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Homomorphisms from C_6 to C_4

  1. Mar 9, 2009 #1
    1. The problem statement, all variables and given/known data

    Show that there are exactly two homomorphisms f:C_(6) --> C_(4)

    2. Relevant equations
    Theorem.
    let f: G -> G1 and h: G -> G1 be homomorphisms and assume that G=<X> is generaed by a subset X. Then f = h if and only if f(x) = h(x) for all x in X.



    3. The attempt at a solution

    C6 = <g>, |g| = 6. The divisors of 6 are 1,2,3,6
    C4 = <g'>, |g'| = 4, the divisors of 4 are 1,2,4
    only 1 and 2 of C6 are the divisors of C4.
    so there are exactly two homomorphism.

    right?
     
  2. jcsd
  3. Mar 10, 2009 #2
    Re: homomorphism

    I'm pretty sure that there aren't any homomorphisms from [tex]C_6\to C_4[/tex]:

    To see this, consider, without loss of generality, the groups [tex]\mathbb{Z}_6:=\{0,1,2,3,4,5\}[/tex] and [tex]\mathbb{Z}_4:=\{0,1,2,3\}[/tex] under addition modulo 6 and 4, respectively. By the fundamental homomorphism theorem for groups, for any homomorphism h from [tex]\mathbb{Z}_6[/tex] to some group H,

    [tex]\mathbb{Z}_6/N\cong H[/tex]​

    for some [tex]N\lhd \mathbb{Z}_6[/tex]. But the only subgroups of [tex]\mathbb{Z}_6[/tex] (all of which are normal, since [tex]\mathbb{Z}_6[/tex] is abelian) are

    [tex]\{0,1,2,3,4,5\}, \{0,2,4\}, \{0,3\}, \{0\}.[/tex]​

    So, taking N to be one of these subgroups,

    [tex]|G/N|=1,2,3,\text{ or }6.[/tex]​

    Thus, the order of any homomorphic image of [tex]\mathbb{Z}_6[/tex] must be one of these. Specifically, the order of any homomorphic image cannot be 4. Thus, there is no homomorphism from [tex]\mathbb{Z}_6\to \mathbb{Z}_4[/tex] and by extension from [tex]C_6\to C_4[/tex].
     
    Last edited: Mar 10, 2009
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