Hello All,

I need help on this one. I'm sure it's simple, but I'm coming to the end off an all-nighter and so my brain has shut down.

Problem:If and are positive integers and is a divisor of , then .

Thanks,

Jhevon

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- Apr 6th 2008, 03:47 AMJhevonAbstract Algebra: Groups
Hello All,

I need help on this one. I'm sure it's simple, but I'm coming to the end off an all-nighter and so my brain has shut down.

**Problem:**If and are positive integers and is a divisor of , then .

Thanks,

Jhevon - Apr 6th 2008, 08:46 AMRebesques
Maybe the statement has an error? If , then I think

...Just a thought. - Apr 6th 2008, 08:48 AMJhevon
- Apr 6th 2008, 09:00 AMRebesques
Just go to sleep and we 'll figure this out later :p

- Apr 6th 2008, 09:09 AMRebesques
Yeap, I am being dumb. If , then the group generated by has elements, and so has elements.

- Apr 6th 2008, 09:11 AMJhevon
hehe, one of the reasons i did not sleep is that i wanted to get this before then :p

are you familiar with the "natural homomorphism" and the "Fundamental homomorphism theorem"? I'm thinking that we have to apply those somehow, but i am having trouble making it all fit - Apr 6th 2008, 09:17 AMThePerfectHacker
Define as . Show this map is well-defined, note, this is were you are going to use the fact that . Now argue that this is a homomorphism. This mapping is onto, and so . The kernel of the homomorphism will be the set (group) , I want to mention here another notation, we can write this also as , which looks a little neater. And so by fundamental homomorphism theorem .

- Apr 6th 2008, 09:25 AMJhevon
i just want to clarify something before i go storming off to do this. we did a theorem that says a homomorphism is one-to-one if and only if the kernel is the set with only the identity. that does not seem to be happening here, but we need one-to-one for the isomorphism

- Apr 6th 2008, 09:38 AMThePerfectHacker
- Apr 6th 2008, 09:52 AMJhevon
Ok, thanks. does this work?

Define by for all . This function is well-defined if and only if by homework problem 21.6 (you don’t expect me to re-invent the wheel, do you?).

We claim that this defines a homomorphism. For if , then .

This mapping is onto, since for any we can find a such that because (should I explain this part more?). So that we have . Clearly the kernel of this homomorphism is .

Thus we can define a mapping, by for each .

This mapping is an isomorphism of onto by the Fundamental Homomorphism Theorem. So that

- Apr 6th 2008, 10:32 AMThePerfectHacker
- Apr 6th 2008, 08:02 PMThePerfectHacker
This can also be proved using the third isomorphism theorem. Note that . And *.

Thus, .

*)Where . - Apr 6th 2008, 08:09 PMJhevon
that is very interesting. it seems you can treat it just like a quotient and actually "cancel" the . very nice.

thanks for the link as well. i'll read through it when i have the time, or maybe we'll get to this in class and i can use it as supplementary material - Apr 6th 2008, 08:28 PMThePerfectHacker
- Apr 6th 2008, 08:52 PMThePerfectHacker
It happens to be a good problem to work on if you want to work on your homomorphism.

**Third Theorem:**If then and .

Try to prove it. Hint: Define as .