Final answer:
A4 is not isomorphic to Z12 because it's not cyclic. Z3 ⊕ Z4 is isomorphic to Z12 because they are both cyclic of order 12 and the component orders are relatively prime. Z6 ⊕ Z2 is not isomorphic to Z12 because it does not contain an element of order 12.
Step-by-step explanation:
To determine if the groups A4, Z3 ⊕ Z4, and Z6 ⊕ Z2 are isomorphic to Z12, we have to consider their group structures and properties. The group Z12 is a cyclic group of order 12. This means that there is an element in Z12 that generates the entire group by its powers, and there are exactly 12 elements in the group.
A4, the alternating group on four elements, is not cyclic and it has 12 elements but does not have an element of order 12, so it cannot be isomorphic to Z12.
Z3 ⊕ Z4 is a direct sum of the cyclic groups Z3 and Z4, which have 3 and 4 elements respectively. Their orders multiply to give 12, and since 3 and 4 are relatively prime, the direct sum Z3 ⊕ Z4 is cyclic and thus is isomorphic to Z12.
Z6 ⊕ Z2 is a direct sum of Z6 and Z2. However, the highest order an element in Z6 ⊕ Z2 can have is lcm(6,2) = 6, thus it cannot be isomorphic to Z12, which requires an element of order 12.