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 $G$ itself.

We first show that $\Psi: G\to G$ is a group homomorphism.
To see this, let $z, w \in G$. Then we have
\begin{align*}
\Psi(zw)=(zw)^p=z^pw^p=\Psi(z)\Psi(w).
\end{align*}
The second equality follows since $G$ is an abelian group.
Thus $\Psi$ is a group homomorphism.

$\Psi$ is surjective

To prove that $\Psi$ is surjective, let $z$ be an arbitrary element in $G$.
Then there exists nonnegative integer $n$ such that $z^n=1$.
Let $w \in \C$ be a $p$-th root of $z$, that is, $w$ is a solution of the equation $x^p-z=0$.
(By the fundamental theorem of algebra, such a solution exists.)

We check that $w \in G$ as follows.
We have
\begin{align*}
w^{p^{n+1}}=(w^p)^{p^n}=z^{p^n}=1.
\end{align*}
Therefore $w$ is a $p$-power root of $1$, hence $w\in G$.

It follows from
\begin{align*}
\Psi(w)=w^p=z
\end{align*}
that $\Psi$ is a surjective homomorphism.

$G$ is isomorphic to the proper quotient

Now by the first isomorphism theorem, we have an isomorphism
\[G/ \ker(\Psi) \cong \im(\Psi)=G.\]
Since we have
\[\ker(\Psi)=\{z \in G \mid z^p=1\},\]
the subgroup $\ker(\Psi)$ consists of $p$-th roots of unity.

There are $p$ $p$-th roots of unity in $\C$ (and hence in $G$), and hence the kernel $\ker(\Psi)$ is a nontrivial subgroup of $G$.
Hence $G/ \ker(\Psi)$ is a proper quotient, and thus $G$ is isomorphic to the proper quotient $G/ \ker(\Psi)$.

Abelian Normal subgroup, Quotient Group, and Automorphism Group
Let $G$ be a finite group and let $N$ be a normal abelian subgroup of $G$.
Let $\Aut(N)$ be the group of automorphisms of $G$.
Suppose that the orders of groups $G/N$ and $\Aut(N)$ are relatively prime.
Then prove that $N$ is contained in the center of […]

Subgroup of Finite Index Contains a Normal Subgroup of Finite Index
Let $G$ be a group and let $H$ be a subgroup of finite index. Then show that there exists a normal subgroup $N$ of $G$ such that $N$ is of finite index in $G$ and $N\subset H$.
Proof.
The group $G$ acts on the set of left cosets $G/H$ by left multiplication.
Hence […]

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 $G'$ […]

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 […]

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 […]

Cyclic Group if and only if There Exists a Surjective Group Homomorphism From $\Z$
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 Group Homomorphism is Injective if and only if Monic
Let $f:G\to G'$ be a group homomorphism. We say that $f$ is monic whenever we have $fg_1=fg_2$, where $g_1:K\to G$ and $g_2:K \to G$ are group homomorphisms for some group $K$, we have $g_1=g_2$.
Then prove that a group homomorphism $f: G \to G'$ is injective if and only if it is […]

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 $h\in H$ […]