In $\mathbb{Z}_{12}$, find non-zero elements $a$, $b$, $c$ with $ab = ac$ but $b \neq c$
$\mathbb{Z}_{12}\setminus \{0\} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$
I have at the moment that $a = 2, b = 6$ and $c = 4$
$2 \times 6 \equiv 0 \pmod {12}$
$4 \times 6 \equiv 0 \pmod{12}$
It would be great if someone could verify this/correct me if I am wrong.
Also is there a shorter way to do this than guess and check?
If you meant $a=6, b=2, $ and $c=4$,
that is indeed an example where $ab\equiv ac$ but $b\not\equiv c\bmod 12$.
To find such examples, note that we need $a(b-c)\equiv 0$ with $a\not\equiv0$ and $b-c\not\equiv0\bmod12$.
Since $12=2^23=2\times6=6\times2=3\times4=4\times3$,
this could be accomplished with $a\in\{2,10\}$ and $b-c=6$, or with $a=6$ and $b-c\in\{2,10\}$,
or with $a\in\{3,9\}$ and $b-c\in\{4,8\}$, or with $a\in\{4,8\}$ and $b-c\in\{3,9\}.$
$$\text{2*2 = 4 = 2*8}$$ $$\text{2*3 = 6 = 2*9}$$ $$\text{2*4 = 8 = 2*10}$$ $$\text{2*5 = 10 = 2*11}$$ $$\text{3*1 = 3*5 = 3*9 = 3}$$ $$\text{3*2 = 3*6= 3*10 = 6}$$ $$\text{3*3 = 3*7 = 3*11 = 9}$$ $$\text{3*4 = 3*8 = 0}$$ $$\text{4*1 = 4*4 = 4*7 = 4*10 = 4}$$ $$\text{4*2 = 4*5 = 4*8 = 4*11 = 8}$$ $$\text{4*3 = 4*6 = 4*9 = 0}$$ $$\text{6*1 = 6*3 = 6*5 = 6*7 = 6*9 = 6*11 = 6}$$ $$\text{6*2 = 6*4 = 6*6 = 6*8 = 6*10 = 0}$$ $$\text{8*1 = 8*4 = 8*7 = 8*10 = 8}$$ $$\text{8*2 = 8*5 = 8*8 = 8*11 = 4}$$ $$\text{8*3 = 8*6 = 8*9 = 0}$$ $$\text{9*1 = 9*5 = 9*9 = 9}$$ $$\text{9*2 = 9*6 = 9*10 = 6}$$ $$\text{9*3 = 9*7 = 9*11 = 3}$$ $$\text{9*4 = 9*8 = 0}$$ $$\text{10*1 = 10*7 = 10}$$ $$\text{10*2 = 10*8 = 8}$$ $$\text{10*3 = 10*9 = 6}$$ $$\text{10*4 = 10*10 = 4}$$ $$\text{10*5 = 10*11 = 2}$$
where $*$ is multiplication modulo 12.
sorry for the bad formatting, I haven't used MathJax in a while.
