2006-06-30 06:32:35 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
I'd solve this one by actually determining every single homomorphism from Z/mZ to Z/nZ completely. Here's a couple of hints to get you going.
(1) Since the group Z/mZ is cyclic, every homomorphism out of Z/mZ is completely determined by where the element 1 is sent (more precisely, the element 1+mZ).
(2) Because m times 1 is 0 in Z/mZ, any homomorphism f : Z/mZ to another group G must send 1 to a number f(1) in G such that m times f(1) is 0 in G.
These two comments will give you a small set of possible homomorphisms from Z/mZ to Z/nZ (hint: the number of such homomorphisms depends upon gcd(m,n) ). It's then not hard to show that everything in that set really works as a homomorphism and that they form a cyclic group.
2006-06-30 06:50:53 · answer #1 · answered by numbertheoryguy 1 · 0⤊ 0⤋
Order is 3. Please do your homework
2006-06-30 13:35:42 · answer #2 · answered by ag_iitkgp 7 · 0⤊ 0⤋