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 $n$ .∎

单词	ProofOfTheConverseOfLagrangesTheoremForFiniteCyclicGroups
释义	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 $n$ .∎
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