Let 𝜎 = (1 2 3 4)(5 6) be a permutation in 𝑆6

Let 𝐻 = ⟨𝜎 ⟩ be the subgroup of 𝑆6 generated by 𝜎.
3. Let 𝜏 = (1 3 5). Write down the elements of the left coset 𝜏𝐻 as products of disjoint
cycles.

To solve this problem, we first need to understand a few concepts related to permutations and cosets. Let's go through them step by step:

### 1. Generating the Subgroup \( H \)
The permutation \(\sigma = (1\ 2\ 3\ 4)(5\ 6)\) generates a subgroup \( H \) of \( S_6 \). To find the elements of \( H \), we determine the powers of \(\sigma\) until the identity permutation is reached.

#### Finding the Order of \(\sigma\)
The order of \(\sigma\) is the least common multiple (LCM) of the lengths of its disjoint cycles. The cycle \((1\ 2\ 3\ 4)\) has a length of 4, and the cycle \((5\ 6)\) has a length of 2. Therefore, the order of \(\sigma\) is:

\[
\text{order}(\sigma) = \text{lcm}(4, 2) = 4
\]

Thus, \( H \) consists of the permutations \(\sigma^0, \sigma^1, \sigma^2,\) and \(\sigma^3\).

#### Elements of \( H \)
- \(\sigma^0 = \text{id}\)
- \(\sigma^1 = (1\ 2\ 3\ 4)(5\ 6)\)
- \(\sigma^2 = (1\ 3)(2\ 4)\)
- \(\sigma^3 = (1\ 4\ 3\ 2)(5\ 6)\)

The subgroup \( H \) is therefore:

\[
H = \{\text{id}, (1\ 2\ 3\ 4)(5\ 6), (1\ 3)(2\ 4), (1\ 4\ 3\ 2)(5\ 6)\}
\]

### 2. Left Coset \(\tau H\)
We now determine the elements of the left coset \(\tau H\), where \(\tau = (1\ 3\ 5)\). The elements of \(\tau H\) are obtained by multiplying \(\tau\) with each element of \( H \).

#### Calculating \(\tau h\) for \( h \in H \)

- \(\tau \text{id} = (1\ 3\ 5)\)
- \(\tau (1\ 2\ 3\ 4)(5\ 6) = (1\ 3\ 5)(1\ 2\ 3\ 4)(5\ 6)\)

  Calculating the product:

  \[
  (1\ 3\ 5)(1\ 2\ 3\ 4) = (1\ 2\ 4\ 5)
  \]

  Thus, \((1\ 3\ 5)(1\ 2\ 3\ 4)(5\ 6) = (1\ 2\ 4\ 5)(5\ 6)\)

- \(\tau (1\ 3)(2\ 4) = (1\ 3\ 5)(1\ 3)(2\ 4)\)

  Calculating the product:

  \[
  (1\ 3\ 5)(1\ 3) = (1\ 5)
  \]

  Thus, \((1\ 3\ 5)(1\ 3)(2\ 4) = (1\ 5)(2\ 4)\)

- \(\tau (1\ 4\ 3\ 2)(5\ 6) = (1\ 3\ 5)(1\ 4\ 3\ 2)(5\ 6)\)

  Calculating the product:

  \[
  (1\ 3\ 5)(1\ 4\ 3\ 2) = (1\ 4\ 2\ 5)
  \]

  Thus, \((1\ 3\ 5)(1\ 4\ 3\ 2)(5\ 6) = (1\ 4\ 2\ 5)(5\ 6)\)

### 3. Elements of \(\tau H\)
The elements of the left coset \(\tau H\) in disjoint cycle notation are:

1. \((1\ 3\ 5)\)
2. \((1\ 2\ 4\ 5)(5\ 6)\)
3. \((1\ 5)(2\ 4)\)
4. \((1\ 4\ 2\ 5)(5\ 6)\)

These represent the unique elements of the coset \(\tau H\).

---

Would you like more details on any of the steps or have any questions?

### Further Questions
1. What is the significance of the order of an element in a permutation group?
2. How can we verify that the list of permutations is indeed a subgroup of \( S_6 \)?
3. What is the difference between left and right cosets in group theory?
4. Can cosets of a subgroup overlap in a permutation group?
5. What is the index of a subgroup, and how is it calculated?
6. How are the elements of a permutation group related to its algebraic structure?
7. How can we determine if two permutations are conjugate in a group?
8. How does the operation of a permutation affect its corresponding cycle structure?

### Tip
When calculating products of permutations, it's helpful to go step-by-step through each element's mapping to ensure accuracy.

Explore the concept of permutations and left cosets in group theory. Learn how to generate subgroups from permutations and calculate left cosets using specific examples. Suitable for advanced mathematics enthusiasts.

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