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 \(\displaystyleφ{\left({3}\right)}=φ{\left({1}+{1}+{1}\right)}=φ{\left({1}\right)}+φ{\left({1}\right)}+φ{\left({1}\right)}={3}φ{\left({1}\right)}={0}\)

since \(\phi\)(1) is an element of \(Z_{3} \times Z_{3}\). This means that 1 is in kernel of \(\phi\), which means that \(\phi\) is not in injective (\(\phi\) is injective if and only if ker \(\phi={0}\), since 0 is the neutral element of \(Z_{9}\), which 1 is not.) This is a contradiction since \(\phi\) 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 \(\geq(a,9)=1\). 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, \(\displaystyle{\left({a},{b},{c}\right)}∈{Z}{9}\cdot{Z}{3}\cdot{Z}{3}\) 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 \(\displaystyle \times :{Z}{9}\cdot{Z}{9}\to{Z}{9}\cdot{Z}{3}\cdot{Z}{3}\) is an isomorphism. Then \(\displaystyle{a}∈{Z}{9}\cdot{Z}{9}\) is of order \(\displaystyle{9}{<}\to \phi{\left({a}\right)}∈{Z}{9}\cdot{Z}{3}\cdot{Z}{3}\) is of order 9 (isomorphisms preserve the order of the element). However, this is impossible, since \(\displaystyle{Z}{9}\cdot{Z}{3}\cdot{Z}{3}\) has less elements of order 9 than \(Z_{9} \times Z_{9}\). Thus, there exists no isomorphism \(\displaystyle{Z}{9}\cdot{Z}{9}\to{Z}{9}\cdot{Z}{3}\cdot{Z}{3}\).