# Can I multiply an element of a group to an element of a different group?

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?

• 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. Feb 6, 2019 at 22:26
• @DerekElkins 4 mod 4 = $\overline 0$. I didn't say that $\overline 4$ exists in the set. Feb 7, 2019 at 9:03
• 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. Feb 7, 2019 at 9:26
• @DerekElkins What do you suggest I should do then? Feb 7, 2019 at 9:44
• 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 ... Feb 7, 2019 at 18:51

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, $$0\cdot a=0$$.