Hi everyone,
I could really use some help with this problem:
Let H be the subset of S3 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 S2
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 S3 and found that the two permutations which kept 2 fixed were the identity permutation, p0 = (1 2 3 /1 2 3) and another permutation m2 = (1 2 3 /3 2 1). Therefore H = {p0, m2}
(i) To start the proof, I wrote out the group table for H and noted that it was closed.
Then for the identity:
p0 * p0 = p0
m2 * p0 = m2
Therefore, the identity element, p0, is also in H
Then for the inverse:
p0 * p0 = p0
m2 * m2 = p0
Therefore, every element has an inverse
Thus, H is a subgroup of S3
(ii) This part of the problem was much trickier for me...
I defined S3 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!