If pi is a group, are the categories of right pi-modules and of 2-sided pi-modules equivalent? The reason I raise this question is that I was looking at Example 5, p. 43 in the TAC reprint of Jon Beck's thesis. He identifies the abelian group objects of Gp/pi as right pi-modules via the following construction. Let M be a right pi-module and let Y = pi x M with multiplication (x,m)(x',m') = (xx',m+m'x). This is an abelian group object in Gp/pi and for any Z --> pi, Hom(Z,Y) = Der(Z,M) the group of derivations d: Z --> M meaning that d(zz') = (dz)z' + d(z'). Here M becomes a right Z-module using the map Z --> Y. For the converse, if Y --> pi is an abelian group object in Gp/pi, then the kernel of the map is an abelian normal subgroup M of Y and the conjugation action of Y on M descends to pi since M is abelian and this conjugation action is a right pi-module structure (see Beck's thesis for all details). But suppose M is a 2-sided pi-module. Now let Y = pi x M and define (x,m)(x',m') = (xx',mx'+xm'). This is just as above an abelian group structure in Gp/pi. For any Z --> pi, Hom(Z,Y) = Der(Z,M) as above but now Der means 2-sided derivations d(zz') = (dz)z' + z(dz'), still an abelian group. But the kernel of Y --> pi is M but now with a new right module structure m*x = x^{-1}mx. So there is a functor from 2-sided to right modules by replacing the 2-sided operation by conjugation which is now a right operation, but could this possibly be an equivalence? I don't see how. Michael [For admin and other information see: http://www.mta.ca/~cat-dist/ ]