$(\implies)$: 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 $f(g)\in f(H)$.
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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).
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Group Homomorphism, Conjugate, Center, and Abelian group
Let $G$ be a group. We fix an element $x$ of $G$ and define a map
\[ \Psi_x: G\to G\]
by mapping $g\in G$ to $xgx^{-1} \in G$.
Then prove the followings.
(a) The map $\Psi_x$ is a group homomorphism.
(b) The map $\Psi_x=\id$ if and only if $x\in Z(G)$, where $Z(G)$ is the […]