Cyclic Group if and only if There Exists a Surjective Group Homomorphism From $\Z$

Problem 225

Show that a group $G$ is cyclic if and only if there exists a surjective group homomorphism from the additive group $\Z$ of integers to the group $G$.

Proof.

$(\implies)$: If $G$ is cyclic, then there exists a surjective homomorhpism from $\Z$

Suppose that $G$ is a cyclic group. Then we have
$G= \langle a \rangle$ for some $a\in G$. Namely, $G$ is generated by $a$. Define a map $f:Z \to G$ by sending $n\in \Z$ to $a^n \in G$.

Since any element of $G$ is of the form $a^n$ for some $n\in Z$, the map $f$ is surjective.
It remains to prove that $f$ is a group homomorphism.
For any $m, n \in \Z$, we have
\begin{align*}
f(m+n)=a^{m+n}=a^ma^n=f(m)f(n).
\end{align*}
Thus $f$ is a group homomorphism.

$(\impliedby)$: If there exists a surjective homomorphism from $\Z$, then $G$ is cyclic

On the other hand, suppose that there exists a surjective group homomorphism $f:\Z \to G$. Define
$a=f(1).$

Then we claim that $G$ is generated by $a$, that is, $G=\langle a \rangle$.
Since $a=f(1)\in G$, we have $\langle a \rangle \subset G$.

On the other hand, for any $g\in G$ there exists $n\in Z$ such that $f(n)=g$ since $f$ is surjective.
Since $f$ is a group homomorphism, we have
\begin{align*}
g&=f(n)\\
&=f(\underbrace{1+\cdots+1}_{n\text{ times}})\\
&=\underbrace{f(1)+\cdots+f(1)}_{n\text{ times}}=nf(1)\\
&=na \in \langle a \rangle.
\end{align*}

Therefore we have $G \subset \langle a \rangle$, hence $G=\langle a \rangle$ and $G$ is a cyclic group.

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