Hi everyone,

I could really use some help with this problem:

Let H be the subset of S_{3}consisting of all the permutations that leave the element 2 fixed.

(i) Prove that H is a subgroup.

(ii) Prove that H is isomorphic to S_{2 }

NOTE: I am unfamiliar with this site, so my work is not written in standard notation. I am sorry it's difficult to read, but thank you for bearing with me.

I began the problem by writing out all of the permutations contained in S_{3}and found that the two permutations which kept 2 fixed were the identity permutation, p_{0}= (1 2 3 /1 2 3) and another permutation m_{2}= (1 2 3 /3 2 1). Therefore H = {p_{0}, m_{2}}

(i) To start the proof, I wrote out the group table for H and noted that it was closed.

Then for the identity:

p_{0 * }p_{0 = }p_{0 }m_{2 * }p_{0 }_{= }m_{2 }_{Therefore, the identity element, }p_{0, is also in H }_{ Then for the inverse: }p_{0 * }p_{0 = }p_{0 }m_{2 * }m_{2 }= p_{0 }Therefore, every element has an inverse

Thus, H is a subgroup of S_{3}

(ii) This part of the problem was much trickier for me...

I defined S_{3}as {e = (1 2 /1 2), a = (1 2 /2 1)}

I then defined a function from H to S2 such that identities were mapped to identities (f(Po) ---> e) and the remaining two permutations were mapped to each other (f(m2) ---> a). As a result, I had a bijective function. Therefore, the only thing left to find is a homomorphism.

Using the above defined function, I computed:

f(p0 * m2) = f(p0) * f(m2)

f(m2) = e * a

a = a

f(m2 * p0) = f(m2) * f(p0)

f(m2) = a * e

a = a

Thus, a homomorphism.

Therefore, H is isomorphic to S2.

Is this a valid proof?

Thank you for any guidance you can provide me!