# 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 $\varphi\colon{\mathbb{Z}}\to G$ by $\varphi(c)=g^{c}$. Since $\varphi(a+b)=g^{a+b}=g^{a}g^{b}=\varphi(a)\varphi(b)$, $\varphi$ is a group homomorphism. If $h\in G$, then there exists $x\in{\mathbb{Z}}$ such that $h=g^{x}$. Since $\varphi(x)=g^{x}=h$, $\varphi$ is surjective.

Note that $\ker\varphi=\{c\in{\mathbb{Z}}\,:\,\varphi(c)=e_{G}\}=\{c\in{\mathbb{Z}}\,:\,g% ^{c}=e_{G}\}$.

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

If $G$ is finite, then let $|G|=n$. Thus, $|g|=|\langle g\rangle|=|G|=n$. If $g^{c}=e_{G}$, then $n$ divides $c$. Therefore, $\ker\varphi=n{\mathbb{Z}}$. 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 ProofThatAllCyclicGroupsOfTheSameOrderAreIsomorphicToEachOther 2013-03-22 13:30:41 2013-03-22 13:30:41 Wkbj79 (1863) Wkbj79 (1863) 9 Wkbj79 (1863) Proof msc 20A05