Exercise 1.6.19 (Surjective homomorphism in the group of roots of unity)

Proof. First $(G,\times )$ is a subgroup of $({ℂ}^{\text{∗}},\times )$. Indeed,

• $1^1=1$, so $1\in G$ and $G\ne \varnothing$.
• If $z\in G$ and $t\in G$, then there are integers $n>0,m>0$ such that $z^n=1$ and $t^m=1$, therefore ${(\mathit{zt})}^{\mathit{mn}}={(z^n)}^m{(t^m)}^n=1$, so $\mathit{zt}\in G$.
• Since $z^n=1$, then ${(z^{-1})}^n=1$, so $z^{-1}\in G$.

Consider, for some fixed integer $k>1$, the map

Then $\phi$ is a homorphism: for all $z\in G,t\in G$, $\phi (\mathit{zt})={(\mathit{zt})}^k=z^k{t}^k=\phi (z)\phi (t).$

We show that $\phi$ is surjective. Let $Z$ be any element of $G$, so that there is some integer $n\ge 1$ such that $Z^n=1$. Since $Z\ne 0$, we may write $Z=\rho e^{\mathit{i𝜃}}$ for some $𝜃\in ℝ$ and some real $\rho >0$. Then $Z^n={\rho }^n{e}^{\mathit{in𝜃}}=1$. Therefore ${\rho }^n=|Z^n|=1$, where $\rho >0$, so $\rho =1$ and $Z=e^{\mathit{i𝜃}}$ satisfies

Put $z=e^{\mathit{i𝜃}∕k}.$ Then $z^{\mathit{nk}}=e^{\mathit{in𝜃}}=Z^n=1$, so $z\in G$, and $\phi (z)=z^k=e^{\mathit{i𝜃}}=Z$:

This shows that $\phi$ is surjective.

Put $u=e^{i2\pi ∕k}$. Since $k>1$, $u\ne 1$, otherwise $\frac{2\pi }{k}\equiv 0(\mathit{mod}2\pi )$, which implies $\mathit{kK}=1$ for some integer $K$, so $k\mid 1$, and this is impossible for $k>1$.

Since $u^k=1$, $u\in G$ and $\phi (u)=1$, so $u\in \ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )$, where $u\ne 1$. This shows that $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )\ne \{1\}$, so $\phi$ is not injective, and $\phi$ cannot be an isomorphism.

In conclusion, $\phi$ is a surjective homomorphism but not an isomorphism. □