To solve Exercise 8, we need to compute $\sigma^{100}$, where $\sigma$ is the permutation $\sigma = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 3 & 1 & 4 & 5 & 6 & 2 \end{pmatrix}$.

Step 1: Express $\sigma$ as a product of disjoint cycles

A permutation can be written as a product of disjoint cycles by following how each element maps under the permutation.

• Start with 1. Under $\sigma$, $1 \mapsto 3$, $3 \mapsto 4$, $4 \mapsto 5$, $5 \mapsto 6$, $6 \mapsto 2$, and $2 \mapsto 1$. Therefore, the cycle representation of $\sigma$ is:

$\sigma = (1\ 3\ 4\ 5\ 6\ 2)$

This is a 6-cycle permutation.

Step 2: Compute $\sigma^{100}$

Since $\sigma$ is a 6-cycle, the order of $\sigma$ is 6. This means $\sigma^6 = \text{id}$, where $\text{id}$ is the identity permutation.

To compute $\sigma^{100}$, notice that:

$100 \mod 6 = 4$

Thus, $\sigma^{100} = \sigma^4$.

Step 3: Compute $\sigma^4$

Now we need to compute $\sigma^4$. We already know $\sigma = (1\ 3\ 4\ 5\ 6\ 2)$, so we can apply the cycle four times.

• After one application of $\sigma$: $1 \mapsto 3$
• After two applications of $\sigma$: $1 \mapsto 4$
• After three applications of $\sigma$: $1 \mapsto 5$
• After four applications of $\sigma$: $1 \mapsto 6$

Thus, $\sigma^4$ is the permutation:

$\sigma^4 = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 6 & 5 & 1 & 2 & 3 & 4 \end{pmatrix}$

In cycle notation, this is:

$\sigma^4 = (1\ 6\ 4\ 2\ 5\ 3)$

Final Answer:

$\sigma^{100} = (1\ 6\ 4\ 2\ 5\ 3)$

Let me know if you want more details on this, and here are five related questions:

1. How do you calculate the inverse of a permutation in cycle notation?
2. What is the order of a permutation, and how is it related to the cycle structure?
3. How do you compute powers of a permutation efficiently?
4. What is the significance of the identity permutation in permutation groups?
5. Can you find the cycle notation for $\mu$?

Tip: Always reduce powers of permutations by their order (modulo the length of their cycles) to simplify the calculation.