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Math Help - Modules

  1. #1
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    Modules

    ok, trying to type this in Latex is frying my brain, so please forgive the typography.

    suppose we have an abelian group G, and an action on G by another group H (which may or may not be abelian). i will write this action h.g for h in H, g in G.

    for the group ring Z[H], does the scalar multiplication:

    (\sum_i a_ih_i)g = \sum_i a_i(h_i.g)

    (where the sum on the right is tacitly appealing to the fact that G is a Z-module)

    make G a Z[H]-module?

    the sticking point for me, is that it seems to me that g-->h.g is not necessarily a group homomorpism (that is h.(g1 + g2) may not be h.g1 + h.g2). so i can't demonstrate that z(g1 + g2) = zg1 + zg2.

    (showing (z1 + z2)g = z1g + z2g isn't hard).

    i am also having trouble "collecting terms" to show that z1(z2g) = (z1z2)g (where z1,z2 are two finite Z-linear combinations of the elements of H). my understanding of the ring product in Z[H] is that it is kind of like "polynomial" multiplication, except that instead of collecting all the (x^i)(x^j) terms where i+j = k, we collect all the "coefficients" of the products (h_i)(h_j) where these equal h_k (if the number of terms in one element is m, and the number of terms in the other element is n, we get mn terms collapsed into max(m,n) terms).

    i think that 1e_H is a unity for Z[H], so showing (1e_H)g = g isn't hard, so i am stuck with just 2 of the 4 module axioms.

    (note: i suppose i should say left ZH-module, but meh.)
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  2. #2
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    Quote Originally Posted by Deveno View Post
    ok, trying to type this in Latex is frying my brain, so please forgive the typography.

    suppose we have an abelian group G, and an action on G by another group H (which may or may not be abelian). i will write this action h.g for h in H, g in G.

    for the group ring Z[H], does the scalar multiplication:

    (\sum_i a_ih_i)g = \sum_i a_i(h_i.g)

    (where the sum on the right is tacitly appealing to the fact that G is a Z-module)

    make G a Z[H]-module?

    the sticking point for me, is that it seems to me that g-->h.g is not necessarily a group homomorpism (that is h.(g1 + g2) may not be h.g1 + h.g2). so i can't demonstrate that z(g1 + g2) = zg1 + zg2.

    (showing (z1 + z2)g = z1g + z2g isn't hard).

    i am also having trouble "collecting terms" to show that z1(z2g) = (z1z2)g (where z1,z2 are two finite Z-linear combinations of the elements of H). my understanding of the ring product in Z[H] is that it is kind of like "polynomial" multiplication, except that instead of collecting all the (x^i)(x^j) terms where i+j = k, we collect all the "coefficients" of the products (h_i)(h_j) where these equal h_k (if the number of terms in one element is m, and the number of terms in the other element is n, we get mn terms collapsed into max(m,n) terms).

    i think that 1e_H is a unity for Z[H], so showing (1e_H)g = g isn't hard, so i am stuck with just 2 of the 4 module axioms.

    (note: i suppose i should say left ZH-module, but meh.)
    G is not necessarily a \mathbb{Z}[H]-module. for example let G =\mathbb{R} and H = \langle \pi \rangle, the multiplicative group generated by \pi. define the action of H on G by h \cdot x = x + \log h, for all h \in H and x \in G. clearly what we defined is a group action but

    h \cdot (x + y) = x+y + \log h \neq x + y + 2 \log h = h \cdot x + h \cdot y.

    so G is not a \mathbb{Z}[H]-module. also the identity z_1(z_2g)=(z_1z_2)g does not hold:

    if h_i \in H and x \in G, then h_1 \cdot ((h_2 +h_3) \cdot x) = 2x + \log(h_1h_2h_3) but (h_1(h_2+h_3)) \cdot x = 2x + \log(h_1^2h_2h_3).
    Last edited by NonCommAlg; June 6th 2011 at 08:45 AM.
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  3. #3
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    ok, so the action has to be a homomorphism. so it's little wonder i got stuck.

    in your example, wouldn't the action described work as long as you used the multiplicative group of non-zero reals?
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  4. #4
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    Quote Originally Posted by Deveno View Post
    ok, so the action has to be a homomorphism. so it's little wonder i got stuck.

    in your example, wouldn't the action described work as long as you used the multiplicative group of non-zero reals?
    i'm looking at the real numbers as an additive group and zero will not cause any problem. the problem is that x^h is not defined for negative values of x.
    anyway, i just fixed this problem by defining the action to be h \cdot x = x + \log h. see my previous post again! i think it's Ok now.
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  5. #5
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    oops. i meant positive reals (what i am thinking of is where you typically make \mathbb{R}^+ into a vector space over \mathbb{R}).

    and thanks for the reply. i can see that certain actions will work (conjugation by elements of H for example), and certain ones won't.

    so instead of h inducing a permutation of G, it has to induce an automorphism, is that correct?
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  6. #6
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    Quote Originally Posted by Deveno View Post
    so instead of h inducing a permutation of G, it has to induce an automorphism, is that correct?
    yes. in fact we only need h to induce a homomorphism. then the action will force it to be an automorphism.
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