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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 $\mathbb{Z}/\ell \mathbb{Z}$?

Thanks for your help.

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

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    $\begingroup$ 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']? $\endgroup$
    – Wave
    Oct 16 at 8:44
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    $\begingroup$ 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. $\endgroup$ Oct 16 at 8:54

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