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Subset G = [8][4][2][10] mod 12 and binary operation $ab=-ab-3a-3b$

I constructed a cayley table but I don't know how to format on this so I'll just type out each line of the 4X4 table:

4 8 10 2

8 4 2 10

10 2 4 8

2 10 8 4

From this I found the identity element to be 4, but the question asks to prove that it is 8.

When I worked out a*(b*c) and (a * b)*c after all the cancelling I got a =? c, thus not associative but Q asked me to prove associativity. Any help would be greatly appreciated

(a*b) * c =? a * (b*c)

(-(ab+3a+3b)) *c =? a * (-(bc+3b+3c))

-c(-(ab+3a+3b)) -3(-(ab+3a+3b))-3c =? -a(-(bc+3b+3c))-3a-3(-(bc+3b+3c))

abc+3ac+3bc+3ab+9a+9b-3c =? abc+3ab+3ac-3a+3bc+9b+9c

12a=?12c

a =/= c

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    $\begingroup$ $[4]$ can’t the identity element: $[4]*[4]=[8]$, and $[4]*[8]=[4]$. Are you sure that you’re using the $*$ operation and not the usual multiplication mod $12$? $\endgroup$ – Brian M. Scott May 20 '20 at 15:16
  • $\begingroup$ That was my bad cause the formatting is poor, but I have [4]*[4]=[4] and [4]*[8]=[8]. I only have the answers in my cayley table and not the "headings" at the side and top (order of which is 8 4 2 10) $\endgroup$ – Galbotrix May 20 '20 at 15:40
  • $\begingroup$ That looks like you’re using ordinary multiplication mod $12$. You need to use the $*$ operation. For instance, $$[4]*[8]=-[4][8]-3[4]-3[8]=-[8]-[0]-[0]=[4]\;.$$ $\endgroup$ – Brian M. Scott May 20 '20 at 15:42
  • $\begingroup$ Looking at your working out and @zipirovich working I now realise I forgot to include the minus sign when working out the mod so I got the wrong answers, thanks a lot :). I'm still struggling on the associativity if you have the time to look at it, it cancels down to a = c for me $\endgroup$ – Galbotrix May 20 '20 at 15:48
  • $\begingroup$ It may be better if you type up your work, i.e. show to us how you simplified $a*(b*c)$ and how you simplified $(a*b)*c$. Chances are there are some arithmetical mistakes or even accidental typos (something that occasionally happens to everyone). So if you show your work, we can find what went wrong. $\endgroup$ – zipirovich May 20 '20 at 15:53
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For associativity we have

$$\begin{align*} (a*b)*c&=(-ab-3a-3b)*c\\ &=-(-abc-3ac-3bc)-3(-ab-3a-3b)-3c\\ &=abc+3ac+3bc+3ab+9a+9b-3c \end{align*}$$

and

$$\begin{align*} a*(b*c)&=a*(-bc-3b-3c)\\ &=-(-abc-3ab-3ac)-3a-(-3bc-9b-9c)\\ &=abc+3ab+3ac-3a+3bc+9b+9c\;, \end{align*}$$

so after cancelling identical terms we have

$$\begin{align*} (a*b)*c-a*(b*c)&=(9a-3c)-(9c-3a)\\ &=12a-12c\\ &=12(a-c)\\ &=0\;, \end{align*}$$

since $[12]=[0]$.

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  • $\begingroup$ I see what you mean now in your comment about X12=X0 now, thanks a lot for that bit, I was getting annoyed at myself for not even being able to do the associativity $\endgroup$ – Galbotrix May 20 '20 at 16:13
  • $\begingroup$ @Galbotrix: You’re welcome. $\endgroup$ – Brian M. Scott May 20 '20 at 16:14
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Apparently, you need to recheck and redo your calculations. For example, $$[8]*[8]=-[8]\cdot[8]-3\cdot[8]-3\cdot[8]=-[64]-[24]-[24]=[-112]=[8],$$ because $-112\equiv8\pmod{12}$, but your table shows $4$ in that place.

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