19
$\begingroup$

I am interested in this paper which I can't read because it's in German:

Frobenius, G., Über die Charaktere der mehrfach transitiven Gruppen., Berl. Ber. 1904, 558-571 (1904). ZBL35.0154.02.

A free online copy is here. I am specifically interested in the character table of $M_{12}$ which appears at the top of p.568.

I would like to know what, exactly, Frobenius proved with regard to this character table. If I understand correctly, at the time of publication, the existence of $M_{12}$ was still somewhat contentious. Mathieu had written down some permutations that generated a 5-transitive group on 12 letters, but there was some uncertainty as to whether or not these permutations generated all of $A_{12}$. I believe the uncertainty continued until the 1930's, at which point Witt used Steiner systems, to show that $M_{12}$ was really a proper subgroup of $A_{12}$.

In light of this, my guess is that Frobenius proved the following theorem:

Theorem. Suppose that $G$ is a sharply 5-transitive subgroup of $S_{12}$. Then the character table of $G$ is as follows...

Note that the theorem as stated does not refer to the specific construction of $M_{12}$ given by Mathieu, but is more a theorem of the kind "Should such a group exist, then...." (a type of theorem that appeared many times in the following 100 years as part of the classification).

However, perhaps I am wrong. So my questions are, specifically:

  1. Does Frobenius' character table calculation pertain to any sharply 5-transitive subgroup of $S_{12}$, or is it specific to $M_{12}$?
  2. Does Frobenius give a proof of his calculations, or is it more of a sketch?
  3. The same questions as above also apply to $M_{24}$, with obvious edits.

Edit: Thank you for the useful comments and answers. Perhaps I should add a fourth question to clarify exactly what I am looking for:

  1. Does Frobenius use any properties of $M_{12}$, apart from sharp 5-transitivity, in his calculation of the character table?

Edit 2 -- 26 June 2018: My MMath student, Sam Hughes, and I have just uploaded a preprint connected to this question. In it we write down the character table of a sharply 5-transitive subgroup of $A_{12}$ without making any reference to the Mathieu groups. So, even though this is not what Frobenius did (as Frieder Ladisch's answer below makes clear), it is interesting to know that he could have done it if he had wanted to! We have cited Frieder's answer below in the preprint, as it was very helpful for clarifying the history of this work. Thanks again!

$\endgroup$
  • 1
    $\begingroup$ Based on the first paragraph of section 5 in the paper, he claims that the only 5-transitive groups on at most 24 elements other than the symmetric and alternating groups are two groups discovered my Mathieu. He then proceeds to calculate their character tables. It sounds like Mathieu had already shown their existence from what Frobenius says in his paper. $\endgroup$ – jwsiegel Feb 27 '18 at 3:18
  • $\begingroup$ @jwsiegel, that's interesting. I wonder, though, what properties of the Mathieu groups he was using. Mathieu "wrote his groups down" by giving permutations that generated them... I wonder if Frobenius made use of the specific given permutations in any way? $\endgroup$ – Nick Gill Feb 27 '18 at 10:18
  • $\begingroup$ @jwsiegel, Let me reiterate, also, that although Mathieu did discover these groups and all the assertions he made about them were correct, there was considerable doubt in the mathematical community as to whether or not these groups were distinct from the alternating groups -- and this doubt was not resolved until Witt came along in the 1930's. If even the order of these groups was not "properly known", it's hard to imagine what other properties Frobenius might have made use of... $\endgroup$ – Nick Gill Feb 27 '18 at 10:19
  • $\begingroup$ I think there was no doubt about $M_{12}$ already in 1904 (maybe on $M_{24}$): E.g., in the 1st edition from 1897 of his book, §109, Burnside describes a "tentative process" by Jordan to determine whether any given $k$-transitive group of degree $n$ is contained in a $(k+1)$-transitive group of degree $n+1$. In a note to §108 at the end of Chapter VIII, he proves Jordan's theorem on sharply transitive groups, and sketches how to get generators of $M_{12}$ by this process. In §153, he gives again generators of $M_{12}$, and an exercise to derive various properties of this group. $\endgroup$ – Frieder Ladisch Mar 1 '18 at 13:52
  • $\begingroup$ The article has since been accepted and can be found ahead of print here: dx.doi.org/10.22108/ijgt.2019.115366.1531 $\endgroup$ – Sam Hughes May 11 at 14:21
16
$\begingroup$

It seems to me that Frobenius is using lots of specific facts about the permutation group $M_{12}$ (and $M_{24}$, respectively) here, and that he has no doubts about the existence of these groups. In particular, he is using specific subgroups of $M_{12}$ and $M_{24}$ (and then using induced characters). Most statements are given without proof, and so the reader would have to verify them by whatever tedious calculations (?).

The first paragraph of §5 reads:

By using the developed theorems, I have calculated the characters of all multiply transitive groups of degree $\leq 24$. Except for the symmetric and alternating groups of the various degrees, no group is known which is more than fivefold transitive, and only two five-transitive groups are known, both discovered by Mathieu, whose characters I want to give here.

In the next paragraph, Frobenius describes the partition of $M_{12}$ into conjugacy classes, without proofs or references, so as if this is generally known or easy to reproduce. He also explains notation in the character table, e.g., $(6)(3)(2)$ denotes the class of elements with cycle lengths $6,3,2,1$, and $(3)^4$ denotes the class of elements which decompose into $4$ cycles of length $3$. (In the linked scan of the paper, the first column of the table denoting the conjugacy classes is partly hidden in the binding.) The third paragraph of §5 explains more notation (second column contains centralizer orders, first row degrees of characters with superscript if there is more than one of the same degree).
The bottom paragraph on p. 567 reads:

By the theorems we have proved, $M_{12}$ has the characters $\alpha-1 : 11^{(1)}$, $\frac{1}{2}(\alpha-1)(\alpha-2)-\beta : 55^{(1)}$ and $\frac{1}{2}\alpha(\alpha-3) + \beta : 54$. [Here, $\alpha$ denotes the number of elements fixed and $\beta$ the number of $2$-cycles.] Formula (5), §2 contains three characters of the third dimension of the symmetric group $S_{12}$, and each of these decomposes into two characters of $M_{12}$, namely $99+55^{(3)}$, $120+45$, $144+176$. The first two characters of Formula (6), §2 decompose into $11^{(2)} + 54 + 66 + 144 $ and $ 66 + 120 + 144 $.

Paragraph following the table on p.568:

The substitutions of $M_{12}$ fixing a symbol constitute a $4$-transitive group $M_{11}$ of degree $11$ and order $11.10.9.8$. Furthermore, $M_{12}$ contains a subgroup which is isomorphic to $M_{11}$, is $3$-transitive of degree $12$ and has order $12.11.10.6$. So one can also use this group to represent $M_{12}$ as a transitive permutation group on $12$ symbols, and thus one obtains an outer automorphism of $M_{12}$, and thereby the classes $(8)(2)$ and $(8)(4)$ and their squares $(4)^2$ and $(4)^2(2)^2$ are interchanged. By this automorphism, the character $11^{(1)}$ is sent to $11^{(2)}$ and the character $55^{(1)}$ is sent to $55^{(2)}$. The other characters are computed using the subgroup $M_{11}$.

Next paragraph:

With benefit, one can also use the following remarkable subgroup of $M_{12}$: Let $(1,2,3,4,5,6)(7,8,9)(10,11)(12)$ be a substitution in the class $(6)(3)(2)$. Then all substitutions $R$ in $M_{12}$, which move the first $6$ (and the last $6$) only among them, constitute a group of order $6!$. Each such substitution $R$ decomposes into two substitutions $R_1$, only permuting the first $6$, and $R_2$, only permuting the last $6$ symbols among themselfs. Both $R_1$ and $R_2$ run through the $6!$ substitutions of the symmetric group $S_6$ of degree $6$, and $R_1$ and $R_2$ correspond by the well known outer autmorphism of this group. In fact, the classes $(6)$, $(3)^2$ and $(2)^3$ of $S_6$ thus correspond to the classes $(3)(2)$, $(3)$ and $(2)$, and their union gives the classes $(6)(3)(2)$, $(3)^3$ and $(2)^4$ of $M_{12}$. The principal character of $S_6$ gives the reducible character $1 + 11^{(1)} + 11^{(2)} +54 +55^{(3)}$ of $M_{12}$, the other linear character of $S_6$ gives the character $11^{(2)} + 55^{(2)} +66$, and thus one gets the two simple characters $55^{(3)}$ and $66$. Then the above formulas give all characters except $16^{(1)}$ and $16^{(2)}$, which can easily be determined from the bilinear relations [=orthogonality relations?].

Last paragraph of §5 (p. 569):

If $\chi(R)$ is a character of $H$, then
$$ \frac{1}{2}(\chi(R)^2 - \chi(R^2)) \:,\quad \frac{1}{2}(\chi(R)^2 + \chi(R^2) ) \tag{1.} $$ are linear combinations of characters with positive integer coefficients. If one chooses for $\chi(R)$ the character $16^{(1)}$, then one gets $120$ and $16^{(2)} + 54 + 66$.

Section 6 on $M_{24}$ is even more sketchy, but has roughly the same structure. Frobenius tells us that he used two other subgroups in addtion to the stabilizer $M_{23}$. One is the setwise stabilizer $M_{16+8}$ of a subset of order $16$, which is the $15+1$-part of an element of cycle structure $15+1+5+3$. The permutations induced on the $16$-subset are said to form the $3$-transitive linear group of order $2^4(2^4-1)(2^4-2)(2^4-2^2)(2^4-2^3)$, and the permutation group induced on the $8$-set is $A_8$.
According to the bottom paragraph on p.~570, on gets another subgroup by taking the stabilizer of a $12$-subset. The stabilizer is isomorphic to $M_{12}$. Each element of the stabilizer decomposes into two permutations $R_1$ and $R_2$ on a $12$-subset and its complement, the correspondence between $R_1$ and $R_2$ is given by the outer automorphism of $M_{12}$ described in §5. Unfortunately, Frobenius gives us no clue how to choose that $12$-subset ("in passender Art")!

$\endgroup$
  • 1
    $\begingroup$ This is a wonderful answer -- thank you so much. Some remarks about some of the properties that Frobenius appears to use, in light of your answer: Some back-of-the-envelope calculations suggest that using sharp 5-transitivity alone one can get a pretty good sense of the conjugacy classes of your putative group $G$. So, for this part, I suspect that the specific construction for $M_{12}$ does not enter (much).... $\endgroup$ – Nick Gill Feb 28 '18 at 17:46
  • 1
    $\begingroup$ Your second extended quote, about characters of dimension 11, 54 and 55, is also the consequence of Frobenius' own theorem stipulating that certain irreducible characters of $S_n$ remain irreducible when restricted to 4-transitive subgroups (see Carlo's answer). So, again, specific properties of $M_{12}$ are not needed here. (And, by the way, Saxl has a lovely paper giving a converse to this theorem of Frobenius.)... $\endgroup$ – Nick Gill Feb 28 '18 at 17:48
  • 1
    $\begingroup$ However, you then quote remarks about a subgroup of $M_{12}$ isomorphic to $S_6$ that fixes a decomposition $6-6$. It is not at all clear to me how one can see this subgroup just using sharp $5$-transitivity. This latter property obviously gives a $S_5$ subgroup preserving a $5-7$ decomposition, but beyond that, I'm not so sure... I guess in your final quote $H$ is this $S_6$-subgroup (?), in which case it is certainly entering his calculations in a definitive way... So that seems a concrete place where he's used something more than sharp 5-transitivity... $\endgroup$ – Nick Gill Feb 28 '18 at 17:59
  • $\begingroup$ @NickGill: have just realized that I had forgotten to translate two sentences in my third quote; I have added them now. $\endgroup$ – Frieder Ladisch Feb 28 '18 at 18:12
  • 1
    $\begingroup$ I think in the final quote, $H$ is meant to be an arbitrary group, and then this is applied to $H=M_{12}$. But this seems not to be part of the calculation anyway, but a "bonus". This is the well known decomposition of $\chi^2$ into symmetric and alternating part, and Frobenius is applying this to one of the characters of degree $16$. But in the first paragraph on p. 568 (my third bock quote), he's using an outer automorphism of $M_{12}$ to get two other characters of $M_{12}$, and this seems to be specific information on $M_{12}$. $\endgroup$ – Frieder Ladisch Feb 28 '18 at 18:19
15
$\begingroup$

This summary of Frobenius' 1904 paper might be of use:

Thomas Hawkins, The Mathematics of Frobenius in Context (page 527-528).

$\endgroup$
  • 5
    $\begingroup$ Hi Carlo, Thank you for this -- it's good to have this summary available in English. I'm still a bit unclear about what is meant by "determine the character tables of the two fivefold transitive Mathieu groups..." I'd really like to know what facts, beyond sharp 5-transitivity, he used in these calculations (if any). If the very definition of the groups was in question, what exactly does it mean to calculate their character table?! $\endgroup$ – Nick Gill Feb 26 '18 at 11:52
7
$\begingroup$

Not really an answer, and probably contained in Frobenius's paper, but a starting point might be that (using a result now usually credited to Blichfeldt) if $G$ is a sharply $5$-transitive group (of any degree) with permutation character $\chi$, and trivial character $\chi_{0},$ then $\chi(\chi - \chi_{0})(\chi-2\chi_{0})(\chi- 3\chi_{0})(\chi-4\chi_{0})$ must be the regular character ( note that the given product of virtual characters vanishes on all non-identity elements of $G$ and takes value $|G|$ on the identity). Hence decomposing each of $\chi,\chi^{2},\chi^{3},\chi^{4},\chi^{5}$ into sums of irreducible characters would yield all non-trivial irreducible characters of $G$ ( and more besides).

$\endgroup$
  • 1
    $\begingroup$ Hi Geoff, thanks for this, that is very interesting. I guess what you are saying is slightly orthogonal to my question as it's not really about what Frobenius did himself... However I am definitely interested in how one might (re)prove this theorem "Suppose that $G$ is a sharply 5-transitive subgroup of $S_{12}$, then its character table is...", and your answer gives a very promising line of enquiry for proving such a theorem directly, and without referencing $M_{12}$ specifically. $\endgroup$ – Nick Gill Feb 28 '18 at 17:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.