Originally Posted by

**Deveno** 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:

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

(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.)

$\displaystyle G$ is not necessarily a $\displaystyle \mathbb{Z}[H]$-module. for example let $\displaystyle G =\mathbb{R}$ and $\displaystyle H = \langle \pi \rangle$, the multiplicative group generated by $\displaystyle \pi$. define the action of $\displaystyle H$ on $\displaystyle G$ by $\displaystyle h \cdot x = x + \log h$, for all $\displaystyle h \in H$ and $\displaystyle x \in G$. clearly what we defined is a group action but

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

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

if $\displaystyle h_i \in H$ and $\displaystyle x \in G$, then $\displaystyle h_1 \cdot ((h_2 +h_3) \cdot x) = 2x + \log(h_1h_2h_3) $ but $\displaystyle (h_1(h_2+h_3)) \cdot x = 2x + \log(h_1^2h_2h_3). $