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.

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.

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a)Notice that 1 is of order 9 in Z9. Every element of Z3*Z3 is of order 1 or 3. truly let $$(a,b)∈ Z3*Z3, (a,b)=/(0,0)$$(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 $$\displaystyleφ{\left({3}\right)}=φ{\left({1}+{1}+{1}\right)}=φ{\left({1}\right)}+φ{\left({1}\right)}+φ{\left({1}\right)}={3}φ{\left({1}\right)}={0}$$
since 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, $$\displaystyle{\left({a},{b},{c}\right)}∈{Z}{9}\cdot{Z}{3}\cdot{Z}{3}$$ is of order 9 if and only if a is of order 9, so there are 6*3*3=54 such elements.
Now suppose that $$\displaystyleφ:{Z}{9}\cdot{Z}{9}\to{Z}{9}\cdot{Z}{3}\cdot{Z}{3}$$ is an isomorphism. Then $$\displaystyle{a}∈{Z}{9}\cdot{Z}{9}$$ is of order $$\displaystyle{9}{<}\toφ{\left({a}\right)}∈{Z}{9}\cdot{Z}{3}\cdot{Z}{3}$$ 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 Z9*Z9. Thus, there exists no isomorphism $$\displaystyle{Z}{9}\cdot{Z}{9}\to{Z}{9}\cdot{Z}{3}\cdot{Z}{3}$$.