Show that (ℚ,+)/(ℤ,+) is an infinite group every element of which has finite order. Here's the question:  "Show that (ℚ,+)/(ℤ,+) is an infinite group every element of which has finite order."  I have found multiple solutions, but there is something I don't understand in them, namely... 
There is a part of the solution that looks like this:
n(m/n + ℤ) = m + ℤ = ℤ
But shouldn't it be m + nℤ?  and even if the m goes away (why is that so) isn't nℤ a different integer then ℤ so how does this show that the order is finite?  You aren't getting back to the identity, which under addition is zero.  Every solution I find has this in it but I can't find an explanation on why this is so.
 A: 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 $$(m/n+m/n+\dots+m/n)+\mathbb{Z}=m+\mathbb{Z}=\mathbb{Z}.$$
A: 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 $a,b \in \mathbb{Q}.$
