Recent content by GatorPower

Indeed, that does help. Thank you everyone for helping out!

That does help, its just that i want to find a better way of doing it! Usually these things turn out to be some number squared or something, so I feel like I am doing something wrong if i compute too much :p

Fermat gives 4^11112 = 1*4^200 mod p, so that gets me a little further, but not quite there.

4 is quadratic residue, so 4^(p-1)/2 = 1 mod p, but then i get 4^11*31 = 1 mod p, and i need to get rid of 31.. Otherwise i get stuck with 4^200 mod p. I'll try to figure something out, thanks for the help!

It is supposedly solvable by hand, but I'm wondering if it was meant to calculate mod p-1 instead of p. Atleast that's what I'll conclude if no one finds another answer. Also, it seems that the answer is 16 mod 683 as well. Perhaps there is some once-in-a-lifetime connection between 682 and...

Im looking through old exams for a course in Cryptography and have beaten my head against the wall for a long time on one of the questions: p = 683 is a prime, p-1 = 2*11*31. What is x = 4^11112 mod p? When i did chinese remainder theorem on primes 2,11,31 i got that x = 16 mod 682, but so...

Hi! I am currently studying homology theory and am using Vicks book "Homology Theory, An introduction to algebraic topology". When I was reading I found a definition that troubles me, I simply cannot get my head around it. Vick defines that: if PHI is a singular p-simplex we define di(PHI), a...

Thank you =)

Another question! Factor group Z6 x Z8 / <(2,0)>. (0,1) + <(2,0)> has order 8, so we have two possible isomorphisms to Z8 x Z2 and Z16. Seeing that the factor group don't have any elements with order 16 we decide it is isomorphic to Z8 x Z2. Correct?

Another problem: Order is now 16 and the factor group is Z6 x Z8 / <(2,0)>. (0,1) + <(2,0)> has order 8, so the only possible isomorphism is to Z8 x Z2 or Z16. But Z16 is cyclic, and the factor group is not so the isomorphism is to Z8 x Z2. Correct?

So it is the second one? What kind of strategies does one use on this kind of problems? My book is pretty bad at explaining this..

I think I have thought about something wrong. There is in fact an element of order 9 which is (0,2) in the factor group. Then the factor group must be isomorphic to the SECOND one.

The two first groups are not cyclic, only Z2 x Z27. One does not have an elt of order 9 since 18/9 = 2 but <(3,0)> + (0,2) does not have order 9 seeing that (27,18) = (3,0) Our group is not cyclic, and does not contain an elt of order 9 so hence it is isomorphic to the first group. Correct?

Homework Statement Determine how many non-isomorphic (and which) abelian groups there are of order 54. Determine which of these groups the factor group Z6 x Z18 / <(3,0)> is isomorphic to. Homework Equations The Attempt at a Solution Fundamental theorem for abelian groups...

Anyone?