# Asked to prove associativity and find that a certain element was identity but I proved the opposite

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

• $[4]$ 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 $12$? – Brian M. Scott May 20 '20 at 15:16
• 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) – Galbotrix May 20 '20 at 15:40
• That looks like you’re using ordinary multiplication mod $12$. You need to use the $*$ operation. For instance, $$[4]*[8]=-[4][8]-3[4]-3[8]=-[8]-[0]-[0]=[4]\;.$$ – Brian M. Scott May 20 '20 at 15:42
• 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 – Galbotrix May 20 '20 at 15:48
• 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. – zipirovich May 20 '20 at 15:53

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 $$[12]=[0]$$.

• 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 – Galbotrix May 20 '20 at 16:13
• @Galbotrix: You’re welcome. – Brian M. Scott May 20 '20 at 16:14

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.