TITLE: Tensor product of abelian group and a free abelian group
QUESTION [6 upvotes]: I am trying to show that if $F,H$ are abelian groups with $F$ free abelian, and if $a \in F$ and $h \in H$ are non-zero, then $a \otimes h \ne 0$ in $F \otimes H$.
This is specifically in a section describing the derived functor Tor. Of course, that doesn't mean the solution has to involve that, but there is probably a way. I know that $F$ free abelian means that $F$ is torsion free and hence $\mbox{Tor}(F,A)=0$. 
I was trying to use a formulation of $\mbox{Tor}$ in terms of exact sequences. If:
$$0 \to R \stackrel{i}{\hookrightarrow} F \to A \to 0$$ is an exact sequence then $\mbox{Tor}(A,B) =\mbox{ker}(i \otimes 1_b)$
Seemed to me if I picked the right sequence I could get that $\mbox{Tor}=0$ implies that the kernel is trivial, which would give the result, but I can't get this to work
Edit It appears that this is false from the answers below. 
Here is a link to the question.

REPLY [2 votes]: Suppose $F=\mathbb Z$ and $H=\mathbb Z/2\mathbb Z$. Let $a=2\in F$ and let $\xi\in H$ be the non-zero element. Then $$a\otimes\xi=(2\cdot1)\otimes\xi=1\otimes(2\cdot \xi)=1\otimes 0=0$$.
It follows that what you want to prove is false.

REPLY [2 votes]: Here is an answer using Tor:
Consider the exact sequence 
$$ 0\rightarrow\mathbb{Z}\rightarrow F\rightarrow K\rightarrow 0,$$
where the map from $\mathbb{Z}$ to $F$ sends the generator $1$ to $a$, and $K$ is the cokernel.  Tensoring with $H$ we get the long exact sequence
$$ \cdots Tor(F,H)\rightarrow Tor(K,H)\rightarrow H\rightarrow F\otimes H\rightarrow H\otimes K\rightarrow 0. $$
Now $Tor(F,H)=0$ since $F$ is torsion-free.  So the kernel of $H\rightarrow F\otimes H$, given by sending $h$ to $a\otimes h$, is $Tor(K,H)$.  This is not always non-zero, and the problem as stated is incorrect.  Consider $H=\mathbb{Z}/2\mathbb{Z}$, $F=\mathbb{Z}$, $a=2$.