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Math Help - semidirect product (sorry the text looks messy)

  1. #1
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    semidirect product (sorry the text looks messy)

    I am working on the semidirect product in group theory. And I got stuck on a question which is not difficult. Here is the question:

    If Q, K are groups and h: Q -->Aut(K) is a homomorphism
    Prove that the semidirect product is a group ( sorry, I can't type the notation of semidirect product). So, I have to let G = (K x Q, *, (1,1))

    The group operation * is
    (k_1,q_1)*(k_2,q_2) =(k_1h(q_1)(k_2), q_1q_2)

    This is my attempt.

    First show that the group is associative:
    [(k_1,q_1)*(k_2,q_2)]*(k_3,q_3) = (k_1h(q_1)(k_2) , q_1q_2)*(k_3,q_3)
    = (k_1h(q_1)(k_2)h(q_1q_2)(k_3),q_1q_2q_3)
    = ...???
    then I dont know how to simplify this. Can we use the fact that h is homomorphism.
    And:
    (k_1,q_1)*[(k_2,q_2)*(k_3,q_3)] = (k_1,q_1)*(k_2h(q_2)(k_3),q_2q_3)
    =(k_1h(q_1)(k_2)h(q_2)(k_3),q_1q_2q_3)
    = ......???

    Second, show that the group G has an identity,

    If: (1,1)*(k,q) = (1.h(1)k,1.q) = (k,q)
    and (k,q)*(1,1) = (k.h(q)(1),q.1) = (k.h(q),q) = (???,q)

    Finally, G has an inverse if (k,q)*(k^-1,q^-1) = (1,1)
    Thus, (k,q)*(k^-1,q^-1) = (k.h(q)(k^-1),q.q^-1) =(h(q), 1)

    Then G is a group

    I dont know what h(q) is equal to, so that I can simplify.

    If you find any mistake please correct it.

    Thank you so much

    P.S: Sorry about the text, it's a bit hard to read if you write down on paper it will be easier.

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  2. #2
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    Quote Originally Posted by knguyen2005 View Post
    first show that the group is associative:
    \begin{aligned}{}[(k_1,q_1)*(k_2,q_2)]*(k_3,q_3) &= (k_1h(q_1)(k_2) , q_1q_2)*(k_3,q_3) \\ &= (k_1h(q_1)(k_2)h(q_1q_2)(k_3),q_1q_2q_3)\end{align  ed}
    ...
    then i don't know how to simplify this.
    Correct. No simplification needed.

    Quote Originally Posted by knguyen2005 View Post
    and:
    \begin{aligned}(k_1,q_1)*[(k_2,q_2)*(k_3,q_3)] &= (k_1,q_1)*(k_2h(q_2)(k_3),q_2q_3) \\ &=(k_1h(q_1){\color{red}\bigl(}(k_2)h(q_2)(k_3) {\color{red}\bigr)},q_1q_2q_3) \end{aligned}

    Remember that h takes an element of Q to an automorphism of K. In this case, h(q_1) acts on the element
    k_2h(q_2)(k_3). And because it is an automorphism, it is multiplicative. Therefore h(q_1)\bigl(k_2.h(q_2)k_3\bigr) = h(q_1)k_2.h(q_1)\bigl(h(q_2)k_3\bigr). Also, h(q_1)h(q_2) = h(q_1q_2). Thus \begin{aligned}(k_1,q_1)*[(k_2,q_2)*(k_3,q_3)] &= (k_1h(q_1)(k_2)h(q_1q_2)(k_3),q_1q_2q_3) \\ &= [(k_1,q_1)*(k_2,q_2)]*(k_3,q_3)\end{aligned}

    Quote Originally Posted by knguyen2005 View Post
    second, show that the group g has an identity,

    if: (1,1)*(k,q) = (1.h(1)k,1.q) = (k,q)
    and (k,q)*(1,1) = (k.h(q)(1),q.1) = (k.h(q),q) = (???,q)

    You can't write h(q) on its own here; h(q) is an automorphism, and it has to act on an element of K. In this case, it acts on the identity element of K, and an automorphism always takes the identity to the identity. Therefore h(q)(1) = 1, and
    (k.h(q)(1),q.1) = (k.1,q.1) = (k,q), as required.

    Quote Originally Posted by knguyen2005 View Post
    finally, g has an inverse if (k,q)*(k^-1,q^-1) = (1,1)
    thus, (k,q)*(k^-1,q^-1) = (k.h(q)(k^-1),q.q^-1) =(h(q), 1)
    Careful here! The inverse of (k,q) has to be an element (j,p) such that (j,p)*(k,q) = (1,1). But (j,p)*(k,q) = (j.h(p)k, pq). If this is equal to (1,1) then pq = 1 (in Q), and so p=q^{-1}. But j.h(p)k = 1 (in K), and so j = \bigl(h(p)k\bigr)^{-1} = \bigl(h(q^{-1})k\bigr)^{-1} = h(q^{-1})k^{-1} (because an automorphism takes inverses to inverses). Therefore (k,q)^{-1} = (h(q^{-1})k^{-1}, q^{-1}). (In other words, the inverse of (k,q) is not (k^{-1},q^{-1}).)

    Afterthought: I think you would find this whole calculation easier to follow if you change the notation for an automorphism. Instead of writing h(q)(k), write it as h_q(k). That will remind you that h_q is an operator, that has to act on an element of K. It satisfies the conditions h_q(k_1k_2) = h_q(k_1)h_q(k_2), \bigl(h_q(k)\bigr)^{-1} = h_q(k^{-1}), and h_{q_1q_2}(k) = h_{q_1}(k)h_{q_2}(k).
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