Cyclic Groups and their Isomorphisms

# Cyclic Groups and their Isomorphisms

We will now look at some very nice theorems regarding cyclic groups and how they are isomorphic to some other groups that we are familiar with.

 Theorem 1: If $(G, *)$ is a cyclic group of infinite order then $(G, *) \cong (\mathbb{Z}, +)$.
• Proof: Since $(G, *)$ is a cyclic group there exists a $g \in G$ such that $G = <g>$ and so that for all $x \in G$ there exists an $n \in \mathbb{Z}$ such that $x = g^n$. Define a function $f : \mathbb{Z} \to G$ by:
(1)
\begin{align} \quad f(n) = g^n \end{align}
• We first show that $f$ is bijective by showing $f$ is injective and surjective. Let $m, n \in \mathbb{Z}$ and suppose that $f(m) = f(n)$. Then $g^m = g^n$ so $g^{m-n} = e = g^0$ where $e$ is the identity element in $G$. But since $G$ has infinite order, this means that $m - n = 0$, so $m = n$. Therefore $f$ is injective.
• Let $x \in G$. Then $x = g^n$ for some $n \in \mathbb{Z}$ since $(G, *)$ is cyclic and $f(n) = g^n = x$. So for every $x \in G$ there exists an $n \in \mathbb{Z}$ such that $f(n) = x$, so $f$ is surjective. Hence $f$ is bijective.
• Now let $m, n \in \mathbb{Z}$. Then:
(2)
\begin{align} \quad f(m + n) = g^{m + n} = g^m * g^n = f(m) * f(n) \end{align}
• Therefore $(G, *)$ and $(\mathbb{Z}, +)$ are isomorphic. $\blacksquare$

We can also state a similar theorem regarding finite order cyclic groups, though, we will omit the proof:

 Theorem 2: If $(G, *)$ is a cyclic group of order $n$ then $(G, *) \cong (\mathbb{Z_n}, +)$.
 Corollary 1: If $(G, *)$ is a group of order $p$ where $p$ is prime then $(G, *) \cong (\mathbb{Z}_p, +)$.
• Proof: Suppose that $\mid G \mid = p$. Then by one of the Corollaries to Lagrange's Theorem we have that $G$ must be a cyclic group of order $p$. But by Theorem 2 this means that $(G, *) \cong (\mathbb{Z}_p, +)$