Tensor product of abelian group with itself is zero

Question

Let $A$ be an abelian group. Assume that $A\otimes A=0$. Does it imply that $A=0$?

I know that the answer is yes if we add an assumption that $A $ is torsion-free.

One more question, related to the above , is whether $\operatorname{Tor} (A,A)=0$ implies that $A $ is torsion-free?

I know that $A $ is torsion-free if and only if $\operatorname{Tor} (A,B)=0$ for all abelian groups $B $. However, the proof uses different groups $B $ and not $B=A $.

Answer 1

What about $A=\Bbb Q/\Bbb Z$? $A$ is both torsion and divisible, so $(1/m)\otimes (1/n)=n(1/nm)\otimes 1/n=(1/nm)\otimes (n/n)=0$.

ADDED IN EDIT

On the category of Abelian groups, Tor is left-exact in both arguments, so if $B$ is a subgroup of $A$ then $\text{Tor}(B,B)$ is a subgroup of $\text{Tor}(A,A)$. If $A$ has nontrivial torsion, then $A$ has a subgroup isomorphic to $C_n$, a cyclic group of finite order $n\ge 2$. Therefore $\text{Tor}(A,A)$ has a subgroup isomorphic to $\text{Tor}(C_n,C_n) \cong C_n$, and so is nonzero.

Answer 2

Another example is to look at $\mathbb Z/n\mathbb Z\otimes_{\mathbb{Z}}\mathbb Z/m\mathbb Z \cong \mathbb Z/(m,n)\mathbb Z$. Clearly both $\mathbb Z/n\mathbb Z$ and $\mathbb Z/m\mathbb Z$ are nontrivial torsion $\mathbb Z$-modules, yet whenever $m$ and $n$ are relatively prime, their tensor product is $0$.