# proof that all cyclic groups of the same order are isomorphic to each other

The following is a proof that all cyclic groups^{} of the same order are isomorphic^{} to each other.

###### Proof.

Let $G$ be a cyclic group and $g$ be a generator^{} of $G$. Define $\phi :\mathbb{Z}\to G$ by $\phi (c)={g}^{c}$. Since $\phi (a+b)={g}^{a+b}={g}^{a}{g}^{b}=\phi (a)\phi (b)$, $\phi $ is a group homomorphism. If $h\in G$, then there exists $x\in \mathbb{Z}$ such that $h={g}^{x}$. Since $\phi (x)={g}^{x}=h$, $\phi $ is surjective^{}.

Note that $\mathrm{ker}\phi =\{c\in \mathbb{Z}:\phi (c)={e}_{G}\}=\{c\in \mathbb{Z}:{g}^{c}={e}_{G}\}$.

If $G$ is infinite^{}, then $\mathrm{ker}\phi =\{0\}$, and $\phi $ is injective^{}. Hence, $\phi $ is a group isomorphism, and $G\cong \mathbb{Z}$.

If $G$ is finite, then let $|G|=n$. Thus, $|g|=|\u27e8g\u27e9|=|G|=n$. If ${g}^{c}={e}_{G}$, then $n$ divides $c$. Therefore, $\mathrm{ker}\phi =n\mathbb{Z}$. By the first isomorphism theorem^{}, $G\cong \mathbb{Z}/n\mathbb{Z}={\mathbb{Z}}_{n}$.

Let $H$ and $K$ be cyclic groups of the same order. If $H$ and $K$ are infinite, then, by the above , $H\cong \mathbb{Z}$ and $K\cong \mathbb{Z}$. If $H$ and $K$ are finite of order $n$, then, by the above , $H\cong {\mathbb{Z}}_{n}$ and $K\cong {\mathbb{Z}}_{n}$. In any case, it follows that $H\cong K$. ∎

Title | proof that all cyclic groups of the same order are isomorphic to each other |
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Canonical name | ProofThatAllCyclicGroupsOfTheSameOrderAreIsomorphicToEachOther |

Date of creation | 2013-03-22 13:30:41 |

Last modified on | 2013-03-22 13:30:41 |

Owner | Wkbj79 (1863) |

Last modified by | Wkbj79 (1863) |

Numerical id | 9 |

Author | Wkbj79 (1863) |

Entry type | Proof |

Classification | msc 20A05 |