Is Z3 ⊕ Z9, isomorpbic to Z27? Decide and answer why exactly?

Question

Is Z3 ⊕ Z9,  isomorpbic to   Z27?  Decide and answer why exactly?

likvau · Accepted Answer

To decide if the first group (the direct sum) is isomorpbic to $Z_{27},$
Description of the groups whose isomorphism is under question. The question arises because both (the direct sum as well as $Z_{27}$ contain the same number of elements:27). We will show that these two groups are NOT isomorphic
Now, $Z_{3} \oplus Z = {(x, y) : x \in Z_{3}, y \in Z_{9}}$
Where, $Z_{3} = {0, 1, 2}$ (addition modulo 3)
$Z_{9} = {0, 1, 2, 3, \dots 8}$ (addition modulo 9) whereas,
$Z_{27} = {0, 1, 2, 3, \dots 26}$ (addition modulo 27)
First observe that the direct sum is not a cyclic group.
Now, $Z_{3} \oplus Z_{9} = {(x, y) : x \in Z_{3}, y \in Z_{9}}$
Now, $9 (x, y) = (9 x, 9 y) = (0, 0), A (x, y) \in Z_{3} \oplus Z_{9}$
So any element has order at most 9
So, $Z_{3} \oplus Z_{9}$ is not cycllc.
(otherwise, $E (x, y) \in Z_{3} o i Z_{9}$ with order 27)
On the other hand $Z_{27}$ is a cyclic group, for example , with 1 as a generator. So the given two groups are not isomorphic.

Is Z_{3} oplus Z_{9}, isomorpbic to Z_{27}? Decide and answer why exactly?

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