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Thread: Permutation Groups

  1. #1
    Apr 2012

    Permutation Groups

    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!
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  2. #2
    Member ModusPonens's Avatar
    Aug 2010

    Re: Permutation Groups

    To prove that it is a homomorphism, you have to check for the product m2*m2 too. For the product e*e is trivial, but in a test I would put it. The rest is correct.
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  3. #3
    MHF Contributor

    Mar 2011

    Re: Permutation Groups

    you can actually prove a STRONGER result: any two groups of order 2 are isomorphic. let G = {e,a} and G' = {e',b} be two groups of order 2.

    we will show that φ:G→G' given by: φ(ak) = bk for k = 0,1 is an isomorphism.

    first, note that by definition, φ(e) = φ(a0) = b0 = e'.

    also, note that since G is a group, a2 must be an element of G, and since a2 = a implies a = e (and a and e are DIFFERENT),

    it must be the case that a2 = e. similarly reasoning show that b2 = e'.

    so in G we have:

    e*e = e
    a*e = a
    e*a = a
    a*a = e

    and in G' we have:

    e'*e' = e'
    b*e' = b
    e'*b = b
    b*b = e'

    it is clear that φ is bijective, and we could argue that since G and G' have "virtually" the same multiplication table, they are isomorphic.

    but let's use the definition of a homomorphism. to do so, we need a more compact description of the multiplication in G and G'. and that is fairly easy:

    (ak)(am) = a(k+m mod 2), and (bk)(bm) = b(k+m mod 2) (you can verify this holds, there are only 4 products to check).

    so φ(akam) = φ(a(k+m mod 2)) = b(k+m mod 2) = (bk)(bm) = φ(ak)φ(am).

    thus φ is a homomorphism.

    now, we simply observe that both {e, (1 3)} and {e, (1 2)} are groups of order 2.

    (here (1 3) is the map:

    3→1 of S3, and

    (1 2) is the map:

    2→1 of S2, just to clarify).

    in fact, you can even generalize THIS result (although i won't prove it here). if p is ANY positive prime integer, then any two groups of order p are isomorphic. so in light of this, it merely suffices to note that 2 is a prime number.
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