Show that (ℚ,+)/(ℤ,+) is an infinite group every element of which has finite order.

Solution 1:

The notation $n(m/n+\mathbb{Z})$ does not mean you are multiplying each element of the set $m/n+\mathbb{Z}$ by $n$ to get a new set. It means you are considering $m/n+\mathbb{Z}$ as an element of the quotient group, and adding $n$ copies of this element together (using the addition operation of the quotient group). And in the quotient group, $(a+\mathbb{Z})+(b+\mathbb{Z})$ is defined as $(a+b)+\mathbb{Z}$: this is just the definition of addition of elements of the quotient group. So $n(m/n+\mathbb{Z})$ is the sum $$(m/n+\mathbb{Z})+(m/n+\mathbb{Z})+\dots+(m/n+\mathbb{Z})$$ where there are $n$ terms, and by definition this sum is $$(m/n+m/n+\dots+m/n)+\mathbb{Z}=m+\mathbb{Z}=\mathbb{Z}.$$

Solution 2:

This is probably too late, but to understand the answer you must understand the group operation. Since $\mathbb{Z}$ is normal in $\mathbb{Q}$, $\mathbb{Q/Z}$ is defined as $\mathbb{Q/Z} = \{q+\mathbb{Z} : q \in \mathbb{Q} \} = \{ \frac{m}{n} +\mathbb{Z} : m,n \in \mathbb{Z}, n \neq 0 \}.$ The group operation is the defined as $(a+\mathbb{Z}) + (b+\mathbb{Z})=(a+b) + \mathbb{Z}$ where $a,b \in \mathbb{Q}.$