𝜎 = (1 3 5 7 8)(4 3 2)(7 6)(8 2 1) be a permutation in 𝑆8
.




Write 𝜎 as a product of disjoint cycles.

With good explanation 

Let's break down the permutation \( \sigma = (1\ 3\ 5\ 7\ 8)(4\ 3\ 2)(7\ 6)(8\ 2\ 1) \) step by step to write it as a product of disjoint cycles. Disjoint cycles are cycles that have no elements in common, meaning they operate on different sets of elements.

### Step 1: Start with the element 1

- In the first cycle \( (1\ 3\ 5\ 7\ 8) \), \( 1 \) is sent to \( 3 \).
- Next, look for \( 3 \) in the following cycles:
  - In the cycle \( (4\ 3\ 2) \), \( 3 \) is sent to \( 2 \).
  - Then in the cycle \( (8\ 2\ 1) \), \( 2 \) is sent to \( 1 \).
- So, starting from 1, we return to 1. Therefore, \( 1 \) is in a 1-cycle by itself.

### Step 2: Move to the element 2

- From the previous analysis, we know that \( 1 \) maps to itself, so we start with \( 2 \):
  - \( 2 \) in \( (4\ 3\ 2) \) maps to \( 4 \).
  - \( 4 \) does not appear in any other cycle, so the cycle is complete.

This gives us the cycle \( (1) \) and \( (2\ 4) \).

### Step 3: Continue with the element 3

- From the earlier trace:
  - \( 3 \) goes to \( 5 \) in the cycle \( (1\ 3\ 5\ 7\ 8) \).
  - \( 5 \) doesn't appear in any other cycle, so it ends here.

This forms the cycle \( (3\ 5) \).

### Step 4: Analyze the element 7

- \( 7 \) maps to \( 6 \) in the cycle \( (7\ 6) \).
- \( 6 \) doesn't appear in any other cycles, forming a 2-cycle \( (6\ 7) \).

### Step 5: Analyze the element 8

- \( 8 \) maps to \( 2 \) in \( (8\ 2\ 1) \) and 2 maps to 4 in \( (4\ 3\ 2) \). But both \( 2 \) and \( 4 \) are already considered in other cycles, so we stop here. However, since \( 8 \) has been considered in both cycles, we map \( 8 \) as a separate cycle.
- It turns out \( 8 \) maps back to itself through the cycles \( (8\ 2\ 1) \), so \( 8 \) is part of its own cycle.

Finally, the disjoint cycles formed are:

\[
\sigma = (1)(3\ 5)(6\ 7)(2\ 4)(8)
\]

### Simplification

Disjoint cycles \( (1) \), \( (8) \) can be ignored as they are 1-cycles, which means they leave elements unchanged. So, the simplified form is:

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

This is the disjoint cycle decomposition of \( \sigma \).

### Summary:

The permutation \( \sigma = (1\ 3\ 5\ 7\ 8)(4\ 3\ 2)(7\ 6)(8\ 2\ 1) \) can be expressed as the product of disjoint cycles:

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

These cycles are disjoint, meaning each cycle operates on a different set of elements, so they can be applied independently.

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Would you like more details on this, or do you have other questions? Here are 5 related questions:

1. How can you determine the order of the permutation \( \sigma \) from its disjoint cycle decomposition?
2. What is the inverse of the permutation \( \sigma \)?
3. Can you combine the cycles to create the original permutation \( \sigma \)?
4. What does it mean for cycles to be disjoint in permutation theory?
5. How do you calculate the effect of a permutation on a specific element?

**Tip:** When working with permutations, it's crucial to track where each element is sent and use the cycles in order to avoid confusion.

Learn how to decompose the permutation 𝜎 = (1 3 5 7 8)(4 3 2)(7 6)(8 2 1) into disjoint cycles with this detailed step-by-step guide. Understand the concept of disjoint cycles in permutation theory. Suitable for advanced high school students and beyond.

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