Existence of a One-One Homomorphism from Z12 to Sn

To determine the value of n for which there exists a one-one homomorphism from Z12 to Sn, we need to understand the properties and characteristics of these two groups.

Definition of Z12:
Z12, also known as the cyclic group of integers modulo 12, is a group of integers under addition with the operation defined as modular arithmetic. It consists of the elements {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}, where the addition is performed modulo 12.

Definition of Sn:
Sn is the symmetric group of order n, which consists of all possible permutations of n elements. The group operation is composition of permutations.

Homomorphism:
A homomorphism is a structure-preserving map between two groups. In this case, we are looking for a one-one (injective) homomorphism, which preserves the group operation and has distinct images for distinct elements.

Conditions for a One-One Homomorphism:
For a one-one homomorphism to exist from Z12 to Sn, the following conditions must be satisfied:
1. The order of the element in Z12 must divide the order of the corresponding element in Sn.
2. The homomorphism must be one-one, meaning distinct elements in Z12 should have distinct images in Sn.

Determining the Value of n:
To find the value of n, we need to consider the possible orders of elements in Z12 and their corresponding elements in Sn.

Possible Orders of Elements in Z12:
The order of an element a in Z12 is the smallest positive integer k such that a^k ≡ 0 (mod 12). The possible orders of elements in Z12 are {1, 2, 3, 4, 6, 12}.

Corresponding Elements in Sn:
For each order of an element in Z12, we need to find the corresponding elements in Sn whose orders are divisible by the order of the element in Z12.

For order 1:
The only element in Z12 with order 1 is 0. In Sn, the identity permutation has order 1. So, there is a one-one homomorphism from Z12 to Sn for this case.

For order 2:
There are no elements in Z12 with order 2. Therefore, there is no one-one homomorphism from Z12 to Sn for this case.

For order 3:
In Z12, the elements with order 3 are {1, 5, 7, 11}. In Sn, the possible permutations with order divisible by 3 are the cycles with length 3, such as (123), (456), etc. So, there is a one-one homomorphism from Z12 to Sn for this case.

For order 4:
In Z12, the elements with order 4 are {1, 5, 7, 11}. In Sn, the possible permutations with order divisible by 4 are the cycles with length 4, such as (1234), (5678), etc.