Question

Let \(\sigma \in S_8\), where \(S_8\) is the permutation group on 8 elements. Suppose \(\sigma\) is the product of \(\sigma_1\) and \(\sigma_2\), where \(\sigma_1\) is a 4-cycle and \(\sigma_2\) is a 3-cycle in \(S_8\). If \(\sigma_1\) and \(\sigma_2\) are disjoint cycles, then the number of elements in \(S_8\) which are conjugate to \(\sigma\) is ..................

Hint:

Remember that disjoint cycles commute. The order of the product doesn't matter (\(\sigma_1\sigma_2 = \sigma_2\sigma_1\)). The key to conjugacy class problems in \(S_n\) is to correctly identify the cycle structure and then apply the counting formula. Be sure to account for all \(n\) elements, including fixed points (1-cycles).

Answer 1

Step 1: Understanding the Concept:
In a symmetric group \(S_n\), two permutations are conjugate if and only if they have the same cycle structure. The question asks for the number of elements conjugate to \(\sigma\), which is equivalent to finding the size of the conjugacy class of \(\sigma\).
Step 2: Key Formula or Approach:
First, we determine the cycle structure of \(\sigma\). Since \(\sigma = \sigma_1 \sigma_2\) where \(\sigma_1\) is a 4-cycle and \(\sigma_2\) is a 3-cycle, and they are disjoint, they operate on \(4+3=7\) distinct elements. As \(\sigma \in S_8\), the remaining \(8-7=1\) element is a fixed point, which is a 1-cycle. Therefore, the cycle structure of \(\sigma\) is (4, 3, 1).
The number of permutations in \(S_n\) with a cycle structure consisting of \(a_1\) cycles of length 1, \(a_2\) cycles of length 2, ..., \(a_n\) cycles of length n is given by the formula: \[ \text{Size of Conjugacy Class} = \frac{n!}{\prod_{k=1}^{n} k^{a_k} a_k!} \] Step 3: Detailed Explanation or Calculation:
For our permutation \(\sigma \in S_8\), the cycle structure is (4, 3, 1).
This means we have:
One cycle of length 4 (\(k=4, a_4=1\)).
One cycle of length 3 (\(k=3, a_3=1\)).
One cycle of length 1 (\(k=1, a_1=1\)).
Zero cycles of other lengths (\(a_k=0\) for \(k \neq 1, 3, 4\)).
The total number of elements is \(n=8\).
Plugging these values into the formula: \[ \text{Number of elements} = \frac{8!}{1^{a_1} a_1! . 2^{a_2} a_2! . 3^{a_3} a_3! . 4^{a_4} a_4! .} \] \[ \text{Number of elements} = \frac{8!}{1^1 . 1! . 3^1 . 1! . 4^1 . 1!} \] \[ \text{Number of elements} = \frac{8!}{1 . 1 . 3 . 1 . 4 . 1} = \frac{8!}{12} \] Step 4: Final Answer: Now, we calculate the value: \[ 8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320 \] \[ \text{Number of elements} = \frac{40320}{12} = 3360 \] Step 5: Why This is Correct: The cycle structure of the permutation \(\sigma\) is correctly identified as (4, 3, 1). The standard formula for the size of a conjugacy class in \(S_n\) is applied with the correct parameters for \(n=8\) and the given cycle structure. The calculation yields 3360, matching the answer key.