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Problem 66 (Every Infinite Cyclic Group Is Isomorphic to $\mathbb{Z}$)

Suppose $G$ is a cyclic group of infinite order. Prove that $ \mathbb{Z}\cong G$. In other words, produce a function $f:\mathbb{Z}\to G$ and show that $f$ is an isomorphism.

Solution

Since $G$ is cyclic, we can pick $a$ such that $\langle a \rangle = G$. We will show that $f:\mathbb{Z} \rightarrow G$ defined by $f(x)=a^x$ is an isomorphism. Let $x,y \in \mathbb{Z}$. We compute $f(xy)=a^{xy}=a^xa^y=f(x)f(y)$. We will now show that $f$ is injective. Suppose $f(b)=f(c)$. This means $a^b=a^c$. Multiplying on the right by $a^{-c}$, we obtain $a^ba^{-c}=a^ca^{-c}$, which simplifies to $a^{b-c}=e$. This implies that $b-c$ is a multiple of the order of $a$. Since $a$ has infinite order, this implies $b-c=0$, or $b=c$. We will now show that $f$ is surjective. Pick $g\in G$. Since $\langle a \rangle = G$, we know we can pick $k\in \mathbb{Z}$ such that $g=a^k$. It follows that $f(k)=a^k=g$.

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