Are Z3 x Z5 and Z15 Isomorphic?

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SUMMARY

Z3 x Z5 is isomorphic to Z15, as both groups contain 15 elements. To establish this isomorphism, one must demonstrate a one-to-one onto function between the two groups. The mapping function can be defined using unique primes, where (a,b) in Z3 x Z5 maps to p^a * q^b. Additionally, it is essential to show that this mapping preserves the group operation, satisfying the homomorphism condition F(a*b) = F(a) # F(b). The discussion emphasizes that the number of elements alone does not determine isomorphism, particularly between abelian and non-abelian groups.

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margaret23
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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
 
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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?