Explain why. (a) Z_9 is not isomorphic to Z_3 times Z_3. (b) Z_9 times Z_9 is not isomorphic to Z_9 times Z_3 times Z_3.

a)Notice that 1 is of order 9 in Z9. Every element of Z3*Z3 is of order 1 or 3. truly let $(a,b)\in Z3\ast Z3,(a,b)=/(0,0)$(so that its order is not 1.) Then 3(a,b)=(a,b)+(a,b)+(a,b)=(3a,3b)=(0,0) so it is of order 3.
Suppose that these groups are isomorphic, and let φ be some isomorphism.
Then ${\displaystyle \phi \left(3\right)=\phi (1+1+1)=\phi \left(1\right)+\phi \left(1\right)+\phi \left(1\right)=3\phi \left(1\right)=0}$
since φ(1) is an element of Z3*Z3. This means that 1 is in kernel of φ, which means that φ is not in injective (φ is injective if and only if ker φ={0}, since 0 is the neutral element of Z9, which 1 is not.) This is a contradiction since φ was supposed to be an isomorphism!
Thus, Z9 an Z3*Z3 are not isomorphic.
b)Notice that a ∈Z9 is of order 9 if and only if geg(a,9)=1. Thus 1,2,4,5,7,8 are of order 9. thus 6 of them.
Now, (a,b) ∈ Z9*Z9 is of order 9 if and only if a is of order 9 and b is of order 9. Thereforem there are ${\displaystyle 9\cdot 6+6\cdot 9=54+54=108}$ element of order 9.
On the order hand, ${\displaystyle (a,b,c)\in Z9\cdot Z3\cdot Z3}$ is of order 9 if and only if a is of order 9, so there are 6*3*3=54 such elements.
Now suppose that ${\displaystyle \phi :Z9\cdot Z9\to Z9\cdot Z3\cdot Z3}$ is an isomorphism. Then ${\displaystyle a\in Z9\cdot Z9}$ is of order ${\displaystyle 9<\to \phi \left(a\right)\in Z9\cdot Z3\cdot Z3}$ is of order 9 (isomorphisms preserve the order of the element). However, this is impossible, since ${\displaystyle Z9\cdot Z3\cdot Z3}$ has less elements of order 9 than Z9*Z9. Thus, there exists no isomorphism ${\displaystyle Z9\cdot Z9\to Z9\cdot Z3\cdot Z3}$.