Let c be a positive integer with 1<=c<=n. Show that the set:

W ={ó exists in Sn such that ó(c)=c} is a subgroup of Sn. (The set W is called the stabilizer of c, because every element in W leaves c fixed.)

To show that the set W = {ó exists in Sn such that ó(c)=c} is a subgroup of Sn, we need to verify three conditions:

1. Closure: For any two elements ó1 and ó2 in W, their composition ó1∘ó2 must also be in W.
2. Identity: The identity element e, which is the permutation that maps every element to itself, must be in W.
3. Inverses: For any element ó in W, its inverse ó^(-1) must also be in W.

Let's go through each of these conditions:

1. Closure:
We need to show that for any two elements ó1 and ó2 in W, their composition ó1∘ó2 is also in W.

Let's assume ó1 and ó2 are permutations in Sn such that ó1(c) = c and ó2(c) = c. To show closure, we need to prove that (ó1∘ó2)(c) = c.

Since ó1(c) = c and ó2(c) = c, we know that ó1(ó2(c)) = ó1(c) = c. Thus, (ó1∘ó2)(c) = c. Therefore, the composition ó1∘ó2 leaves c fixed, and (ó1∘ó2) is in W.

2. Identity:
We need to show that the identity element e is in W.

The identity permutation e maps every element to itself, so e(c) = c. Therefore, e is in W.

3. Inverses:
We need to show that for any element ó in W, its inverse ó^(-1) is also in W.

Let's assume ó is a permutation in Sn such that ó(c) = c. To show inverses, we need to prove that ó^(-1)(c) = c.

Since ó(c) = c, applying the inverse operator on both sides of the equation gives us ó^(-1)(ó(c)) = ó^(-1)(c). Simplifying, we get (ó^(-1)∘ó)(c) = c.

Since the composition of a permutation and its inverse is the identity permutation, we have e(c) = c. Therefore, ó^(-1)(c) = c, and ó^(-1) is in W.

Since we have verified all three conditions (closure, identity, and inverses), we can conclude that the set W = {ó exists in Sn such that ó(c)=c} is a subgroup of Sn.