Thread: Number of homomorphisms between two groups

Hi,

The exact question in the book is "Determine the number of homomorphisms from the additive group Z15 to the additive group Z10" (Zn is cyclic group of integers mod n under addition)

Now if the question asks to find the number of homomorphisms from Z15 onto Z10, then by the First Isomorphism Theorem I can prove that none exit. But if homomorphisms from Z15 into Z10 are allowed to be counted, how do I do it?

You've got a pretty good start - the idea is to use the First Isomorphism Theorem. You more or less identified that needs to map onto some kind of subgroup of . It can't be , because has no integer solution. So there are three other possibilities: maps into (the subgroup isomorphic to) , , or .

Try applying a similar argument.

Originally Posted by sashikanth

Hi,

The exact question in the book is "Determine the number of homomorphisms from the additive group Z15 to the additive group Z10" (Zn is cyclic group of integers mod n under addition)

Now if the question asks to find the number of homomorphisms from Z15 onto Z10, then by the First Isomorphism Theorem I can prove that none exit. But if homomorphisms from Z15 into Z10 are allowed to be counted, how do I do it?

Let f be a homomorphism from Z15 to Z10. The image of f has to be a subgroup of Z10. Thus, the possible cases are |im f|=1, 2, 5, 10 by Lagrange's theorem. The kernel of f also has to be a subgroup of Z15. Thus, by the first isomophism theorem, |im f| has to be either 1 or 5.

Case 1: |im f|=1. There is only one such homomorphism, a trivial one.
Case 2: |im f|=5. There are four such homomorphisms, i.e., \eulerphi(5) = 4.

For case 2, f(1) has to generate the group, im f. It follows that the number of choices for f(1) is \eulerphi(5) = 4.

Thus, the total number is 5.