$(\implies)$: 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.

Image of a Normal Subgroup Under a Surjective Homomorphism is a Normal Subgroup
Let $f: H \to G$ be a surjective group homomorphism from a group $H$ to a group $G$.
Let $N$ be a normal subgroup of $H$. Show that the image $f(N)$ is normal in $G$.
Proof.
To show that $f(N)$ is normal, we show that $gf(N)g^{-1}=f(N)$ for any $g \in […]

Abelian Groups and Surjective Group Homomorphism
Let $G, G'$ be groups. Suppose that we have a surjective group homomorphism $f:G\to G'$.
Show that if $G$ is an abelian group, then so is $G'$.
Definitions.
Recall the relevant definitions.
A group homomorphism $f:G\to G'$ is a map from $G$ to $G'$ […]

Group of $p$-Power Roots of 1 is Isomorphic to a Proper Quotient of Itself
Let $p$ be a prime number. Let
\[G=\{z\in \C \mid z^{p^n}=1\} \]
be the group of $p$-power roots of $1$ in $\C$.
Show that the map $\Psi:G\to G$ mapping $z$ to $z^p$ is a surjective homomorphism.
Also deduce from this that $G$ is isomorphic to a proper quotient of $G$ […]

Surjective Group Homomorphism to $\Z$ and Direct Product of Abelian Groups
Let $G$ be an abelian group and let $f: G\to \Z$ be a surjective group homomorphism.
Prove that we have an isomorphism of groups:
\[G \cong \ker(f)\times \Z.\]
Proof.
Since $f:G\to \Z$ is surjective, there exists an element $a\in G$ such […]

Group Homomorphism from $\Z/n\Z$ to $\Z/m\Z$ When $m$ Divides $n$
Let $m$ and $n$ be positive integers such that $m \mid n$.
(a) Prove that the map $\phi:\Zmod{n} \to \Zmod{m}$ sending $a+n\Z$ to $a+m\Z$ for any $a\in \Z$ is well-defined.
(b) Prove that $\phi$ is a group homomorphism.
(c) Prove that $\phi$ is surjective.
(d) Determine […]

Group Homomorphism, Preimage, and Product of Groups
Let $G, G'$ be groups and let $f:G \to G'$ be a group homomorphism.
Put $N=\ker(f)$. Then show that we have
\[f^{-1}(f(H))=HN.\]
Proof.
$(\subset)$ Take an arbitrary element $g\in f^{-1}(f(H))$. Then we have $f(g)\in f(H)$.
It follows that there exists $h\in H$ […]

Fundamental Theorem of Finitely Generated Abelian Groups and its application
In this post, we study the Fundamental Theorem of Finitely Generated Abelian Groups, and as an application we solve the following problem.
Problem.
Let $G$ be a finite abelian group of order $n$.
If $n$ is the product of distinct prime numbers, then prove that $G$ is isomorphic […]

Group Homomorphism Sends the Inverse Element to the Inverse Element
Let $G, G'$ be groups. Let $\phi:G\to G'$ be a group homomorphism.
Then prove that for any element $g\in G$, we have
\[\phi(g^{-1})=\phi(g)^{-1}.\]
Definition (Group homomorphism).
A map $\phi:G\to G'$ is called a group homomorphism […]