Proof that H is isomorphic to K

To prove that H is isomorphic to K, we need to show that there exists a bijective function between the two sets that preserves the operation. In other words, we need to find a function that maps elements from H to K in a one-to-one manner, while preserving the function composition.

Definition of H and K

Let's first define the sets H and K based on the given conditions:

H = {b ∈ S5 | b(1) = 1}
K = {b ∈ S5 | b(2) = 2}

Defining the function

We can define a function f: H → K as follows:

For any b ∈ H, let f(b) be the permutation obtained by swapping the elements 1 and 2 in b.

In other words, if b(1) = 1 and b(2) = 2, then f(b)(1) = 2 and f(b)(2) = 1. For any other element j ≠ 1 or 2, f(b)(j) = b(j) since the function only swaps the elements 1 and 2.

Proving the bijection

To show that f is a bijection, we need to prove two properties:

1. Injectivity: Suppose f(b1) = f(b2) for some b1, b2 ∈ H. We need to show that b1 = b2.

Assume f(b1) = f(b2). Then, by definition of f, we have f(b1)(1) = f(b2)(1) = 2 and f(b1)(2) = f(b2)(2) = 1.

Since b1 and b2 are permutations in S5, their values at positions other than 1 and 2 must be the same. Therefore, b1(j) = b2(j) for all j ≠ 1 or 2.

But this implies b1 = b2 since the permutations are defined by their values at each position.

2. Surjectivity: For any b ∈ K, we need to find an element b' ∈ H such that f(b') = b.

Let b ∈ K. We define b' as the permutation obtained by swapping the elements 1 and 2 in b. By definition, b' ∈ H.

Now, we have f(b') = b since f swaps the elements 1 and 2 in any permutation.

Therefore, f is a bijection between H and K.

Preserving function composition

To show that the function f preserves the operation, we need to prove that for any b1, b2 ∈ H, f(b1 ∘ b2) = f(b1) ∘ f(b2).

Let b1, b2 ∈ H. We have:

f(b1 ∘ b2)(1) = f(b1(1 ∘ b2))(1) = f(b1(b2(1)))(1)

Since b2(1) = 1 (since b2 ∈ H), we have:

f(b1(b2(1)))(1) = f(b1(1))(1) = f(b1)(1) = 2

Similarly, we can prove that f(b1 ∘ b2)(2