Z/nZ is Isomorphic to Zn
 Table of Contents

Z/nZ is Isomorphic to Zn

Recall that $(\mathbb{Z}/n\mathbb{Z}, +)$ denotes the group of integers $\{0, 1, 2, ..., n - 1\}$ modulo $n$, and $\mathbb{Z}_n$ denotes the cyclic subgroup of order $n$. We have already noted that $\mathbb{Z}/n\mathbb{Z}$ is isomorphic to $\mathbb{Z}_n$ via an explicit isomorphism. We will now prove this fact against using The First Group Isomorphism Theorem.

 Proposition 1: For each $n \in \mathbb{N}$, $\mathbb{Z}/n\mathbb{Z}$ is isomorphic to $\mathbb{Z}_n$.
• Proof: Let $\mathbb{Z}_n = \langle x \rangle$ where $x^n = 1$. Let $\phi : \mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}_n$ be defined for all $x \in \mathbb{Z}$ by:
(1)
\begin{align} \quad \phi(m) = x^{m \pmod n} \end{align}
• Observe that $\phi$ is indeed a homomorphism from $\mathbb{Z}$ since for all $m_1, m_2 \in \mathbb{Z}$ we have that:
(2)
\begin{align} \quad \phi(m_1 + m_2) = x^{m_1 + m_2 \pmod n} = x^{m_1 \pmod n} x^{m_2 \pmod n} = \phi(m_1) \phi(m_2) \end{align}
• Now observe that:
(3)
\begin{align} \quad \ker(\phi) = \{ m \in \mathbb{Z} : x^{m \pmod n} = 1 \} = \{ m \in \mathbb{Z} : n \mid m \} = n\mathbb{Z} \end{align}
• So by The First Group Isomorphism Theorem we have that:
(4)
\begin{align} \quad \mathbb{Z}/n\mathbb{Z} = \mathbb{Z}/\ker(\phi) \cong \phi(G) \end{align}
• But $\phi$ is surjective, since for all $x^t \in G = \langle x \rangle$, $1 \leq t \leq n$, we have that $\phi(t) = x^t$. So $\phi(G) = \mathbb{Z}_n$ and thus from above:
(5)
\begin{align} \quad \mathbb{Z}/n\mathbb{Z} \cong \mathbb{Z}_n \quad \blacksquare \end{align}
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