Group homomorphism

1. Aug 9, 2008

dJesse

I'm looking for a formula isomorphism from the set of integers under multiplication mod 13 to the set of integers under addition mod 12. I know the other way around it's easily expressed as a power of class 2. But this way I have no idea if its expressible as a formula.

2. Aug 9, 2008

HallsofIvy

That should be straightforward. Have you written down the operation tables for the two sets?

3. Aug 9, 2008

dJesse

I have now, but actually now i'm even doubting the isomorphism at itself. The only thing i can think of is something with a logarithm, base 2 cause that converts a 2 into 1 and a product into a sum. But i cant verify it cause it doesnt invert the mapping of the power of class 2.

4. Aug 9, 2008

5. Aug 9, 2008

dJesse

Okay i'm sorry, the title is wrong, i'm looking for a isomorphism. (In the end i'm looking for a homomorphism from the integers under multiplication mod13 to the complex numbers length 1.)

6. Aug 9, 2008

HallsofIvy

Well, that looks straight forward. When I write down the two operation tables I see several things: first the identity of {Z13,*} is 1 while the identity of {Z12, +} is 0. Any isomorphism must map 1 into 0. I notice that 12*12= 1 mod 13 and that 6+ 6= 0 mod 12. That is, that 12 and 6 have the property that they are there own inverses in the respective groups. Since they are the only values that have that property, 12 must map into 6: f(12)= 6 for an isomorphism from {Z13,*} to {Z12, +}. I then notice that 5*5= 12 mod 13 and that 3+ 3= 6 mod 12. Again, that tells me that f(5)= 3. Unfortunately, there is no n such that n*n= 5 mod 13 but I do see that 12*5= 8 mod 13 and 6+ 3= 9 mod 12. We must have f(8)= 12. Continuing in that way you should be able to identify f(n) for all n in Z13.

7. Aug 11, 2008

dJesse

Hey thanks for the stated reasoning. Now, I'm asking if it is also correct to just take the inverse of all the couples I get from the relation k->2^k, but then I get a different mapping from yours: f={(1,0),(2,1),(3,4),(4,2),(5,9),(6,5),(7,11),(8,3),(9,8),(10,10),(11,7),(12,6)}

8. Aug 11, 2008

dJesse

okay i found out it is not okay, but can you explain me why not? i thought the decomposition theorem for group homomorphisms sais it should be correct?

ps: i assume you meant f(8)=9

Last edited: Aug 11, 2008