1

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.

Dana Hill
  • 715
  • 1
    When you write "isn't $n$ a different integer then [sic] $\mathbb Z,$" you appear to think that $\mathbb Z$ is an integer. But it is a set of integers. $\qquad$ – Michael Hardy Apr 05 '17 at 18:33
  • Duplicates for the title questions: this question, and this question. – Dietrich Burde Apr 05 '17 at 18:44
  • @DietrichBurde : Neither of those is an exact duplicate. – Michael Hardy Apr 05 '17 at 18:47
  • I don't have any notes, I have a textbook. This is an online class with no lectures, no notes, no discussion forum, not a whole lot of anything. I'm having to teach this to myself. (and I will not be taking another class from this professor) – Dana Hill Apr 05 '17 at 18:51
  • @MichaelHardy As I said, certainly a duplicate for the title question, but not for the actual question in the body. – Dietrich Burde Apr 05 '17 at 18:56
  • I'm taking two classes this semester. One provides weekly real-time lectures where we can ask questions, a very active discussion forum and prompt helpful responses by email. Then there is this class. When I ask questions by email it takes days for him to respond, and there have been several times he did not respond at all. (con't) – Dana Hill Apr 05 '17 at 19:04
  • I had to ask him three times about the grade on an assignment and after the third time he finally responded and said he didn't have it (I verified that it was sent). I am also a teacher (high school) and I don't feel that this prof is making any attempt to teach, only to present the material with the least amount of effort. – Dana Hill Apr 05 '17 at 19:08
  • Possible duplicate of http://math.stackexchange.com/questions/2218479/contruct-a-homomorphism-phi-of-mathbbq-such-that-mathrmker-phi. – lhf Apr 05 '17 at 22:03
  • @lhf Not a duplicate. That question was about finding the answer. I was having difficulty understanding the answer. But thanks for contributing to the conversation. – Dana Hill Apr 06 '17 at 16:04

2 Answers2

6

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 $$(m/n+m/n+\dots+m/n)+\mathbb{Z}=m+\mathbb{Z}=\mathbb{Z}.$$

Eric Wofsey
  • 330,363
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 $a,b \in \mathbb{Q}.$

CrypticParadigm
  • 599