TITLE: What does the group ring $\mathbb{Z}[G]$ of a finite group know about $G$?
QUESTION [43 upvotes]: The group algebra $k[G]$ of a finite group $G$ over a field $k$ knows little about $G$ most of the time; if $k$ has characteristic prime to $|G|$ and contains every $|G|^{th}$ root of unity, then $k[G]$ is a direct sum of matrix algebras, one for each irreducible representation of $G$ over $k$, so $k[G]$ only knows the dimensions of the irreducible representations of $G$. 
The group ring $\mathbb{Z}[G]$, on the other hand, knows at least as much as $k[G]$ for every $k$ (which one can obtain by tensoring with $k$); this information should let you deduce something about the entries in the character table of $G$, although I'm not sure exactly what. If the characteristic of $k$ divides $|G|$ then $k[G]$ knows something about the modular representation theory of $G$, although again, I'm not sure exactly what. So:

How strong of an invariant is $\mathbb{Z}[G]$? More precisely, what group-theoretic properties of $G$ do I know if I know $\mathbb{Z}[G]$? What are examples of finite groups $G_1, G_2$ which are not isomorphic but which satisfy $\mathbb{Z}[G_1] \cong \mathbb{Z}[G_2]$? 

If in addition to $\mathbb{Z}[G]$ we are given the augmentation homomorphism $\mathbb{Z}[G] \to \mathbb{Z}$ then it looks like we can recover the group cohomology and homology of $G$, so it seems plausible that $\mathbb{Z}[G]$ contains quite a lot of information. 
At a minimum, setting $k = \mathbb{R}$ I think we can compute the Frobenius-Schur indicator of every complex irreducible representation of $G$. 
Edit: It seems the second problem is known as the isomorphism problem for integral group rings and is quite hard.

REPLY [3 votes]: The Finite abelian case-
Let $G_1 $ be a finite abelian group. 
If $\Bbb{Z}G_1\cong \Bbb{Z}G_2$, and suppose $\phi$ is the isomorphism, then you can define another isomorphism $\Phi: \Bbb{Z}G_1\to \Bbb{Z}G_2$ such that for $\alpha= \sum_{i=1}^nr_ig_i \in \Bbb{Z}G_1$, $\Phi(\alpha)=\sum_{i=1}^{n}\varepsilon(\phi(g_i))^{-1}r_i\phi(g_i)$.
$\Phi$ is normalized in the sense that $\varepsilon(\Phi(\alpha))=\varepsilon(\alpha)$ .
Now as $G_1$ is finite abelian, $\Bbb{Z}G_1$ is commutative $\implies $ $G_2$ is abelian. Now as $\Bbb{Z}$ has IBN, it implies $|G_1|=|G_2|$.
Now as each $g \in G_1$ is invertible in $\Bbb{Z}G_1$, so $\Phi(g)$ is a normalized unit in $\Bbb{Z}G_2$, and it is not so hard to check that all units of integral group ring of an abelian group are trivial (easier if we have finite abelian, which is our case), so $\Phi(g)\in \pm G_2$ and $\Phi$ is normalized implies $\Phi(g)\in G_2$. This shows that $\Phi(G_1) \subset G_2$ but $|G_1|=|G_2|$, we get $\Phi(G_1)=G_2$. $\hspace{7cm} \blacksquare$
It is also true for finite metabelian groups. 
Moreover, your hunch was correct that you can deduce something about enteries in character table, to be precise,  by using a theorem by Glauberman, you can also prove that  

if $G_1$ is finite such that $\Bbb{Z}G_1\cong \Bbb{Z}G_2$, then character tables of $G_1$ and $G_2$ are equal (after a possible rearrangement). 

See Sehgal, Milies book " An Introduction to group rings "