Thread: Number of different homomorphisms

How many homomorphisms exist from the abelian group to .

My answer is - none.

Since for every we have then we must have ( is an homomorphism). so must be or or but since a product of any of the above is not of order 2 then would not be a homomorphism.

Does it seem right or am I fundamentaly wrong?

Any help appreciated

Originally Posted by skyking

How many homomorphisms exist from the abelian group to .

My answer is - none.

Since for every we have then we must have ( is an homomorphism). so must be or or but since a product of any of the above is not of order 2 then would not be a homomorphism.

Does it seem right or am I fundamentaly wrong?

Any help appreciated

You are wrong but I think not fundamentally: your argument about the order of the

image is right, but you oversaw the fact that the order of any of the three 2-cycles

you wrote there indeed is 2, so you can define , and

the same can be done with any of the other two elements of order 2 in that direct sum.

Of course, the above is not the whole story since you still have

to figure out how to define f on the other elements of that sum...

Tonio

Also, a "trivial" point: the zero homomorphism exists between any two groups.

Edit: After rereading your post it seems that you got that already.

Originally Posted by tonio

You are wrong but I think not fundamentally: your argument about the order of the

image is right, but you oversaw the fact that the order of any of the three 2-cycles

you wrote there indeed is 2, so you can define , and

the same can be done with any of the other two elements of order 2 in that direct sum.

Of course, the above is not the whole story since you still have

to figure out how to define f on the other elements of that sum...

Tonio

OK, so this is what I get. Define , so in order for to be a homomorphism we must define (where are 2-cycles in ). But the we get since is not a 2-cycle. similarly we can show the same for other combinations of . Therefore is not a homomorphism.

So concluding we get that the only homomorphism is the trivial one.

Am I missing something here?

What happens if you send a single nonidentity element of Z2+Z2 to a 2-cycle, and send everything else to the identity?

Originally Posted by Tinyboss

What happens if you send a single nonidentity element of Z2+Z2 to a 2-cycle, and send everything else to the identity?

It still would not be a homomorphism. for example if we define and everything else goes to , then we get .

But if we choose 2 nonidentity elemnts from and send them to the same 2-cycle (for example ) then we will get a homomorphism. For instance , then and and similarly for other combinations.

So now in addition to the trivial homomorphism we get choices for different homomorphisms.

I think I am not missing anything now.

Thanks