If you find any mistakes, please make a comment! Thank you.

Q/Z is isomorphic to the group of complex roots of unity

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.1 Exercise 3.1.14

Consider the additive quotient group $\mathbb{Q}/\mathbb{Z}$.

(1) Show that every coset of $\mathbb{Z}$ in $\mathbb{Q}/\mathbb{Z}$ has exactly one representative $q \in \mathbb{Q}$ such that $0 \leq q < 1$.
(2) Show that every element of $\mathbb{Q}/\mathbb{Z}$ has finite order, but that there exist element of arbitrarily large order.
(3) Show that $\mathbb{Q}/\mathbb{Z}$ is the torsion subgroup of $\mathbb{R}/\mathbb{Z}$.
(4) Prove that $\mathbb{Q}/\mathbb{Z}$ is isomorphic to the multiplicative group of roots of unity in $\mathbb{C}^\times$.


(1) Suppose we have a coset represented by two rationals, say $q + \mathbb{Z} = r + \mathbb{Z}$, such that $0 \leq q,r < 1$. In particular, we have $q \in r + \mathbb{Z}$ so that $q = r+n$ for some integer $n$. If $n < 0$, then $q = n+r 0$, then $q = n+r \geq 1$, a contradiction. Thus $n = 0$, so that $q = r$.

(2) Let $a/b + \mathbb{Z}$, with $a,b$ integers. Note that $$b(a/b + \mathbb{Z}) = a+\mathbb{Z} = \mathbb{Z},$$ so that $|a/b + \mathbb{Z}| \leq b$. So every element of $\mathbb{Q}/\mathbb{Z}$ has finite order.

Now let $b \in \mathbb{Z}$ and consider $1/b + \mathbb{Z}$. Note that $k(1/b + \mathbb{Z}) = 0$ if and only if $b$ divides $k$. The smallest such $k$ with $k > 0$ is $b$, so that $|1/b + \mathbb{Z}| = b$. Since $b$ is an arbitrary integer, there exist elements in $\mathbb{Q}/\mathbb{Z}$ of arbitrarily large order.

(3) We have $\mathbb{Q}/\mathbb{Z} \leq T(\mathbb{R}/\mathbb{Z})$ by the previous point. Now suppose $x + \mathbb{Z} \in \mathbb{R}/\mathbb{Z}$ has finite order $k$; then $kx \in \mathbb{Z}$, so that $kx = n$ for some integer $n$. Thus $x = n/k$ is rational, and $x + \mathbb{Z} \in \mathbb{Q}/\mathbb{Z}$.

(4) Define a mapping $\varphi : \mathbb{Q}/\mathbb{Z} \rightarrow \mathbb{C}^\times$ by $q \mapsto 1^q$.

(Well defined) Suppose $q + \mathbb{Z} = r + \mathbb{Z}$; then $q = r+n$ for some integer $n$. Thus $$\varphi(q + \mathbb{Z}) = 1^q = 1^{r+n} = 1^r = \varphi(r + \mathbb{Z}),$$ and $\varphi$ is well defined.

(Homomorphism) We have \begin{align*}\varphi((q+\mathbb{Z})+ (r + \mathbb{Z})) =&\ \varphi((q+r) + \mathbb{Z})\\ =&\ 1^{q+r} = 1^q1^r\\ =&\ \varphi(q + \mathbb{Z})\varphi(r + \mathbb{Z}),\end{align*} so $\varphi$ is a homomorphism.

(Injective) Suppose $\varphi(q+\mathbb{Z}) = \varphi(r+\mathbb{Z})$. Then $1^q = 1^r$, so that $1 = 1^{r/q}$. Now either $r = 0$, so that either $q = 0$ or $q = 1$, or $r/q = 1$. In either case we have $q + \mathbb{Z} = r + \mathbb{Z}$. Hence $\varphi$ is injective.

(Surjective) Every root of unity in $\mathbb{C}^\times$ has the form $1^q$ for some rational $q$ such that $0 \leq q < 1$, and $\varphi(q) = 1^q$. Thus $\varphi$ is surjective.


This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.

This Post Has One Comment

  1. Three comments about your solution to (1):
    - There seems to be a typo when you write $q=n+r0$, and
    - Don't we also need to show that the case $n > 0$ results in a contradiction to conclude that $n=0$?
    - Your proof shows the uniqueness of the solution assuming p,r exists and are bounded by $[0,1)$, but it doesn't prove their existence. This part might be obvious but I suppose for completeness you'd have to prove it too.

Leave a Reply

Close Menu