Number of non-isomorphic subgroups of G

To find the number of non-isomorphic subgroups of a cyclic group G of order 12, we can use the fact that the number of subgroups of order d divides the order of the group for any positive divisor d. This is known as Lagrange's theorem.

Determining the divisors of 12

The divisors of 12 are 1, 2, 3, 4, 6, and 12. We can examine each of these divisors to determine the number of subgroups of G with that order.

Subgroups of order 1

Every group has at least one subgroup of order 1, which is the trivial subgroup consisting of just the identity element. Therefore, G has 1 subgroup of order 1.

Subgroups of order 2

To determine the number of subgroups of order 2, we need to find the number of elements of order 2 in G. Since G is cyclic, it must have at least one element of order 2. Let's call this element a.

The subgroup generated by a, denoted 〈a〉, consists of the identity element and a. Since the order of a is 2, this subgroup has 2 elements. It is the only subgroup of order 2 in G.

Subgroups of order 3

Similarly, we need to find the number of elements of order 3 in G. Let's call an element of order 3 as b. The subgroup generated by b, denoted 〈b〉, consists of the identity element, b, and b^2. Since the order of b is 3, this subgroup has 3 elements.

There are 4 elements left in G that are not in 〈b〉. Let's call one of these elements c. The subgroup generated by c, denoted 〈c〉, consists of the identity element, c, c^2, and c^3. Since the order of c is 4, this subgroup has 4 elements.

Therefore, there are no subgroups of order 3 in G.

Subgroups of order 4

To determine the number of subgroups of order 4, we need to find the number of elements of order 4 in G. Let's call an element of order 4 as d. The subgroup generated by d, denoted 〈d〉, consists of the identity element, d, d^2, and d^3. Since the order of d is 4, this subgroup has 4 elements.

There are 2 elements left in G that are not in 〈d〉. Let's call one of these elements e. The subgroup generated by e, denoted 〈e〉, consists of the identity element, e, e^2, e^3, e^4, and e^5. Since the order of e is 6, this subgroup has 6 elements.

Therefore, there are no subgroups of order 4 in G.

Subgroups of order 6

To determine the number of subgroups of order 6, we need to find the number of elements of order 6 in G. Let's call