If $a\otimes(b\otimes c)=0$ then $(a\otimes b)\otimes c=0$ I'm trying to prove the identity above, while the tensor product is between members of abelian groups $A,B,C$.
This seemed trivial to me at first but since the tensor products are quotient groups I can't deduce equality at each term.
Edit: even more trivially, I can't understand why $a\otimes b=0$ implies $b\otimes a=0$
 A: Note that every abelian group can be seen as a $\Bbb{Z}$-module and that a $\Bbb{Z}$-linear map between abelian groups is just a group homomorphism.
Then this answer shows you how to prove that $A \otimes (B \otimes C) \simeq (A \otimes B) \otimes C$ as modules, hence as groups, where the isomorphism is the obvious one:
$$
\varphi \colon \;\, a \otimes (b \otimes c) \mapsto (a \otimes b) \otimes c
$$
In particular $\varphi$ is a group homomorphism, hence if $a \otimes (b \otimes c) = 0$ we have
$$
0 = \varphi(0) = \varphi(a \otimes (b \otimes c)) = (a \otimes b) \otimes c
$$

Similarly, you can use the universal property of tensor product and the map $A \times B \to B \otimes A$ given by $(a,b) \mapsto b \otimes a$ to show that the map
$$
a \otimes b \mapsto b \otimes a
$$
is a group isomorphism between $A \otimes B$ and $B \otimes A$. Like we did above, this implies that $a \otimes b = 0$ in $A \otimes B$ if and only if $b \otimes a = 0$ in $B \otimes A$.

Note that $a \otimes b = 0$ does not imply that $a$ or $b$ must be $0$, let alone both. For example, consider $A = \Bbb{Z}$, $B = \Bbb{Z}/4 \Bbb{Z}$. Then
$$
2 \otimes 2 = (2 \cdot 1) \otimes 2 = 1 \otimes (2 \cdot 2) = 1 \otimes 4 = 1 \otimes 0
$$
(here $\cdot$ stands for additive power), even though $2$ is non-zero in both $\Bbb{Z}$ and $\Bbb{Z}/4 \Bbb{Z}$.
