# Group of $p$-Power Roots of 1 is Isomorphic to a Proper Quotient of Itself

## Problem 221

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.

## Proof.

### $\Psi: G\to G$ is a group homomorphism

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 $G/ \ker(\Psi)$.

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