theorem for finite cyclic groups ProofOfTheConverseOfLagrangesTheoremForFiniteCyclicGroups 2013-03-22 13:30:27 2013-03-22 13:30:27 Wkbj79 (1863) Wkbj79 (1863) 10 Wkbj79 (1863) Proof msc 20D99 CyclicRing3 ProofThatGInGImpliesThatLangleGRangleLeG