Question

Give the order of the element in the factor group:
26+<12> in Z60/<12>

Answer 1

if i am reading this correctly, you have \(26+<12>\) that is the coset which includes 26
since \(26=2\times 12+2\) this coset is the same as \(2+<12>\)

Answer 2

as \(Z_{60}/<12>\cong Z_5\) you can keep adding until you get 1

Answer 3

the answer given:
<12> = {0, 12, 24, 36, 48}. We prefer to compute sums of the element 2 in the coset 26 + <12>, rather
than the element 26. Computing, 2 + 2 = 4, 4 + 2 = 6, 6 + 2 = 8, 8 + 2 = 10, and 10+2 = 12 \[12 \in <12>\].
Thus 26 + <12>has order 6 in the factor group..
why we just find the order of 2?and why, then we conclude it is the order of 26+<12>?

Answer 4

ok i was wrong, i thought this was
\[\mathbb{Z}_{60}/\mathbb{Z}_{12}\]

Answer 5

so ignore my answer completely

Answer 6

as for why you compute with 2, it is because it is easier
26 + <12> is a coset, all elements that look like 26 + 12k for some k
it doesn't matter which coset representative you use, so since \(26=12\times 2+2\) you might as well use 2 i.e \(26+12k=2+12(k+2)=2+12j\) so it is the same coset

Answer 7

if you wanted you could keep adding 26 to itself until you get a multiple of 12
it will still be order 6, it is just more annoying

Answer 8

how we want to know a factor group isomorphic to what?