# On the Definition of multiplication in an abelian group

In class we had the following

Definition: Let $(A,+)$ be an abelian Group with $a \in G$. We define: $$na:= \begin{cases}na, \ \forall n \in \mathbb{N} \\ |n|(-a), \ \forall n \in \mathbb{Z}\setminus\mathbb{N} \end{cases}$$

Fair enough, doesn't seem too complicated. But when introducing a beginning student like me to the notion of a Module this becomes incredible confusing.

Apparently the above definition is the only and the most natural way to show that for an $(A,+)$ abelian group the mapping $$\cdot : \begin{cases}\mathbb{Z} \times A & \longrightarrow A \\ (r,a) & \longmapsto r\cdot a \end{cases} \\$$ follows the laws of a $\mathbb{Z}$-Module. While it makes sense to me that the above definition comes in handy when trying to verify this, I am still absolutely puzzled about how to work with it.

Let me try to elaborate on my confusion (this might get tedious for a trained Mathematician). I am aware that in (higher) mathematics the simple notation of $na$ has absolutely no meaning whatsoever until we give meaning to it by a definition.

Since we have done this with the above, one could mean that it is possible to work with it and show certain properties.

One of my tutors however meant that it's not clever to write $na$ as in the definition above, it would be better to define it recursively by a sum, because $$na= \underbrace{a+\dots + a}_{n-\text{times}} \tag{*}$$ isn't mathematically rigorous and again, has in that context no meaning. Note that (*) is what they use in most papers I find online when exhausting google about this topic, see for example this paper here and the same thing for the Wikipedia entry here.

So in order to rigorously proof that the above definition of multiplication follows the laws of a $\mathbb{Z}$-Module I would have to write (I suppose) $$\sum_{i=1}^{n+1} x:= \begin{cases}(\sum_{i=1}^n x) +x, \ \forall n \in \mathbb{N} \\ (\sum_{i=1}^{|n|}(-x)) + (-x), \forall n \in \mathbb{Z}\setminus \mathbb{N} \\0\cdot x = 0 \cdot(-x) =0 \text{ for } n=0 \end{cases}$$ I hope that this is right. However with this definition I manage to show for example the distributivity law only for $n \in \mathbb{N}$ (by induction).

For negative values of $n$ I would say that $|n|= -n$ which is again a positive number and try induction over $p=|n|$, but there nothing seems to work.

Let $p=1$ then $$\sum_{i=1}^1 (x+y)=\left(\sum_{i=1}^0(-(x+y)\right)+(-(x+y)) = (-(x+y))$$ and I have no chance to continue with this. Is my definition wrong or do I just apply it in a flawed way? Also, despite to what my tutor said, might it be better to work with the original definition at the very top of this post and if so how?

If you plan on answering this question you don't have to show me all the properties of a $\mathbb{Z}$-Module, I'd be happy to see exemplary calculations of lets say the distributivity.

• It's probably worth pointing out that in the above definition as you've written it, you have, for example, for $n \in \mathbb N$ "$na = na$". Implicitly the definition is assuming what your tutor told you, that we define $na$ to be the obvious thing - i.e. $a + a + \ldots + a$ Commented Nov 5, 2014 at 20:11
• you're absolutely right about that @Mathmo123, which is another point of what confuses me, we're defining something we 'don't know' by something we don't know. However it is the correct definition taken out from my class. Commented Nov 5, 2014 at 20:21
• These abstract definitions become meaningful when you extend them to previously unknown situations. It's hard to see the value in parsing what $na$ means if you're only ever working with integers. You're asking good questions, but I recommend working with more examples of modules ($\Bbb{Z}_n$ as a $\Bbb{Z}$-module and $\Bbb{R}^n$ is an $\Bbb{R}[x]$-module, etc.) and then revisiting the definitions. Commented Nov 5, 2014 at 20:27
• @Spaced I think that's the point - in the definition, we are taking $na$ to be exactly what you think it should be - i.e. $a + \ldots + a$. The definition is assuming you are acquainted with basic arithmetic, so doesn't need to define anything for $n \in \mathbb N$. The definition is really about $n \in \mathbb Z \setminus \mathbb N$ Commented Nov 5, 2014 at 20:28

I would suggest working in stages. Assuming we already know all about integer arithmetic, first define multiplication for naturals only, with the recursion equations $$f(0_{\mathbb N},a) = 0_G \qquad f(n+1,a) = f(n+a)+_G a$$ and now prove that it associates and distributes as we want to, as long as the numbers involved are nonnegative. Your question indicates that you know how to do this by induction.
Then, extend $f$ from $\mathbb N\times G\to G$ to $\mathbb Z\times G\to G$ by requiring $$\tag{*} f(-n,a) = -_G\,f(n,a)$$ Declare this equation to be the definition of the left-hand side when $n\ge1$; with this definition in place the cases for $n\le 0$ then become an (easy) theorem.
We can then also prove: $$f(-n,a) = f(n,-_G\, a)$$ by induction on $n$ for $n\ge 1$, directly from the definition for $n=0$ and by $(*)$ for $n<0$.
Finally prove the associative and distributive laws in the cases where one or more of the numbers are negative. There's a somewhat tedious amount of case-by-case work here according to which of the numbers have which signs and so forth, but it is not difficult. For example one of the cases of the distributive law, for $0\le m \le n$, goes
\begin{align} f(-n, a) +_G f(m,a) &= -_G f(n,a) +_G f(m,a) \\ &= -_G (f(m,a)+f(n-m,a)) +_G f(m,a) \\ &= f(m,a) -_G f(m,a) -_G f(n-m,a) \\ &= -_G f(n-m, a) \\ &= f(-(n-m), a) \\ &= f((-n)+m, a) \end{align}