More Abstract-Homomorphism&quotient

[TheForumLord]

Homework Statement

The last parts of the problems were:
1. prove that for each natural n, the additive quotient group Q/Z contains a one and only subgroup of order n and that sub-group is cyclic.
2. let G,H be two finite sub-groups of Q/Z. prove that G is contained in H if and only if
o(G)|o(H).
3. find all homomorphisms from Z/nZ to Q/Z.
but I've no clue how to start with the 4th part:
4. find all homomorphisms from Q/Z to Z...

TNX for all the helpers...

Homework Equations

none.

The Attempt at a Solution

none...

 

[Dick]

For number 4, consider a nonzero element of Q/Z. You can represent it by a rational n/m with n and m integers. If you add n/m to itself m times, you get 0 in Q/Z. What does that tell you about the image of n/m under the homomorphism?

 

[foxjwill]

For the first one, try finding a more familiar group that's isomorphic to Q/Z (hint: think complex)

 

[TheForumLord]

For number 4, consider a nonzero element of Q/Z. You can represent it by a rational n/m with n and m integers. If you add n/m to itself m times, you get 0 in Q/Z. What does that tell you about the image of n/m under the homomorphism?

Let n/m+Z be a nonzero element of the quo. group. the order of this element if m (if we assume that gcd (n,m)=1) . the image of n/m under the homo. has to be an element in Z of order m. But in Z, which element has order m?? how come this is the only homo. possible?

Help is needed :(

TNX !

 

[Dick]

If phi is your homomorphism, it doesn't really say phi(n/m) has order m. It says phi(n/m)*m=0. The only integer satisfying that is 0. So phi(n/m)=0, for all elements of Q/Z. That's why it's the only homomorphism possible.

 

[TheForumLord]

so the homo is:
each and every element in Q/Z goes to 0?

TNX

 

[Dick]

so the homo is:
each and every element in Q/Z goes to 0?

TNX

What else could it be?

 

[TheForumLord]

right...Tnx a lot! Hope you'll be able to help me in my next thread too...

 

[TheForumLord]

Can you please give me directions about the 3rd one?I
I'm pretty lame in all the Homomorphism theorems...

TNX again

 

[Dick]

Just start thinking about it. If you know what the image of the element '1' in Z/nZ is, then you know the rest of the homomorphism. What are the possibilities?

 

[TheForumLord]

The image of "1" in Z/nZ (which is of course "n" or "0" ) can be each and every one of the generators of the sub-groups from order n...I've proved that there's only one sub-group of order n in Q/Z and it's cyclic...Each one of it's generators are from the form i/n... Am I on the right track?

TNX a lot!

 

[Dick]

Z/nZ={0,1,2,...n-1}. '1' doesn't mean 0 or n. It means 1. But yes, every element of Z/nZ has order n, so the image of an element i under the homomorphism has to satisfy phi(i)*n=0 in Q/Z. That makes it one of the i/n, sure.

 

[TheForumLord]

TNX a lot man...You were very helpful!