Cyclic groups are isomorphic to Z or Zn
Dependencies:
A cyclic group of infinite order is isomorphic to $\mathbb{Z}$. A cyclic group of order $n$ is isomorphic to $\mathbb{Z}_n$.
Proof
Let $\phi(k) = a^k$.
\[ \phi(k_1 + k_2) = a^{k_1 + k_2} = a^{k_1}a^{k_2} = \phi(k_1)\phi(k_2) \]
\[ \phi(k_1) = \phi(k_2) \Rightarrow a^{k_1} = a^{k_2} \Rightarrow a^{k_1 - k_2} = e \]
When $\langle a \rangle$ is infinite, this is only possible when $k_1 = k_2$. When $\langle a \rangle$ has order $n$, this means $n \mid (k_1 - k_2)$. This means $k_1 = k_2$, because $k_1, k_2 \in \mathbb{Z}_n$. Therefore, $\phi$ is one-to-one.
Every element in $\langle a \rangle$ can be written as $a^k = \phi(k)$. When $\langle a \rangle$ has infinite order, $k \in \mathbb{Z}$. When $\langle a \rangle$ has finite order, $k \in \mathbb{Z}_n$. Therefore, $\phi$ is onto.
Therefore, $\phi$ is an isomorphism from $\mathbb{Z}$ or $\mathbb{Z}_n$ to $\langle a \rangle$.
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Info:
- Depth: 4
- Number of transitive dependencies: 7