What does the group ring $\mathbb{Z}[G]$ of a finite group know about $G$?

Question

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.

Answer 1

You might consult Chapter 9 in Milies, Sehgal, "An Introduction to Group Rings". It discusses how several invariants of groups are encoded in their integral group rings, and how this leads to an affirmative answer of the integral group ring isomorphism problem for some classes of groups.

Answer 2

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 "

Answer 3

If you know that $G$ and $H$ are at least abelian, then $\mathbb{Z}[G]\cong\mathbb{Z}[H]$ implies $G\cong H$.

proof: The homology groups of $G$ and $H$ are independent of the choice of resolution up to canonical isomorphism, and the groups are defined by $H_iG=H_i(F_G)$ where $F$ is a projective resolution of $\mathbb{Z}$ over $\mathbb{Z}G$ (similary for $H$). Since $\mathbb{Z}G\cong\mathbb{Z}F$, the $G$-modules $F_i$ can be regarded as $H$-modules via restriction of scalars and hence the projective resolution $F$ for the homology of $G$ can also be used for the homology of $H$. We have $F_G\cong F_H$ since $\mathbb{Z}\otimes_{\mathbb{Z}G}F\cong \mathbb{Z}\otimes_{\mathbb{Z}H}F$ by the obvious map $1\otimes f\mapsto 1\otimes f$ [using the isomorphism $\varphi:\mathbb{Z}G\rightarrow \mathbb{Z}H$ we have $1\otimes f=1\otimes gf\mapsto 1\otimes \varphi(g)f=1\otimes f$], and so $H_i(G)\cong H_i(H)$ for all $i$. In particular, $G/[G,G]\cong H_1(G)\cong H_1(H)\cong H/[H,H]$. Since $G$ and $H$ are abelian groups, $[G,G]=0=[H,H]$ and hence $G\cong H$.

If we knew the augmentation map then we can recover $G_{ab}$ as $I/I^2$, where $I$ is the augmentation ideal (kernel of the augmentation map). But even if we don't know the augmentation map, $\mathbb{Z}[G]\cong\mathbb{Z}[H]$ would imply $I_G\cong I_H$ and hence $G_{ab}\cong H_{ab}$, reproducing my above statement.

And in general there does exist counterexamples, I forgot the sources on them, but this MO question refers to one of them:
https://mathoverflow.net/questions/60609/strong-group-ring-isomorphisms

(I first claimed, without thinking, that $G$ sits as the group of units in $\mathbb{Z}[G]$, which is what Qiaochu refers to in his comment below).