There is an example on pg-155-156 of Pinter's abstract algebra book which is:
Let and be any groups and consider their product .
Remember that consists of all the ordered pairs as ranges over and ranges over . You multiply ordered pairs by multiplying corresponding
components; that is, the operation on is given by
Now, let be the function from onto given by
It is easy to check that is a homomorphism. Furthermore, is in the kernel of if and only if . This means that the kernel of consists of all the ordered pairs whose
second component is . Call this kernel ; then
We symbolize all this by writing
By the fundamental homomorphism theorem, we deduce that .
[It is easy to see that is an isomorphic copy of ; thus identifying with , we have shown that, roughly speaking, .]
My question is: why and how is that isomorphic to ?
Can anyone kindly tell me: why's that here isomorphic to ?