HW #1 due Friday Apr 13

1. In class we lamented that for G acting on a finite set X, we could hardly figure out anything about the action knowing only X^G.

Let Z^X be the free abelian group generated by X; the action of G on X extends in a unique way to an action on Z^x by group automorphisms.

What can you determine about the action of G on X, knowing (Z^X)^G?

2. Define an R-supermodule as...
oh, this sounds so exciting! ...
... a pair (M₀,M₁) of modules. And a morphism thereof is just a pair of morphisms of R-modules. Pretty pedestrian really.
There are several forgetful functors R-SuperMod -> R-Mod, but the one we want is (M₀,M₁) |-> M₀ + M₁ (direct sum).

Slightly more interesting: if R is commutative, so that the tensor product of two R-modules is naturally an R-module again, define
(M₀,M₁) @_R (N₀,N₁) = (M₀ @_R N₀ + M₁ @_R N₁, M₀ @_R N₁ + M₁ @_R N₀).

a. Make (detailed) sense of the following statement, and prove it (not much detail):
"The functors Forget o @ and @ o Forget are naturally isomorphic."

Define an R-superalgebra as an R-supermodule M plus an associative "multiplication" M @_R M -> M.

b. Given an R-module N, define a superalgebra structure on the supermodule (R,N).

Call a superalgebra (M₀,M₁) supercommutative if M₀ commutes with everything, and M₁ anticommutes with itself, i.e. ab = -ba for a,b in M₁. (Cohomology rings from topology are examples, where M₀ is the sum of the even-degree parts, M₁ the odd.)

c. Given a supermodule M, define the "free supercommutative superalgebra" on M. It should be a left adjoint to the forgetful functor from supercommutative superalgebras to supermodules.

3. Let R = Q[x] / < x² - n >, where Q is the rationals and n is an integer. Assume R is not a field. Describe R in terms of more familiar rings.

4. Let H be a subgroup of G. Show there is a unique largest normal subgroup of H and give a way to check whether h in H is an element thereof.