Question

Prove that the set of elements of finite order in the group $\mathrm{ℝ}/\mathrm{\mathbb{Z}}$ is the subgroup of $\mathrm{\mathbb{Q}}/\mathrm{\mathbb{Z}}$ .

Short answer

It is proved that the set of elements of finite order in the group $\mathrm{ℝ}/\mathrm{\mathbb{Z}}$ is the subgroup of $\mathrm{\mathbb{Q}}/\mathrm{\mathbb{Z}}$.

Step-by-step solution

Refer to the result of Exercise 25

• Every element of $\mathrm{\mathbb{Q}}/\mathrm{\mathbb{Z}}$has finite order, and
• $\mathrm{\mathbb{Q}}/\mathrm{\mathbb{Z}}$contains an element of every possible finite order.

Prove that the set of elements of finite order in the group ℝ/ℤ  is the subgroup of ℚ/ℤ 

Suppose $\left[\mathrm{q}\right]\in \mathrm{ℝ}/\mathrm{\mathbb{Z}}$is an element of order $\mathrm{n}\in \mathrm{N}$ . Here, n is the smallest possible integer such that role="math" localid="1654515513635" $\underset{\mathrm{n}-\mathrm{terms}}{\mathrm{q}+\mathrm{q}........+\mathrm{q}}=\mathrm{nq}$, where role="math" localid="1654515517767" $\mathrm{nq}\in \mathrm{\mathbb{Z}}$.

Consider for role="math" localid="1654515581498" $\mathrm{nq}=\mathrm{m}$for role="math" localid="1654515584948" $\mathrm{m}\in \mathrm{\mathbb{Z}}$. We get

role="math" localid="1654515631742" $q=\frac{m}{n}$for $\mathrm{m}\in \mathrm{\mathbb{Z}}$

This implies $q\in \mathbb{Q}/\mathbb{Z}$, which means that every element of finite order in role="math" localid="1654515773918" $ℝ/\mathbb{Z}$ also belongs to $\mathrm{\mathbb{Q}}/\mathrm{\mathbb{Z}}$.

We already know from exercise 25 that every element of role="math" localid="1654515769391" $\mathrm{\mathbb{Q}}/\mathrm{\mathbb{Z}}$ has finite order, and it contains elements of every possible finite order.

Therefore, we can conclude that the set of elements of finite order in the group $\mathrm{ℝ}/\mathrm{\mathbb{Z}}$ is the subgroup of $\mathrm{\mathbb{Q}}/\mathrm{\mathbb{Z}}$.