Introduction:
In this question, we are asked to determine the total number of non-isomorphic groups of order 142. To solve this problem, we need to understand the concept of group isomorphism and apply it to the given order.

Group Isomorphism:
Group isomorphism is a relation between two groups that preserves the group structure. Two groups, G and H, are isomorphic if there exists a bijective function (one-to-one correspondence) f from G to H such that for any two elements a and b in G, the operation on a and b is the same as the operation on f(a) and f(b) in H.

Key Point:
The number of non-isomorphic groups of a given order can be determined by finding the number of distinct isomorphism classes of groups of that order.

Prime Factorization:
To determine the number of non-isomorphic groups of order 142, we first need to factorize 142 into its prime factors. The prime factorization of 142 is 2 × 71.

Key Point:
The number of non-isomorphic groups of order p × q, where p and q are distinct prime numbers, is equal to the number of non-isomorphic groups of order p multiplied by the number of non-isomorphic groups of order q.

Step 1: Number of Non-Isomorphic Groups of Order 2:
The number of non-isomorphic groups of order 2 is 1. This is because the only group of order 2 is the cyclic group of order 2, which is isomorphic to itself.

Step 2: Number of Non-Isomorphic Groups of Order 71:
To determine the number of non-isomorphic groups of order 71, we can use the fact that groups of prime order are cyclic. Therefore, the number of non-isomorphic groups of order 71 is also 1.

Step 3: Total Number of Non-Isomorphic Groups:
Using the key point mentioned earlier, the total number of non-isomorphic groups of order 142 is equal to the number of non-isomorphic groups of order 2 multiplied by the number of non-isomorphic groups of order 71, which is 1 × 1 = 1.

Conclusion:
The total number of non-isomorphic groups of order 142 is 1.