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Math Help - Abstract Algebra homomorphism proof

  1. #1
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    Cool Abstract Algebra homomorphism proof

    Let ϕ: ℤ70 →ℤ5 be the group homomorphism that sends a class [a]70 to [a]5
    (a) Determine the subgroup lattice of ℤ70
    (b) Identify the subgroup of ℤ70 which is the kernel of ϕ
    (c) Is the kernel a cyclic group?
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  2. #2
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    Re: Abstract Algebra homomorphism proof

    (a) the order of a subgroup must _____ the order of the group
    (b) 5 (mod 70) is an element. can yuo think of others? what is the general pattern?
    (c) use the fact that φ is a homomorphism.
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  3. #3
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    Re: Abstract Algebra homomorphism proof

    This is what I have so far


    (a) Z₇₀ has lattice corresponding to the factors of 70:

    (Dropping coset notation for simplicity,)
    .........Z₇₀ = <1>
    ......./......|......\
    ....<2>..<5>..<7>
    ..../....X......X....\
    <10>..<14>..<35>
    ........\.....|....../
    ......<70> = {0}

    (b) ker ϕ = {[x]₇₀ in Z₇₀ : ϕ([x]₇₀) = [x]₅ = [0]₅} = <14>.

    (c) Yes; subgroups of a cyclic group are cyclic.
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  4. #4
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    Re: Abstract Algebra homomorphism proof

    14 = 4 (mod 5), not 0.

    by the FIT, Z70/ker(φ) has order 5. but |Z70/ker(φ)| = 70/|ker(φ)| = 5, so |ker(φ)| = 70/5 = 14.

    does |<14>| = 14?
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