# 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.

Contents

## 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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