Consider $26 + <12>$ in $\mathbb{Z}_{60} / <12>$, I need to find the order of the element in this quotient group.
So I basically wrote out what the group might look like
$\dfrac{\mathbb{Z_{60}}}{<12>} =\{<12>,1+<12>, 2+<12>,3+<12> , \dots\ 11 + <12> \}$
So $26 + <12> = \{26,38, 50,2,14 \}$, so $2$ is in the coset and therefore $26+<12> = 2+<12>$.
So to ask for the element of the quotient group is the same as asking the order of cyclic subgroup of the quotient group. So I need to find all $n(2+<12>)$ for some integer $n$
So for instance $n=0$, I get $0(2 + <12>) = 0 + <12> = <12>$.
Now aftering trying out $n = 6$, I get back $<12>$,does this show that $<12>$ is identity? Because I tried the definition of identity $x \star e = x$ and $2 + <12> + e = 2 + <12> \iff e = 2 + <12> - (2 + <12>) = 2 + <12> - 2 - <12> = 0$?
Does the negative sign distribute over to the $<12>$?
Questions
Does the negative sign distribute over to the $<12>$?
My book just skips all the details. I am not even sure how this is obvious, but how do they immediately know that $26 + <12> = 2 + <12>$ without even writing out $26 + <12>$ first?
3.When I wrote that
$n=0$, I get $0(2 + <12>) = 0 + <12> = <12>$.
Is this even right? Can I even write out like this? Why doesn't $0$ distribute over?
- Please give me your harshest criticism of how I can write this out better.