Are Z3 x Z5 and Z15 Isomorphic?

  • Thread starter Thread starter margaret23
  • Start date Start date
Click For Summary

Homework Help Overview

The discussion revolves around the isomorphism between the groups Z3 x Z5 and Z15, both of which involve modular arithmetic. Participants are exploring the properties and structures of these groups to determine if they are isomorphic.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Some participants attempt to establish isomorphism by counting elements and suggesting mappings. Others raise the need for a one-to-one onto function and discuss the importance of homomorphism in proving isomorphism.

Discussion Status

The discussion is ongoing, with participants providing various approaches to demonstrate isomorphism. There is recognition that simply having the same number of elements does not suffice for isomorphism, and some participants suggest exploring the order of elements as a potential avenue for proof.

Contextual Notes

Participants are considering the definitions and properties of groups, including the implications of being abelian versus non-abelian. There is also mention of unique factorization and the need for a well-defined mapping between the groups.

margaret23
Messages
11
Reaction score
0
Show that Z3 X Z5 is isomorphic to Z15 (where Zn is like the intergers mod n)

i m not sure if i m proving it right.. if i first right out Z3 x Z5 ={(1x1), (1X2),(1x3),(1x4), (1X5), (2X1),(2x2),(2x3), (2x4), (2X5), (3x1), (3x3),(3x3),(3x4), (3X5)}
Z15={1,2,3...15}

both of which have 15 elements... there for they are in the same form?? therefore isomorphic??

i would appreciate any help
thanks
 
Physics news on Phys.org
To show that Z3 X Z5 is isomorphic to Z15 you have to come up with a one-to-one onto function between the two. I think something like this might work if (a,b) is in Z3 X Z5 then the mapping p^a*q^b where p and q are unique primes is one-to-one by the unique factorization thm. However, it's not onto but we do know it's well ordered so we have elements a_1<a_2<...<a_15 then you can take a_n to n and this would make it onto.
 
buzzmath said:
To show that Z3 X Z5 is isomorphic to Z15 you have to come up with a one-to-one onto function between the two. I think something like this might work if (a,b) is in Z3 X Z5 then the mapping p^a*q^b where p and q are unique primes is one-to-one by the unique factorization thm. However, it's not onto but we do know it's well ordered so we have elements a_1<a_2<...<a_15 then you can take a_n to n and this would make it onto.

To show they're isomorphic
You also need to show that if F is the mapping between them, * is the operation on Z3xZ5, and # is the operation on Z15

that
F(a*b) = F(a)#F(b)
where a and b are elements of Z3xZ5 if the function maps Z3xZ5 to Z15.
 
Sorry, I forgot to say that the mapping also has to be a homomorphism.
 
margaret23 said:
Show that Z3 X Z5 is isomorphic to Z15 (where Zn is like the intergers mod n)

i m not sure if i m proving it right.. if i first right out Z3 x Z5 ={(1x1), (1X2),(1x3),(1x4), (1X5), (2X1),(2x2),(2x3), (2x4), (2X5), (3x1), (3x3),(3x3),(3x4), (3X5)}
Z15={1,2,3...15}

both of which have 15 elements... there for they are in the same form?? therefore isomorphic?

The number of elements does not determine the group. This is clear, since abelian groups cannot be isomorphic to non-abelian groups.

It is easier to map elements of Z15 to the other group, as it happens, from the description you've given.

Of course, if you can show that some element of Z3xZ5 has order 15 you're also done, but do you understand why?