**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 $\mathbb{C}^\times$.

Solution:

(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.

## Jimmy

2 Aug 2022Three 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.