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

a)Notice that 1 is of order 9 in Z9. Every element of $Z_3\times Z_3$ is of order 1 or 3; truly let $(a,b)\in Z_3\times Z_3,(a,b)=(0,0)$(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 $\varphi$(1) is an element of $Z_3\times Z_3$. This means that 1 is in kernel of $\varphi$, which means that $\varphi$ is not in injective ($\varphi$ is injective if and only if ker $\varphi =0$, since 0 is the neutral element of $Z_9$, which 1 is not.) This is a contradiction since $\varphi$ was supposed to be an isomorphism!
Thus, $Z_9$ an $Z_3\times Z_3$ are not isomorphic.
b)Notice that a $\in Z_9$ is of order 9 if and only if $\ge (a,9)=1$. Thus 1,2,4,5,7,8 are of order 9; thus 6 of them.
Now, (a,b) $\in Z_9\times Z_9$ 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 $6\times 3\times 3=54$ such elements.
Now suppose that ${\displaystyle \times :Z9\cdot Z9\to Z9\cdot Z3\cdot Z3}$ is an isomorphism. Then ${\displaystyle a\in Z9\cdot Z9}$ is of order ${\displaystyle 9<\to \varphi \left(a\right)\in Z9\cdot Z3\cdot Z3}$ 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 $Z_9\times Z_9$. Thus, there exists no isomorphism ${\displaystyle Z9\cdot Z9\to Z9\cdot Z3\cdot Z3}$.