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The question goes like this:

Write down a group table for the groups C4 and C2 x C2. For every element a in C4 and C2 x C2 determine the smallest positive integer m such that $ m\cdot a$ equals the identity element

I managed to sketch the tables for both groups and then I said something like this:

We know that the set of elements in C4 are {$ \overline 0, \overline1, \overline2 , \overline3 $} and that m $\in$ $\mathbb Z$ and m > 0.

Then for a = $\overline 1$ the right m would be 4 because m $\cdot$ a = $\overline 1$ $\cdot$ 4 = $\overline4$ = $\overline 0$ = identity element. And here m would be the smallest integer.

For a = $\overline 2$ the right m would be 2 because m $\cdot$ a = $\overline 2$ $\cdot$ 2 = $\overline4$ = $\overline 0$ = identity element. And here m would be the smallest integer.

Is this wrong? I mean to multiply an element of a group, i.e C4 with an element of a different group, i.e $\mathbb Z$? Can you guys provide me with some hints if this is wrong?

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  • $\begingroup$ There is no element $\bar 4$, so your last two equalities in each example don't make sense. You are also basically asserting all the equalities without argument, and also asserting that the $m$ you provide is minimal without argument. Admittedly, this whole thing is just an exercise in a completely straightforward and small computation, so the argument is basically going to be "compute and see" anyway. $\endgroup$
    – Derek Elkins left SE
    Feb 6, 2019 at 22:26
  • $\begingroup$ @DerekElkins 4 mod 4 = $\overline 0$. I didn't say that $\overline 4$ exists in the set. $\endgroup$
    – Riccardo A.
    Feb 7, 2019 at 9:03
  • $\begingroup$ If $\bar 4=\bar 0$ then it certainly is in the set. The real point of that comment is that you haven't defined what $\bar 4$ means at all. Maybe you have a definition written down, but it is not written in your question. Admittedly, I was interpreting your question as whether the quoted part constitutes a reasonable proof, but you don't actually explicitly ask that. If your question is just whether $m=4$ and $m=2$ are the solutions to those two examples, then yes, they are. $\endgroup$
    – Derek Elkins left SE
    Feb 7, 2019 at 9:26
  • $\begingroup$ @DerekElkins What do you suggest I should do then? $\endgroup$
    – Riccardo A.
    Feb 7, 2019 at 9:44
  • $\begingroup$ If your goal is to make an informal proof, then, in general, proofs work from definitions, so my suggestion is to define what $\bar 4$ means or realize that it's unnecessary. More generally, apply definitions everywhere. Your title question does indeed ask for clarity on a definition, but if you didn't know what e.g. $4\cdot\bar 1$ meant, on what basis did you come to the conclusion that it should be $\bar 4$ or $\bar 0$? If you did want to spell out the calculations using the definition given by David Hill, it would look something like ... $\endgroup$
    – Derek Elkins left SE
    Feb 7, 2019 at 18:51

1 Answer 1

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Let $m>0$. You should interpret $$ m\cdot a=\underbrace{a+a+\cdots+a}_{\mbox{$m$ times}} $$ and to extend to negative numbers set $ (-m)\cdot a= m\cdot(-a)$. Of course, $0\cdot a=0$.

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  • $\begingroup$ If that is so it means that my approach would be a legit one, right? $\endgroup$
    – Riccardo A.
    Feb 6, 2019 at 21:55
  • $\begingroup$ That is correct. $\endgroup$
    – David Hill
    Feb 6, 2019 at 21:56
  • $\begingroup$ Thank you, that was helpful! $\endgroup$
    – Riccardo A.
    Feb 6, 2019 at 21:58

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