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.

vestirme4 2021-05-08 Answered

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

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Expert Answer

un4t5o4v
Answered 2021-05-09 Author has 4896 answers

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 $$\displaystyle{Z}{9}\cdot{Z}{9}\to{Z}{9}\cdot{Z}{3}\cdot{Z}{3}$$.

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