Let $R$ be a commutative ring with unit. If $G$ is a finite (or in general, discrete) group, let $R[G]$ be the group $R$-algebra associated to $G$. The isomorphism problem for group rings asks for condiions on groups $G$ and $H$ such that $R[G]\simeq R[H]$.

Group rings are not a complete invariant, even of finite groups; in a 2001 Annals paper, Hertweck discovered two finite groups $G$ and $H$ with $\mathbb{Z}[G]\simeq \mathbb{Z}[H]$ and $G\not\simeq H$. In general, group algebras over a field are even weaker invariants; for example, if $G$ and $H$ are any two finite abelian groups of order $n$, then $\mathbb{C}[G]\simeq \mathbb{C}[H]\simeq \mathbb{C}^n$, by e.g. the Chinese Remainder Theorem or Artin-Wedderburn.

I am curious about a slight strengthening of the isomorphism problem for group rings. Namely, if $(S, +, \cdot)$ is a (not necessarily commutative) ring with unit, let the opposite ring $S^{op}=(S, +, \times)$ be the ring whose underlying Abelian group under addition is the same as that of $S$, but with the multiplicative structure reversed, i.e. $a\times b=b\cdot a$; the formation of the opposite is clearly functorial. Note that if $R[G]$ is a group ring, it is naturally isomorphic to its opposite through the map $\phi_G: g\mapsto g^{-1}$.

The Problem

Now if $G, H$ are groups and $\psi: R[G]\to R[H]$ is an isomorphism of group rings, we may ask if it is compatible with the formation of the opposite ring---that is, does $\phi_H\circ \psi=\psi^{op}\circ \phi_G$? Say that $G, H$ have strongly isomorphic group rings if such a $\psi$ exists.

What is known about groups with strongly isomorphic group rings over commutative rings $R$? Are there non-isomorphic finite groups $G, H$ with $\mathbb{Z}[G]$ strongly isomorphic to $\mathbb{Z}[H]$, for example? More weakly, when is $\mathbb{C}[G]\simeq \mathbb{C}[H]$?

Strong Isomorphism is Strong

Just to convince you that strong group ring isomorphism is in fact a stronger condition that group ring isomorphism, note that $\mathbb{C}[\mathbb{Z}/4\mathbb{Z}]$ is isomorphic to $\mathbb{C}[\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}]$ but not strongly isomorphic. This is because $\phi_{\mathbb{Z}/4\mathbb{Z}}$ is not the identity, but $\phi_{\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}}$ is the identity on the underlying set of $\mathbb{C}[\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}]$.

Addendum (4/6/2011)

Andreas Thom points out in his excellent answer that the case of finite abelian groups over $\mathbb{C}$ is not much harder than the case of usual group ring isomorphisms. Unfortunately the question over e.g. $\mathbb{Z}$ is likely to be extremely difficult, since the usual isomorphism problem over $\mathbb{Z}$ is apparently quite hard---I don't yet understand Hertweck's construction well enough, for example, to tell if the groups he constructs have strongly isomorphic group rings. In any case, I would accept as an answer a summary of the current state of the art for strong isomorphism over $\mathbb{Z}$ (for example, does Hertweck's construction admit a strong isomorphism?), or any relatively recent reference addressing the more general question (as Qiaochu Yuan points out in a comment, the question is equivalent to asking when the group rings of $G, H$ are isomorphic as $*$-algebras, which suggests to me that the question must have been studied by someone).

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    $\begingroup$ I think this is equivalent to asking that the two be isomorphic as *-algebras (over $R$ considered as a *-ring with trivial involution). $\endgroup$ Apr 5, 2011 at 6:54
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    $\begingroup$ If you would even demand that RG and RH be isomorphic as Hopf rings, and if R is an integral domain, then G = group-like-elts(RG) = group-like-elts(RH) = H. So your question is in between the original isomorphism problem and this Hopf-ring observation. $\endgroup$ Apr 5, 2011 at 14:11
  • $\begingroup$ @Qiaochu: Yes, that seems right. $\endgroup$ Apr 5, 2011 at 15:43

1 Answer 1


If $G$ is a finite abelian group, then $\mathbb C[G] = \lbrace f \colon \hat G \to \mathbb C \rbrace$, where $\hat G$ is the Pontrjagin dual of $G$. The isomorphism $g \mapsto g^{-1}$ translates into the same map on the Pontrjagin dual (basically multiplication by $-1$ on $\hat G$), but now it is a bit easier to analyze. Note also, that there is a non-canonical isomorphism $G \cong \hat G$.

Hence, in order to find two non-isomorphic abelian groups which have strongly isomorphic complex group rings, we just have to analyze (in addition to the cardinality of the group) the orbit structure of multiplication by $-1$ on the group. Indeed, we are now just talking about an algebra of function on a set with some $\mathbb Z/2 \mathbb Z$-action.

Example: In $\mathbb Z/8\mathbb Z \times \mathbb Z/2\mathbb Z$, there are precisely $4$ elements, which are fixed under multiplication by $-1$, namely $(0,0),(4,0),(0,1)$ and $(4,1)$. The same is true for $\mathbb Z/4\mathbb Z \times \mathbb Z/4\mathbb Z$. Here, one has $(0,0),(2,0),(0,2)$ and $(2,2)$. Hence, there exists an isomorphism between $\mathbb C[\mathbb Z/8\mathbb Z \times \mathbb Z/2\mathbb Z]$ and $\mathbb C[\mathbb Z/4\mathbb Z \times \mathbb Z/4\mathbb Z]$, which respect the isomorphism which is induced by $g \mapsto g^{-1}$.

I do not know about an example with coefficients in $\mathbb Z$.

  • $\begingroup$ This is very nice! Unfortunately, $\mathbb{Z}$ coefficients are likely to be much harder, since the isomorphism problem itself was open til 2001. $\endgroup$ Apr 5, 2011 at 4:45
  • $\begingroup$ On the other hand strong isomorphism of integral group rings is more than halfway there. In fact $\{\pm g \mid g\in G\}=\{x\in\mathbb{Z}G^\times \mid x^{-1} = \phi_G(x)\}$: One inclusion is trivial. Now write $x=\sum a_g g$. The coefficient of $1$ in $x\phi(x)$ equals $\sum a_g^2$. Since we are over $\mathbb{Z}$ exactly one of the coefficients must be $\pm1$ and all others zero. Corollary: Since $\{\pm g\} = C_2\times G$ any pair of groups with trivial $O_2$ is isomorphic if their integral group rings are strongly isomorphic. $\endgroup$ Mar 15, 2014 at 0:37

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