Let σ =

1 2 3 4 5 6 7
5 4 2 3 6 7 1 
∈ S7
1. Write σ as a product of disjoint cycles.
2. Determine ord(σ).
3. Determine sgn(σ).
4. Compute the number of elements in the conjugacy class of σ, |C(σ)|.

Given the permutation \(\sigma\) in \( S_7 \):

\[
\sigma = \begin{pmatrix} 
1 & 2 & 3 & 4 & 5 & 6 & 7 \\
5 & 4 & 2 & 3 & 6 & 7 & 1
\end{pmatrix}
\]

Let's solve each part of the problem.

### 1. Write \(\sigma\) as a product of disjoint cycles.

To find the disjoint cycle representation of \(\sigma\), follow where each element is mapped under \(\sigma\):

- Start with 1: \(\sigma(1) = 5\), \(\sigma(5) = 6\), \(\sigma(6) = 7\), and \(\sigma(7) = 1\). This forms the cycle \((1 \, 5 \, 6 \, 7)\).
- Next, consider the smallest number not yet in a cycle, which is 2: \(\sigma(2) = 4\), \(\sigma(4) = 3\), and \(\sigma(3) = 2\). This forms the cycle \((2 \, 4 \, 3)\).

Thus, \(\sigma\) can be written as:

\[
\sigma = (1 \, 5 \, 6 \, 7)(2 \, 4 \, 3)
\]

### 2. Determine \(\text{ord}(\sigma)\).

The order of a permutation is the least common multiple (LCM) of the lengths of its disjoint cycles. The cycle lengths in \(\sigma\) are:

- \((1 \, 5 \, 6 \, 7)\) has length 4.
- \((2 \, 4 \, 3)\) has length 3.

The order of \(\sigma\) is:

\[
\text{ord}(\sigma) = \text{LCM}(4, 3) = 12
\]

### 3. Determine \(\text{sgn}(\sigma)\).

The sign (or parity) of a permutation is given by \(\text{sgn}(\sigma) = (-1)^k\), where \(k\) is the number of transpositions in the permutation's cycle decomposition.

- The cycle \((1 \, 5 \, 6 \, 7)\) can be decomposed into 3 transpositions: \((1 \, 7)(1 \, 6)(1 \, 5)\).
- The cycle \((2 \, 4 \, 3)\) can be decomposed into 2 transpositions: \((2 \, 3)(2 \, 4)\).

Thus, the total number of transpositions is \(3 + 2 = 5\). Since 5 is odd, the sign of \(\sigma\) is:

\[
\text{sgn}(\sigma) = (-1)^5 = -1
\]

### 4. Compute the number of elements in the conjugacy class of \(\sigma\), \(|C(\sigma)|\).

The size of the conjugacy class of a permutation \(\sigma\) in \(S_n\) depends on the cycle type. The formula is:

\[
|C(\sigma)| = \frac{n!}{z_1^{m_1} m_1! \times z_2^{m_2} m_2! \times \dots \times z_k^{m_k} m_k!}
\]

Where:
- \(n = 7\) (since \(\sigma \in S_7\))
- \(z_i\) is the length of the \(i\)-th cycle.
- \(m_i\) is the number of cycles of that length.

In our case:
- There is one cycle of length 4 (\(z_1 = 4, m_1 = 1\)).
- There is one cycle of length 3 (\(z_2 = 3, m_2 = 1\)).

So,

\[
|C(\sigma)| = \frac{7!}{4^1 \cdot 1! \times 3^1 \cdot 1!} = \frac{5040}{4 \times 3} = \frac{5040}{12} = 420
\]

Thus, the number of elements in the conjugacy class of \(\sigma\) is 420.

---

Do you have any questions, or would you like more details on any part of this solution?

### Related Questions:

1. How do you find the order of a permutation given its cycle structure?
2. What is the significance of the sign (sgn) of a permutation?
3. Can a permutation have an order that is not the LCM of its cycle lengths? Why or why not?
4. What is the relationship between conjugacy classes and cycle structures in symmetric groups?
5. How does the formula for the size of a conjugacy class apply to permutations with cycles of the same length?

### Tip:
When working with permutations, visualizing the cycle structure can make it easier to compute properties like order, sign, and conjugacy classes.

Explore the permutation σ = (1 5 6 7)(2 4 3) in the symmetric group S7. Learn how to determine its disjoint cycle representation, order, sign, and the number of elements in its conjugacy class.

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