Determine whether $D_4/Z(D_4)\otimes Z(D_4)$ is isomorphic to $D_4$.
My attempt
I found that the center $Z(D_4)$ is comprised of $I$ and $R^{2}$, and is normal to $D_4$, so $D_4/Z(D_4)$ is the group of the cosets: $\{IZ,rZ,RZ,R^2Z,R^3Z,RrZ,R^2rZ,R^3rZ\}$. Computing the cosets we find that some are duplicates, resulting in $D_4/Z(D_4)=\{Z,rZ,RZ,RrZ\}$ so $$D_4/Z(D_4)\otimes Z(D_4)=\\\{\{I,R^2\}\otimes I,\{I,R^2\}\otimes R^2,\{r,rR^2\}\otimes I,\{r,rR^2\}\otimes R^2,\\\{R,R^3\}\otimes I,\{R,R^3\}\otimes R^2,\{Rr,rR\}\otimes I,\{Rr,rR\}\otimes R^2\}$$ has 8 elements like $D_4$.
Now I don't know how to associate to each and every element of $D_4/Z(D_4)\otimes Z(D_4)$ an element of $D_4$, knowing the properties $r^2=R^4=I$ and $RrR=r$.
If by $\otimes$ you mean the classical tensor product over $\mathbb{Z}$ (it could be, since the groups you are tensoring are abelian), then the answer is clearly NO, since the tensor product of abelian groups is an abelian group. Moreover, your tensor product will have four distinct elements, and not eight (you forgot that you have some relations in the tensor product..) . In fact $D_4/Z(D_4)\simeq (\mathbb{Z}/2\mathbb{Z})^2$ and $Z(D_4)\simeq\mathbb{Z}/2\mathbb{Z}$, so your tensor product is isomorphic to $(\mathbb{Z}/2\mathbb{Z})^2\otimes \mathbb{Z}/2\mathbb{Z}\simeq (\mathbb{Z}/2\mathbb{Z})^2$.