# How do I find my composition law?

I have the following question:

Let $$\ell \ge 1$$ be an integer and consider the cyclic group $$(\mathbb{Z}/\ell \mathbb{Z},+)$$.

Show that there is a well defined composition law $$\times$$ on $$\mathbb{Z}/\ell \mathbb{Z}$$ s.t. $$[m]\times[n]=[m\times n] \,\,\,\,\forall m,n \in \mathbb{Z}$$

I'm a bit confused since in the exercise they gave us the composition law $$+$$ and now they denoted it by $$\times$$. Which one should I take? Isn't it the composition law $$+$$ which is well defined on $$\mathbb{Z}/\ell \mathbb{Z}$$?

• Actually both operations are inherited from $\Bbb Z$ the same way. Also, I guess, $l>1$. Oct 16 '21 at 8:36

$$\newcommand{\Z}{\mathbb{Z}} \newcommand{\zl}{\Z/\ell \Z} \newcommand{\def}{\stackrel{\text{def}}{=}}$$It's a little notationally confusing, so I understand the struggle.

It's that, inside the brackets, you're dealing with the operation as you would for elements of $$\Bbb Z$$. It might be more intuitive if you use a second notation for the one on $$\Bbb Z / \ell \Bbb Z$$.

So, for instance, we define addition $$\oplus$$ in $$\zl$$ by

$$[m] \oplus [n] \def [m + n]$$

In other words, addition of equivalence classes in $$\zl$$ gives you the same equivalence class, as you would get if you found the equivalence class of $$m+n$$ in $$\Z$$ first.

Similarly, we're now looking at multiplication $$\otimes$$ in $$\zl$$ defined by

$$[m] \otimes [n] \def [m \times n]$$

where $$\times$$ is the usual multiplication in $$\Z$$. We want to show this is well-defined. You've already verified that $$\oplus$$ is well-defined, but now you want to endow $$\zl$$ with a multiplicative operation as well, $$\otimes$$.

(We just often use the same notation for both since one naturally induces the other, but, as you've seen, it can be confusing.)

• Oh wow, now it makes much more sense. As I understood it, I need to show two things. First that this expression $[m]\times [n]=[m\times n]$ makes sense, i.e. if I take $[m],[n]$ then $[m]\times [n]=mn+l\mathbb{Z}$ and that it is well defined right? So I need to show that if $[m]=[m']$ and $[n]=[n']$ then [m\times n]=[m'\times n']?
– Wave
Oct 16 '21 at 8:44
• Effectively, yeah, that's the main thing you need to show -- that the choice of representative of the equivalence class doesn't matter, in other words. Oct 16 '21 at 8:54