Isomorphic Group with $G=(\mathbb Z_{2^\infty}\oplus \frac{\mathbb Q}{\mathbb Z}\oplus \mathbb Q)\otimes_{\mathbb Z}\mathbb Q $ Let 
$$G=\left(\mathbb Z_{2^\infty}\oplus\mathbb Q/\mathbb Z\oplus \mathbb Q\right)\otimes_{\mathbb Z}\mathbb Q $$ 
Now $G$ isomorphic with which case: $0$ ? , or $\mathbb Q \, $  ? , or $\mathbb Q/\mathbb Z\,$ ? , or$\,\mathbb Z_{2^\infty}$?
 A: Three very useful facts below, apply them (in order) to quickly simplify the isomorphism class:


*

*Tensor products distribute over arbitrary direct sums, e.g. $\rm (A\oplus B)\otimes C\cong (A\otimes C)\oplus (B\otimes C).$

*Tensoring a divisible module against a torsion module yields zero.

*Tensoring ${\rm Frac}(R)$-modules over $R$ is the same as tensoring them over ${\rm Frac}(R)$.



Here are some explanations of what the above mean and why they're true. I like to be wordy and tangential and a bit general, so you could skip the background and just utilize the above points for their utility if you wished or if the below looks daunting.
An $R$-module $M$ is essentially an abelian group equipped with an "$R$-action." Whereas a group action on some object $X$ can be described as a group homomorphism $G\to{\rm Aut}(X)$, the action of a ring on a module can be described similarly as a ring homomorphism $R\to{\rm End}(M)$ (recall that the endomorphisms of an abelian group form a ring under pointwise addition and composition).
It is often said that an $R$-module can be thought of as a vector space over a ring $R$ instead of necessarily a field. Modules can be vastly more pathological (not necessarily free, not necessarily admitting a basis, nontrivial annihilation may exist, etc.), so ultimately I'd recommend being wary of this perspective on modules. I much prefer the "action" way of thinking.
That is actually a left module. A right module has stuff acting from the right, so technically it could be described by a homomorphism $R\to{\rm End}(M)^{\rm op}$ to the opposite ring. A $R$-$S$ bimodule has both left $R$ and right $S$ actions. To tensor two modules $M,N$ over a ring $R$ we need $M$ to be a right $R$-module and $N$ a left $R$-module for the sliding rule $mr\otimes n=m\otimes rn$ to make sense. Recall that the tensor product is the module spanned by the symbols $m\otimes n$ subject to linearity in both arguments of the $\otimes$ symbol and the "sliding scalars" rule.
Distributivity: This property is analogous to arithmetic where multiplication distributes over addition, and holds for various types of objects (vector spaces, algebras, modules) regardless of what is being tensored over. Any pure tensor in $(A\oplus B)\otimes C$ will be of the form $(a,b)\otimes c$, which by linearity is equal to $(a,0)\otimes c+(0,b)\otimes c$, which we may view as sitting inside $(A\otimes C)\oplus(B\otimes C)$.
It is possible to extend this all the way to an isomorphism. By induction, then, we may use this property for expressions involving a large finite number of tensor products and direct sums. The very same argument can be carried out for the case of arbitrary direct sums. Only a finite number of scalars can be moved about the $\otimes$ symbol at a time, which precludes tensor products distributing over arbitrary products, but direct sums have the nice condition that all but finitely many of the coordinates of any element are zero, so only a finite number of scalar passes over $\otimes$ are relevant.
Divisibility v. Torsion: Suppose $M$ is an $R$-module. It is divisible if every element in $M$ has a nonempty preimage under the multiplication-by-$r$ map for every $r\in R$. It is a torsion module if every element of $M$ is annihilated by some regular element of $R$ (i.e. $rm=0$ for some $r\in R$ which is not a zero divisor from either side - I sort of glossed over left/right stuff in my module primer).
Suppose $D$ is a divisible $R$-module and $T$ is a torsion $R$-module. Pick any pure tensor $d\otimes t$ inside the tensor product $D\otimes_RT$; there is an $r\in R$ such that $rt=0$, and in parallel there is an $h\in D$ such that $d=hr$. Then $d\otimes t=hr\otimes t=h\otimes rt=h\otimes0=0$. Every pure tensor and thus every tensor is zero, so we have $D\otimes_RM$. Even if $D$ is only divisible with respect to some multiplicative subset $S\subset R$ the same argument works (so in particular this applies to $p$-divisible groups).
I will point out the the Prüfer groups ${\bf Z}(p^\infty)$ are ($p$-)torsion and $p$-divisible. (Note divisible implies $p$-divisible but not vice versa, whereas $p$-torsion implies torsion but not vice versa.) Also ${\bf Q}/{\bf Z}$ is both divisible and torsion, and has the Prüfer / $p$-primary / partial fraction decomposition $$\frac{\bf Q}{\bf Z}\cong\bigoplus{\bf Z}(p^\infty).$$ Note that ${\bf Z}(p^\infty)$ may alternately be described as ${\bf Z}[p^{-1}]$ or ${\bf Q}_p/{\bf Z}_p$ (look up $p$-adic numbers).
Fractions & Sliding: Suppose $R$ is a domain and $F={\rm Frac}(R)$ its fraction field. If $V,W$ are $F$-modules, then $V\otimes_RW\cong V\otimes_F W$. Intuitively, it should suffice to see that arbitrary fractions may be slid across the $\otimes_R$ symbol, for which it suffices to see that reciprocals specifically can be engineered across. Suppose $1/r\in F$; then moving $1/r$ to the right is the same as moving $r$ to the left (which is permissible across $\otimes_R$); IOW $a/r\otimes b=a/r\otimes r(b/r)=(a/r)r\otimes (b/r)=a\otimes b/r$.
More generally, if $D$ is a division algebra and $D\ge R$ a subring and $D\ge S\ge R$ any intermediate ring and $M,N$ are $S$-modules, then $M\otimes_RN\cong M\otimes_SN$. Same argument applies. It could work in this case, for example, with $\bf Q$ and tensoring over $\bf Z$.
A: Take any basic tensor:
$$\left(\gamma\,,\,\frac{a}{b}+\Bbb Z\,,\,c\right)\otimes d\;,\;\;c,d,\in\Bbb Q\;,\;\;a,b,\in\Bbb Z\,\,,\,\,b\ne 0\,\,,\;\;\gamma=e^{\frac{2\pi i}{2^k}}\,,\;\;k\in\Bbb N$$
Since the tensor is over the integers (abelian groups=$\,\Bbb Z-$modules) , we can factor out stuff:
$$\left(\gamma\,,\,\frac{a}{b}+\Bbb Z\,,\,c\right)\otimes d=\left(\gamma\,,\,\frac{a}{b}+\Bbb Z\,,\,c\right)\otimes \frac{bd}{b}=b\left(\gamma\,,\,\frac{a}{b}+\Bbb Z\,,\,c\right)\otimes \frac{d}{b}=$$
$$=\left(b\gamma\,,\,\frac{ba}{b}+\Bbb Z\,,\,bc\right)\otimes \frac{d}{b}=\left(b\gamma\,,\,0\,,\,bc\right)\otimes \frac{d}{b}=\ldots$$
