Understanding Homomorphism: Z24 to Z18 with Phi(1)=16 Explained

[gonzo]

Excuse me if I get the English names wrong here, I hope my question is clear.

I would be happy if someone could explain this problem to me ... I'm having trouble with some aspects of homomorphisms.

The question is:

Is there a homomorphism Phi: Z24 --> Z18 which fullfills Phi(1)=16?

The book just gives the answer as "no", but I don't really know how to figure this out in any direct way.

We had an earlier similar problem where we were given a similar homomorphism that we were told was a homomorphism: Z18-->Z12 where phi(1)=8, though no more details than that. We were only asked to find I am phi, and ker phi, which I understood how to do. But it seems to me to be about the same for the above problema and would give similar results for ther image and kernel of phi, though obviously all phi will fail some test of being a homomorphism.

I would be most happy if someone could explain this to me. Thanks.

 

[gonzo]

Okay, I had a thought, so I would like to know if this is correct.

We know that I am phi is a subgroup of Z18, which means it's size divides 18.

We also know that the size of I am phi divides the size of Z24, which means it divides 24.

We also know that the I am phi(H) of any H subgroup of Z24 is going to be a subgroup of Z18.

so from phi(1)=16, we have that 16 is in at least one image of phi. However, the only subgroups 16 can be a member of are of size 9 or 18, but neither 9 nor 18 divide 24, so 16 can't belong to any image of phi, so the answer is no.

Is that on the right track? I think I confused myself typing all of that.

 

[lurflurf]

I think the thing to use here is
phi(1)=1
as is easy to see from
phi(1x)=phi(1)phi(x)=phi(x)
phi(1)=1
I am confused by your phi is it just a maping? Then you want to find out if it can be a homomophism?

 

[gonzo]

it's "addition" when you count in the Zn sets. so you are thinking of phi(0)=0, which isn't helpful that I can see.

I have a related question for someone who can help.

I thought I might be starting to understand this a bit, then found another discrepancy with the book answer.

The question is how many homomorphisms exist from Z8 to Z24

So, using the idea of the size of image phi having to divide both Z8 and Z24, then the I am phi can be of size 1, 2, 4 or 8, each of which produces a unique subgroup in Z24 that I can see.

So as far as I can see, there are 4 possible homomorphisms from Z8 to Z24, but the book puts it at 8 instead, can someone point out what I'm missing? Thanks.

 

[lurflurf]

it's "addition" when you count in the Zn sets. so you are thinking of phi(0)=0, which isn't helpful that I can see.

in that case
phi(-1)=phi(23)=8
-phi(1)=-16=2
These should be equal, but are not.
In addition the order of phi(Z24) is 9, clearly inconsistant.
9 does not divide 24

 

[matt grime]

FACT: if f is a homomorphism from G to H then the order of f(x) divides the oreder of x. Note, haven't chekced that this is the correct thing to examine, but I can't see it being anything else.