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

vestirme4

Answered question

2021-05-08

Explain why (a) Z9 is not isomorphic to Z3×Z3;
(b) Z9×Z9 is not isomorphic to Z9×Z3×Z3.

Answer & Explanation

un4t5o4v

un4t5o4v

Skilled2021-05-09Added 105 answers

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 φ(3)=φ(1+1+1)=φ(1)+φ(1)+φ(1)=3φ(1)=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 (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 96+69=54+54=108 element of order 9.
On the order hand, (a,b,c)Z9Z3Z3 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 ×:Z9Z9Z9Z3Z3 is an isomorphism. Then aZ9Z9 is of order 9<ϕ(a)Z9Z3Z3 is of order 9 (isomorphisms preserve the order of the element). However, this is impossible, since Z9Z3Z3 has less elements of order 9 than Z9×Z9. Thus, there exists no isomorphism Z9Z9Z9Z3Z3.

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