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:
(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.)
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?
oops. i meant positive reals (what i am thinking of is where you typically make into a vector space over ).
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?
yes. in fact we only need h to induce a homomorphism. then the action will force it to be an automorphism.
Originally Posted by Deveno