# Abstract algebra module

Dick
Homework Helper
Those are the other possible permutations of S3. They should be in the other one coset, right?

(x3 + 2x + 2) / (2x+2)

In getting the first term, I multiplied 5*2x = 10x = x.
Right, the cosets are {e,(123),(132)} and {(12),(23),(13)}. And for the other one the first term of the quotient is x^3 divided by 2x mod 3. What's that?

Right, the cosets are {e,(123),(132)} and {(12),(23),(13)}. And for the other one the first term of the quotient is x^3 divided by 2x mod 3. What's that?
I'm not sure what to make of that. Is it 2x congruence modulo 3? 2x - 1 = 3k, k is an integer.

Dick
Homework Helper
I'm not sure what to make of that. Is it 2x congruence modulo 3? 2x - 1 = 3k, k is an integer.
If you doing it in the real numbers x^3 divided by 2x is (1/2)x^2. If you are doing it mod 3 you want (1 divided by 2 mod 3)*x^2.

Deveno
ok, we have a subgroup N = nZ. let's write N (somewhat incompletely) as:

N = nZ = {0,n,-n,2n,-2n,3n,-3n,......}

one possible (left, or right, it really doesn't make any difference because "+" in Z is commutative) coset is:

1+N = 1+nZ = {1,n+1,1-n,2n+1,1-2n,3n+1,1-3n,.....}

now, do 1+N and N have any elements in common? if not, then they are distinct cosets. what about k+N and (k+1)+N? are those different?

what might be the smallest positive integer k such that:

k+N = N?

after answering that question, see if you can figure out when k+N and m+N have the same elements. actually DO this for some small value of n (6 works pretty well, most people have the multiples of 6 memorized).