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Math Help - some abstract algebra help needed

  1. #1
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    some abstract algebra help needed

    Hi, I have a few questions on my abstract algebra homework that is bugging me, and wondered if anyone can send me some help!

    1. Let G, G' and G'' be groups. let φ: G > G' and γ: G' > G' be homomorphisms. Prove γ(φ(x)) is a homomorphism from G to G''.

    2. If G is an abelian group, prove that the mapping θ: G > G' defined by θ(x)=x^-1 is an automorphism.

    3. Suppose φ:G > G' is an isomorphism from G to G'. Prove that if H is a subgroup of G, then φ(H) (the image set of H under the map) is a subgroup of G''.

    any help would be appreciated, proofs always have me stumped!
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  2. #2
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    Quote Originally Posted by harbong View Post
    1. Let G, G' and G'' be groups. let φ: G > G' and γ: G' > G' be homomorphisms. Prove γ(φ(x)) is a homomorphism from G to G''.
    First the composition \gamma \circ \phi maps G\to G''.
    We need to show,
    \gamma \circ \phi (xy)=\gamma \circ \phi(x) \gamma \circ \phi(y)
    Now,
    \gamma \circ \phi(xy)=\gamma (\phi (xy))=\gamma(\phi(x) \phi(y)) because \phi is a homomorphism, and \phi(x),\phi(y)\in G' thus, \gamma(\phi(x)) \gamma(\phi(y)) because \gamma is also a homomorphism. Q.E.D.
    2. If G is an abelian group, prove that the mapping θ: G > G' defined by θ(x)=x^-1 is an automorphism.
    There is a problem with this question.
    Because an automorphism is defined between the same groups but you did not mention that. Thus, I assume the groups are the same.
    We need to show that,
    \theta: G\to G is an isomorpism.
    Simple, first this is a function between these two sets.
    Next,
    \theta (x)=\theta (y)
    Thus,
    x^{-1}=y^{-1}
    Thus,
    x=y
    This show the map is one-to-one.

    Next, for any g\in G we can choose g^{-1}\in G (for it has an inverse) then, \theta (g^{-1})=g. Thus this map is onto.

    Finally we show the homomorphism part.
    \theta(xy)=(xy)^{-1}
    In general the inverse for any group is,
    y^{-1}x^{-1}
    But this group is abelian thus,
    x^{-1}y^{-1}=\theta(x)\theta(y)
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  3. #3
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    Quote Originally Posted by harbong View Post
    3. Suppose φ:G > G' is an isomorphism from G to G'. Prove that if H is a subgroup of G, then φ(H) (the image set of H under the map) is a subgroup of G''.
    It is not necessary to state the map is a group isomorphism, saying that it is a group homomorphism is sufficient to show that condition.

    If,
    H\leq G
    Then,
    \phi [H]=\{\phi (h)|h\in H\}

    Associative: That part is simple, it uses the same binary operation as does G' which is a group and hence is associative.

    Identity: We know that, ge=eg=g for all g\in G then its map,
    \phi(ge)=\phi(eg)=\phi(g)
    But because this is a homomorphism we have,
    \phi(g)\phi(e)=\phi(e)\phi(g)=\phi(g)
    Thus we see that \phi(e) is the identity element.

    Inverse: Let \phi(h)\in \phi[H] then,
    hh^{-1}=h^{-1}h=e
    Thus, the map,
    \phi(hh^{-1})=\phi(h^{-1}h)=\phi(e)
    But because we have a homomorphism,
    \phi(h)\phi(h^{-1})=\phi(h^{-1})\phi(h)=\phi(e)
    Where, \phi(e) is the identity element.
    Thus, we see by definition of an inverse that,
    [\phi(h)]^{-1}=\phi(h^{-1}).

    Closure: Let \phi(h_1),\phi(h_2)\in \phi[H].
    Then,
    \phi(h_1)\phi(h_2)=\phi(h_1h_2)
    Where, h_1h_2\in H for it is closed.
    The by definition of a functional image we see that,
    \phi(h_1h_2)=\phi(h_1)\phi(h_2)\in \phi[H]
    And hence we have,
    \phi[H]\leq G'
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  4. #4
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    thank you so much!
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