Let $\mu:A\to B$ be mono of abelian groups and let $G$ be torsion-free abelian. Prove $\mu_*:A⊗_\Bbb ZG\to B⊗_\Bbb ZG$ is mono This is Exercise 4.2.8 of Robinson's "A Course in the Theory of Groups (Second Edition)". According to Approach0, it is new to MSE.
I asked a question on the notation here earlier:
The notation $A\otimes_{\Bbb Z}G$ in Robinson's "A Course in the Theory of Groups (Second Edition)".
Robinson did not define tensor products  prior to the exercise.
(NB: I have not included the category-theory tag for a reason: the tools available here are almost entirely group theoretic.)
The Details:
A monomorphism is an injective homomorphism.
A group is torsion-free if all its elements, apart from the identity, have infinite order.
The Question:

(Dieudonné). Let $\mu:A\to B$ be a monomorphism of abelian groups and let $G$ be a torsion-free abelian group. Prove that $\mu_*:A\otimes_{\Bbb Z}G\to B\otimes_{\Bbb Z}G$ is a monomorphism where $(a\otimes g)\mu_*=(a\mu)\otimes g.$ [Hint: Reduce first to the case where $G$ is finitely generated and then to the case $G=\Bbb Z$.]

Thoughts:
My understanding of tensor products is poor. If I recall correctly, the last time I handled them was back in 2013, when I undertook a module in vector calculus for my MMath degree. I saw them a year before that, I believe, in a linear algebra module.
As such, this is not a question I believe I can answer myself any time soon.

The hint doesn't help me.

I think I understand the definition of a tensor product of abelian groups given here:

Suppose $G$ and $H$ are abelian groups (possibly equal, possibly distinct). Their tensor product as abelian groups, denoted $G\otimes_{\Bbb Z}H$ or simply as $G\otimes H$, is defined as the quotient of the free abelian group on the set of all symbols $\{ g\otimes h\mid g\in G, h\in H\}$ by the following relations:

*

*$(g_1+g_2)\otimes h=(g_1\otimes h)+(g_2\otimes h)$ for all $g_1,g_2\in G, h\in H$,


*$g\otimes (h_1+h_2)=(g\otimes h_1)+(g\otimes h_2)$ for all $g\in G, h_1,h_2\in H$.


The fact that a monomorphism is defined by its action on the generators of its domain is clearly of relevance.

I need to show that for all $r,s\in A\otimes_{\Bbb Z}G$, we have that $r\mu_*=s\mu_*$ implies that $r=s$. I also need to show that for all $x,y\in A\otimes_{\Bbb Z}G$, we have $(x+y)\mu_*=(x\mu_*)+(y\mu_*)$.

The kind of answer I'm hoping for is a full solution.

Please help :)
 A: If $\mu_*$ is not a monomorphism, then there exists a nonzero element that lies in the kernel. But that nonzero element can be expressed as a finite linear combination of pure tensors (elements of the form $a\otimes g$ for $a\in A$ and $g\in G$). Thus, this witness will also exist if we replace $G$ with the subgroup generated by the $g$ that occur in this witness. Thus, we can reduce to $G$ finitely generated.
Since $G$ is torsion free and finitely generated, it is isomorphic to a direct sum of a finite number of copies of $\mathbb{Z}$. But the tensor product distributes over (finite) direct sums
:
$$A\otimes_{\mathbb{Z}}(G_1\oplus G_2) \cong (A\otimes_{\mathbb{Z}}G_1)\oplus (A\otimes_{\mathbb{Z}}G_2).$$
Thus, we may decompose the map $\mu_*\colon A\otimes_{\mathbb{Z}}G \to B\otimes_{\mathbb{Z}}G$ into a family of maps with domain $A\otimes_{\mathbb{Z}}\mathbb{Z}$ and codomain $B\otimes_{\mathbb{Z}}\mathbb{Z}$ (using thee universal property of the direct sum as both a coproduct and a product for abelian groups). If $\mu_*$ is not injective, then at least one of these maps is not injective. Thus, we are reduced to the case of $G\cong \mathbb{Z}$.
So now we have a monomorphism $\mu\colon A\to B$, and the induced map $\mu_*\colon A\otimes_{\mathbb{Z}}\mathbb{Z}\to B\otimes_{\mathbb{Z}}\mathbb{Z}$.  We want to verify that this is a monomorphism.
Lemma. For all abelian groups $A$, $A\otimes_{\mathbb{Z}}\mathbb{Z}\cong A$.
Proof. First note that every element of $A\otimes_{\mathbb{Z}}\mathbb{Z}$ can be written as a pure tensor of the form $a\otimes 1$: indeed, given an element
$$\sum_{i=1}^n m_i(a_i\otimes k_i)$$
we can first use bilinearity to rewrite $a_i\otimes k_i=(k_ia_i)\otimes 1$, and then rewrite $m(k_ia_i\otimes 1) = (m_ik_i)a\otimes 1$. Then we have
$$\sum_{i=1}^n m_i(a_i\otimes k_i) = \sum_{i=1}^n (m_ik_ia_i)\otimes 1 = \left(\sum_{i=1}^n m_ik_ia\right)\otimes 1.$$
Consider the map $f\colon A\otimes_{\mathbb{Z}}\mathbb{Z}\to A$ induced by the bilinear map $A\times \mathbb{Z}\to A$ given by $(a,n)=na$. Assume that $x\in A\otimes_{\mathbb{Z}}\mathbb{Z}\in\ker(f)$. Then we write $x=a\otimes 1$ for some $a\in A$ and we conclude that $0=f(x)=f(a\otimes 1)=a$. But $0\times 1=0_{A\otimes_{\mathbb{Z}}\mathbb{Z}}$, so $f$ is one-to-one. Finally, $f$ is surjective, since $a\in A$ is the image of $a\otimes 1$. $\Box$
Now consider the map $A\to B$ given by
$$A\stackrel{\cong}{\to}A\otimes_{\mathbb{Z}}\mathbb{Z}\stackrel{\mu_*}{\to}B\otimes_{\mathbb{Z}}\mathbb{Z}\stackrel{\cong}{\to}B.$$
The map is
$$a \longmapsto a\otimes 1\longmapsto \mu(a)\otimes 1\longmapsto \mu(a).$$
That is, this map is equal to $\mu$. Since $\mu$ is 1 to 1 and both the first and last maps are isomorphisms, $\mu_*$ is also one-to-one.
Which is what we wanted to prove.
