Results 1 to 7 of 7

Math Help - The number of homomorphisms

  1. #1
    Newbie
    Joined
    Jul 2008
    Posts
    5

    The number of homomorphisms

    Find the number of homomorphisms between:
    #Hom(S_4, C_4) (intuitively something tells me this is two)
    #Hom(C_4, S_4)
    #Hom(C_6, S_3)
    #Hom(D_3, A_4) (D_3 has order 6)

    Any help would be appreciated!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by Aedolon View Post
    Find the number of homomorphisms between:
    #Hom(S_4, C_4) (intuitively something tells me this is two)
    If \theta : S_4 \to \mathbb{Z}_4 is a homomorphism then \theta is completely determined by its value on a generating set.
    We will use the generating set \left< (12),(1234) \right>.
    Let \sigma \in S_4 and k = \text{ord}(\sigma).
    Then 0 = \theta ( \text{id} ) = \theta (\sigma^k) = \theta(\sigma)^k.
    Thus, it is necessary that \text{ord} (\theta(\sigma)) divides \text{ord}(\sigma).
    Since \text{ord}(12) = 2 it means \theta (12) = 0,2.
    Note that (234)=(12)(1234) thus \theta(234) = \theta(12) + \theta(1234).
    Since \text{ord}(234)=3 it means (by above) \theta(234)=e.
    Thus, \theta(12) + \theta(1234) = e.
    If \theta(12) = e then \theta(1234)=e.
    This is the trivial homomorphism i.e. everything is mapped to 0.

    The other possibility is \theta(12) = 2 and then \theta(1234) = 2.
    We will prove this defines a homomorphism.
    Begin with \theta(234) = 0.
    Conjugating* (12) by powers of (234) gives (13) and (14).
    Therefore, \theta(12) = \theta(13) = \theta(14) = 2.
    Note that (x,y) = (1,x)(1,y)(1,x).
    It follows that \theta (x,y) = 2+2+2=2.

    This means if \sigma is even then it is a product of even number of transpositions and so \theta (\sigma) = 0 since 2+...+2 even number of times is 0. If \sigma is odd then \sigma is a product of odd number of transpositions and so \theta(\sigma) = 2 since 2+...+2 odd number of times is 2. Therefore, \theta is simply the mapping which sends even permutations into 0 and odd permutations into 2. It is trivial to proof now that it is indeed a homomorphism.

    Thus, \text{hom}(S_4,\mathbb{Z}_4) \simeq \mathbb{Z}_2.
    And like you guessed it does indeed have only two elements.

    *)Definition, a conjugated by b means the product bab^{-1}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by Aedolon View Post
    #Hom(C_4, S_4)
    Let \theta: \mathbb{Z}_4 \to S_4 be a homomorphism.
    Since \mathbb{Z}_4 = \left< 1 \right> it means \theta is determined by its value on 1.
    Similarly to above \text{ord}(\theta(1)) must divide \text{ord}(1) = 4.
    However, in case of cyclic group this condition is also sufficient to define a homomorphism.

    Any \sigma \in S_4 can be written in terms of disjoint cycles.
    Therefore, for \text{ord}(\sigma) = 1,2,4 the possibilities are:
    \sigma = (xy), \sigma = (xy)(zw), \sigma = (xyzw).
    Thus, the non-trivial values for \theta (1) can be:
    (12),(13),(14),(23),(24),(34)
    (12)(34),(13)(24),(14)(23)
    (1234),(1243),(1324),(1342),(1423),(1432)
    This gives us a total of 15+1=16 elements.
    ----

    A faster way to count the # of such elements is to know the conjugacy class equation for S_4.
    It is 24 = 1+3+6+6+8.
    The 8 are the conjugacy classes of 3-cycles, while everything else is what we are looking for thus altogether we have 1+3+6+6 = 16 elements.
    But if you never learned this then just avoid it.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by Aedolon View Post
    #Hom(C_6, S_3)
    Same idea, \mathbb{Z}_6 = \left< 1 \right>.
    We also require \text{ord}(\theta (1)) to divide \text{ord}(1)=6.
    But if \sigma \in S_3 then \text{ord}(\sigma) divides |S_3| = 3! = 6.
    Thus, \theta(1) can be any element in S_3.
    Thus, |\text{hom}(\mathbb{Z}_6,S_3)| = 6.

    (I need more time on the last one).
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jul 2008
    Posts
    5
    Great! You have helped me a lot on this one :-)!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Jul 2008
    Posts
    5
    Anyone for the last one?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by Aedolon View Post
    Anyone for the last one?
    How about you try doing the last one?
    First D3=S3. And S_3 = \{ \text{id},(123),(123)^2,(12),(123)(12),(123)^2(12)  \}. Thus (123),(12) generate S_3.
    Can you continue from here ?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Number of different homomorphisms
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: March 15th 2011, 07:45 AM
  2. Number of homomorphisms between two groups
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: January 3rd 2011, 05:10 AM
  3. Number of Homomorphisms from Zn to Zm
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 9th 2010, 04:20 PM
  4. Number of group homomorphisms
    Posted in the Math Challenge Problems Forum
    Replies: 0
    Last Post: June 8th 2010, 11:17 AM
  5. Homomorphisms
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: April 23rd 2007, 11:53 AM

Search Tags


/mathhelpforum @mathhelpforum