(I) Let $L^*=L \otimes \mathbb{Q}$ where $L$ is an abelian torsion-free group and $\mathbb{Q}$ is the additive group of rationals. Since $L$ is torsion-free the mapping $g \mapsto g \otimes 1$ is a monomorphism from $L$ to $L^*$.
(II)
I aim at proving (I), which is part of a proof I am reading. Two questions: (a) How could I use (II) to prove (I)? (b) Which of the two propositions can more directly be proved?
As to (a): as $g$ is left invariant under $\mu_*$, I think I should have $L$ in the role of $G$ and $\mathbb{Q}$ in the role of $B$ (I know, I get $\mathbb{Q} \otimes L$ instead of $L \otimes \mathbb{Q}$ but they are isomorphic). And I need $a\mu = 1$ and $a \otimes g=g$. Let's put $A=\mathbb{Q}$ and $\mu$ the identity function. Then now $a$ has to be $a=1$. But then I need $1 \otimes g=g$.
All I now about tensor products comes from P.Halmos, Finite Dimensional Vector Spaces. He only defines tensor products for finite dimentional vector spaces, but I thought that for the few places where the tensor product is used in the book I am reading (group theory), enough of the properties I learned would remain when passing to infinite dimension and to modules. The definition Halmos gives of tensor products is the following: Let $V$ and $U$ be finite dimensional vector spaces. Then $V\otimes U$ is the dual space of the space of all bilinear forms over $V\oplus U$.
What is $1\otimes g$? Let $w$ be a bilinear form over $\mathbb{Q} \oplus G$. $(1\otimes g)(w)= w(1,g)$. If I fix $g$ and put $f=w(x,g)$ with $x\in \mathbb{Q}$ then $f$ is a linear functional from $\mathbb{Q}$ to the scalar ring $\mathbb{Z}$. Hum... I give up. For $1\otimes g=g$ to make sense I need to identify $g$ in the right side of the equallity with a linear functional on the space of the bilinear forms over $\mathbb{Q}\oplus G$, which is a problem.