Z15 can be represented as a pair of quantities a and b so that a has a range of 0 to 4 (Z5) and b has a range of 0 to 2 (Z3). So we have a scheme of representation:
(0,0)=0
(1,1)=11
(2,2)=7
(0,3)=3
(1,4)=14
(2,0)=10
(0,1)=6
(1,2)=2
(2,3)=13
(0,4)=9
(1,0)=5
(2,1)=1
(0,2)=12
(1,3)=8
(2,4)=4.
In general (X,Y) maps to (5X+6Y) modulo 15, where X is confined to modulo 3 and Y to modulo 5.
EXAMPLES:
ADDITION:
(2,3)+(1,2)=(13+2) mod 15 = 15 mod 15 = 0
(2,3)+(1,2)=((2+1) mod 3, (3+2) mod 5)=(0,0)=0
(1,4)+(0,2)=(14+12) mod 15 = 26 mod 15 = 11
(1,4)+(0,2)=((1+0) mod 3, (4+2) mod 5)=(1,1)=11
SUBTRACTION:
(2,3)-(1,2)=(13-2) mod 15 = 11
(2,3)-(1,2)=((2-1) mod 3, (3-2) mod 5)=(1,1)=11
(1,4)-(0,2)=(14-12) mod 15 = 2
(1,4)-(0,2)=((1-0) mod 3, (4-2) mod 5)=(1,2)=2
MULTIPLICATION
(2,3)*(1,2)=(13*2) mod 15 = 26 mod 15 = 11
(2,3)*(1,2)=(2,3)+(2,3)=(1,1)=11
(1,4)*(0,2)=(14*12) mod 15 = 168 mod 15 = 3
(1,4)*(0,2)=((1,4)+(1,4)+(1,4))+((1,4)+(1,4)+(1,4))+((1,4)+(1,4)+(1,4))+((1,4)+(1,4)+(1,4))
=(0,2)+(0,2)+(0,2)+(0,2)=(0,3)=3
Or is this cheating?!
The point is that the mapping shows for each of all the combinations of Z3 and Z5 there is one and only one Z15 element, and no elements of Z15 have been omitted, so demonstrating isomorphism.