# Groups determined by their group ring and direct products

In the paper [W. Kimmerle - R. Lyons - R. Sandling - D.N. Teague: Composition factors from the group ring and Artin's theorem on orders of simple groups, Proc. London Math. Soc. (3) 60 (1990), no. 1, 89-122] I found the following statement:

" It is well known that if the groups $G_1$ and $G_2$ are determined by their integral group rings, then $G_1 \times G_2$ is determined by $\mathbb{Z}(G_1 \times G_2)$ where $\mathbb{Z}(G_1 \times G_2)$ is the group ring.

I can not prove the above statement and I can not find where it has been proved. I would be grateful if you could give me a proof.

This is the link to the aforementioned paper.

• Could you provide a reference of the paper? Feb 18 '17 at 17:30
• @SalvatoreSiciliano Dear Salvatore, I added the reference at the end of question. Please see the paper. Feb 18 '17 at 18:18
• When you reference a paper, good scholarship mandates that you name the author, the title and enough information to identify the publication. That applies in a paper and here. Feb 18 '17 at 18:29

You are asking a reasonable question. I am not an expert on these things, but I think I can help with understanding the missing lemma and with the paper as a whole.

We are considering the integral group ring $\mathbb Z[G]$ of the finite group $G$. Then one can fix a map $\mathbb Z[G] → \mathbb Z$ which will play the role of the augmentation. There is an old result of Glauberman (and possibly someone else) that asserts that one can find the set $S$ of class sums in $Z[G]$. Then one takes sums of elements of $S$. It is easy to see which sums correspond to $\hat N$, the sum of all elements in $N$, for a normal subgroup $N$ of $G$. Since it is easy to see which normal subgroups contain others (they involve a bigger subset of $S$), we wind up describing the lattice of all normal subgroups of $G$ in this manner. An easy proof can be found in Passman’s book “The algebraic structure of group rings” (Lemma 2.3(v)(vi), pages 664–665).

If we look at a particular $\hat N$, then its square is itself times $\lvert N\rvert$, so $\lvert N\rvert$ is determined. Also its annihilator, say $A$, in $\mathbb Z[G]$ is easily seen to be what is called $\Delta(N,G)$, the augmentation ideal of $\mathbb Z[N]$ times $\mathbb Z[G]$. But this is the kernel of the natural epimorphism $\mathbb Z[G] \to \mathbb Z[G/N]$, so $\mathbb Z[G/N] \cong Z[G]/A$. We see that $\hat N$ determines $\lvert N\rvert$ and also $\mathbb Z[G/N]$.

Now for the well known fact. Since we have described the lattice of normal subgroups of $G$ we can look at all pairs $\hat M$, $\hat N$ such that $M \cap N = 1$ and $M N = G$. For each such pair, $G = N \times M \cong G/M × G/N$ and we know $\mathbb Z[G/M]$ and $\mathbb Z[G/N]$. By assumption, there will be at least one “good” pair where both $G/M$ and $G/N$ are determined. Hence G is determined. I guess this is the only way I can interpret the missing result.

Now for the paper as a whole. We are looking for the chief factors of $G$. So start with $\hat N$, where $N$ is a minimal normal subgroup. Since we know $\mathbb Z[G/N]$, induction will give us the chief factors of $G/N$. So we need only determine the isomorphism class of $N$. Of course, we know $\lvert N\rvert$. If $\lvert N\rvert = p^n$, then $N$ is an elementary abelian $p$-group of that order. So assume not. Then $N$, being characteristically simple, must be a finite direct product of isomorphic nonabelian simple groups, say isomorphic to $H$. We know that any finite simple group has some Sylow $p$-subgroup of order $p$. Thus $\lvert N\rvert$ will tell us first the number of factors of $H$ and then $\lvert H\rvert$. The number of factors is the smallest exponent of a prime in $\lvert N\rvert$. If we are lucky, $\lvert H\rvert$ determines $H$, since there is just one family of counterexamples (I mean two infinite families with the same orders). If not, then more work is needed and I am not sure where to go.

One thought is to let $C$ be the centralizer of $N$ in $G$. Since $N$ has trivial center, we have $N \cap C = 1$. Indeed, $C$ is the unique largest normal subgroup with this property. If $C$ is not $1$, then $N$ lives in $G/C$. By induction on $\lvert G\rvert$, we can determine the chief factors of $G/C$ and hence find $N$. On the other hand, if $C = 1$, then $G$ embeds in the automorphism group of N and somehow the paper uses this information. Presumably there exists an old paper which studies the isomorphism question for simple groups and where one could start. I hope this helps.

• Please use LaTeX. Also use paragraphs. Feb 26 '17 at 6:41
• Although I have seen some people insistently not use TeX, and reluctantly accept that this is a valid choice, note that in this case it actually created an error ($\lvert N\rvert = p^n$ became |N| = pn). I have submitted an edit that adds TeX, and restores paragraph breaks (which seem to have been in the original document, but to have gotten lost on cut and paste). Feb 27 '17 at 2:00