To write the permutation $\sigma = (24)(1759)(428)(735)$ as a product of disjoint cycles, we need to follow the action of each element in the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ under the permutation $\sigma$.

Step 1: Identify the action of each element

• Start with $1$:

  • In $(1759)$, $1 \mapsto 7$.
  • In $(735)$, $7 \mapsto 3$.
  • In $(428)$, $3$ is not included, so it stays $3$.
  • In $(24)$, $3$ is not included, so it stays $3$.
  • Thus, $1 \mapsto 3$.

• Next, follow $3$:

  • In $(1759)$, $3$ is not included, so it stays $3$.
  • In $(735)$, $3 \mapsto 5$.
  • In $(428)$, $5$ is not included, so it stays $5$.
  • In $(24)$, $5$ is not included, so it stays $5$.
  • Thus, $3 \mapsto 5$.

• Next, follow $5$:

  • In $(1759)$, $5 \mapsto 9$.
  • In $(735)$, $9$ is not included, so it stays $9$.
  • In $(428)$, $9$ is not included, so it stays $9$.
  • In $(24)$, $9$ is not included, so it stays $9$.
  • Thus, $5 \mapsto 9$.

• Next, follow $9$:

  • In $(1759)$, $9 \mapsto 1$.
  • In $(735)$, $1$ is not included, so it stays $1$.
  • In $(428)$, $1$ is not included, so it stays $1$.
  • In $(24)$, $1$ is not included, so it stays $1$.
  • Thus, $9 \mapsto 1$.

So, we have the cycle $(1, 3, 5, 9)$.

Step 2: Continue with remaining elements

• Next, follow $2$:

  • In $(24)$, $2 \mapsto 4$.
  • In $(428)$, $4 \mapsto 8$.
  • In $(1759)$, $8$ is not included, so it stays $8$.
  • In $(735)$, $8$ is not included, so it stays $8$.
  • Thus, $2 \mapsto 8$.

• Next, follow $8$:

  • In $(24)$, $8$ is not included, so it stays $8$.
  • In $(428)$, $8 \mapsto 2$.
  • In $(1759)$, $2$ is not included, so it stays $2$.
  • In $(735)$, $2$ is not included, so it stays $2$.
  • Thus, $8 \mapsto 2$.

So, we have the cycle $(2, 8, 4)$.

• Next, follow $7$:
  • In $(1759)$, $7 \mapsto 1$, but we already handled $1$ in the first cycle.
  • In $(735)$, $7 \mapsto 3$, but we already handled $3$ in the first cycle.
  • Thus, $7$ is already in the cycle we found earlier.

Step 3: List the cycles

The permutation $\sigma$ can now be expressed as the product of disjoint cycles:

$\sigma = (1, 3, 5, 9)(2, 8, 4)(7)$

Since $(7)$ is just the identity on 7, it can be omitted:

$\sigma = (1, 3, 5, 9)(2, 8, 4)$

This is the final disjoint cycle decomposition.

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Would you like more details on this, or do you have any questions? Here are some related questions you might consider:

1. How do you determine the order of a permutation given in disjoint cycle form?
2. What is the significance of disjoint cycles in group theory?
3. How do you compose two permutations written in cycle notation?
4. How do you convert a permutation from cycle notation to two-line notation?
5. What is the inverse of a permutation in disjoint cycle notation?

Tip: When working with permutations, always ensure that cycles are disjoint (i.e., they do not share any common elements) when decomposing them into products of cycles.