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$\prod_{i=1}^\infty \mathbb{Z}$ is a counterexample. It is torsion free, not free (Why isn't an infinite direct product of copies of $\Bbb Z$ a free module?), and every nonzero element is only divisible by finitely many integers, so your extra hypotheses hold.
It only holds for the trivial group. As it was mentioned in the comments, we consider $\mathbb{Z}$ as a trivial $\mathbb{Z}G$-module and in this way the augmentation map $\epsilon \colon \mathbb{Z}G \to \mathbb{Z}$ is a morphism of $\mathbb{Z}G$-modules. Let $f \colon \mathbb{Z} \to \mathbb{Z}G$ be a morphism of $\mathbb{Z}G$-modules and say $f(1) = ... 4$\mathbb{R}[x]$-submodules of$\mathbb{R}^2$are vector subspaces$V$such that$T(V)\subset V$. In particular,$V$other than$0$or$\mathbb{R}^2$would be a line. Now can you find a line fixed by a rotation? 3 One reason why one needs the bilinear map (or multilinear in general), instead of just going ahead to define a map on the tensor product, is that one needs to show that the latter is well-defined. So it may be easy to define some$A \otimes B \rightarrow C$, but it actually may be quite hard to prove that it is well-defined. On the other hand, once a ... 3 How do you use the splitness of your exact sequence? Once done, it is easy. So, let me call the middle term with basis$e_1,e_2$and the map on the left by$i$. Then$i(1)=ae_1+be_2$. This splits says we have a map$j:Ae_1\oplus Ae_2\to A$with$j\circ i$identity. If$j(e_i)=c_i$, we see that$ac_1+bc_2=1$. Let$v_1=i(1), v_2=-c_2e_1+c_1e_2$. Then using the ... 2 Here's a counterexample. Take$A=k[x]/(x^2)$and$M=N=A/(x)$. Then$\operatorname{Hom}(M,A)\cong M$, generated by the map$f:1\mapsto x$, and$\operatorname{Hom}(A,M)\cong M$, generated by the map$g:1\mapsto 1$. The tensor product$\operatorname{Hom}(M,A)\otimes \operatorname{Hom}(A,M)$is then also isomorphic to$M$, generated by$f\otimes g$. But ... 2 For every x in M, there is a unique left R-module homomorphism from R to M that sends 1 to x. If x is nonzero, this homomorphism is surjective (why?). The kernel is precisely the annihilator of x. Now use the so-called correspondence theorem to see this is a maximal left ideal. Can you finish the argument from here? 2 Let$U$and$V$be free modules over a nontrivial commutative ring$R$and let$\alpha:U\to R$and$\beta:V\to R$be two$R$-linear maps which have$1$in their image; such things are easily seen to exist using freeness. Then using the properties of tensor products you can show that there is a morphism of abeelian groups$f:U\otimes_RV\to R$such that for ... 2 First you have that$x=5k+1$for$k\in \mathbb{N}$by your first equation. Then,$5k+1\equiv 0\ (\mathrm{mod}\ 66)$, so, as you said in the second part,$k\equiv 13+66q \ (\mathrm{mod}\ 66)$. Now$x=5k+1=5(13+66q)+1=66+330q$. Finally you use the third equation ($66\equiv 3$and$330\equiv 1$both mod$7$):$x\equiv -1 \ (\mathrm{mod}\ 7)$, so ... 1 No, and the axiom of choice is irrelevant. For example, take$R = \mathbb{Z}, M = \mathbb{Q}/\mathbb{Z}$.$M$is a divisible$R$-module but cannot be made into a module over the fraction field$\mathbb{Q}$: modules over$\mathbb{Q}$correspond to uniquely divisible abelian groups. 1 Let$R$be a ring with unit 1 as you say. By$R$is "free as a bimodule" it often means$R$is free as a module over$R\otimes R^{op}$, where$(a\otimes b)*r=arb$(conventions differ in literature). Recall the definition of a free module is a direct sum of$R$. Thus in general$R$is not free as an$R$bimodule. In practice, often one work with the case$R$... 1 Not really an answer, but I wanted to post this here in case anyone else ends up thinking about this thing. It might help set you on the right track. Anyway, after browsing through this collection of slides (see "Ingredients of Construction" slide) I now know that the morphisms in this category can be described using walled Brauer diagrams (see e.g. page ... 1 Let$R$be an$k$-algebra and$M$a simple$R$-module of dimension$n$, such that$M$is also a finite vector space over$k$. Then the space $$Hom_{k}(M,M)\cong Hom(k^{n},k^{n})$$ has dimension$n^2$over$k$as a$k$-vector space, thus the dimension of the result above. To see that$Hom_{k}(M,M)\cong M^{n}$as an$R$-module, it suffice to construction ... 1 The group$\mathbb{Z}^\mathbb{N}$satisfies your hypotheses but is not free. For a much smaller counterexample, let$\hat{\mathbb{Z}}$be the profinite completion of$\mathbb{Z}$and let$\alpha\in\hat{\mathbb{Z}}$be an element such that$a\alpha+b$is divisible by only finitely many integers whenever$a,b\in\mathbb{Z}$with$a\neq 0$(such an$\alpha\$ can ...