If g = Z4 × Z2 and h = Z6, we want to determine what the quotient group g/h is isomorphic to.

1. Understanding the Notation:
First, let's understand the notation used in the question:
- Z4 represents the cyclic group of integers modulo 4: {0, 1, 2, 3}, where the group operation is addition modulo 4.
- Z2 represents the cyclic group of integers modulo 2: {0, 1}, where the group operation is addition modulo 2.
- Z6 represents the cyclic group of integers modulo 6: {0, 1, 2, 3, 4, 5}, where the group operation is addition modulo 6.
- × denotes the direct product of groups.

2. Direct Product of Groups:
The direct product of two groups, G and H, denoted as G × H, is a new group formed by taking all possible pairs of elements from G and H and defining a group operation on them. In this case, g = Z4 × Z2 means that the elements of g are ordered pairs (a, b) where a is from Z4 and b is from Z2.

3. Quotient Group:
The quotient group g/h is formed by taking the elements of g and identifying them based on their equivalence modulo h. In other words, elements of g that are congruent modulo h are considered the same in the quotient group.

4. Determining the Quotient Group:
To determine the quotient group g/h, we need to find the equivalence classes of g under the relation defined by h.

i. Equivalence Classes:
Let's examine the elements of g to determine their equivalence classes under h:
- The elements of g can be represented as {(0, 0), (0, 1), (1, 0), (1, 1), (2, 0), (2, 1), (3, 0), (3, 1)}.
- Modulo h = Z6, the elements of g are equivalent to each other if their sum modulo 6 is the same.
- We can list the equivalence classes:
- [(0, 0), (3, 1)]
- [(0, 1), (3, 0)]
- [(1, 0), (4, 1)]
- [(1, 1), (4, 0)]
- [(2, 0), (5, 1)]
- [(2, 1), (5, 0)]

ii. Quotient Group:
The quotient group g/h consists of the above equivalence classes. We can denote it as follows:
g/h = {[(0, 0), (3, 1)], [(0, 1), (3, 0)], [(1, 0), (4, 1)], [(1, 1), (4, 0)], [(2, 0), (5, 1)], [(2, 1), (5, 0)]}

5. Isomorphism:
To determine what the quotient group g/h is isomorphic to, we need to find a familiar group structure that exhibits the same properties as g/h.