Show that the group $(\Bbb R,+)/(\Bbb Z,+)$ has non-trivial finite subgroups.

Question

Show that the group $(\Bbb R,+)/(\Bbb Z,+)$ has non-trivial finite subgroups.

The group is

$$(\Bbb R,+)/(\Bbb Z,+) = \{a + \Bbb Z \mid a \in \Bbb R\},$$

but I think all of the cosets are infinite? For example with $3+\Bbb Z = \{\dots,3,4,5,6, \dots \}$.

How can I get a finite subgroup when the cosets are of the form $a + \Bbb Z$ for $a \in \Bbb R$?

Answer 1

Hint: Let $n\in\Bbb N.$ We have

$$n\left(\frac{1}{n}+\Bbb Z\right)=\Bbb Z.$$

Answer 2

The fact that the cosets are infinite is irrelevant: if you consider the group $G=\mathbb{Q}\setminus\{0\}$ under multiplication and its subgroup $H=\mathbb{Q}_{>0}$ of the positive rationals, then $G/H$ is finite, notwithstanding that the cosets are infinite.

If you consider $f\colon\mathbb{R}\to\mathbb{C}\setminus\{0\}$ defined by $$ f(x)=\cos(2\pi x)+i\sin(2\pi x)=e^{2\pi ix} $$ (the codomain is the multiplicative group of nonzero complex numbers) then $f$ is a group homomorphism and its kernel is $\mathbb{Z}$. Its image is $\mathbb{T}$, the group of modulus $1$ complex numbers. Thus $$ \mathbb{R}/\mathbb{Z}\cong\mathbb{T} $$ and the group $\mathbb{T}$ has several finite subgroups, for instance the set of $n$th roots of $1$, where $n>0$.