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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.

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    $\begingroup$ Could you provide a reference of the paper? $\endgroup$ – Salvatore Siciliano Feb 18 '17 at 17:30
  • $\begingroup$ @SalvatoreSiciliano Dear Salvatore, I added the reference at the end of question. Please see the paper. $\endgroup$ – Amir Baghban Feb 18 '17 at 18:18
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    $\begingroup$ 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. $\endgroup$ – Mariano Suárez-Álvarez Feb 18 '17 at 18:29
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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.

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    $\begingroup$ Please use LaTeX. Also use paragraphs. $\endgroup$ – Avi Steiner Feb 26 '17 at 6:41
  • $\begingroup$ 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). $\endgroup$ – LSpice Feb 27 '17 at 2:00

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