User geoff robinson - MathOverflow

Answer by Geoff Robinson for How much of character theory can be done without Schur's lemma or the Artin-Wedderburn theorem?

2013-05-23T19:49:41Z

I spent a long time writing an answer to this question, but MO did not believe I was a human being ( I did mis-spell one of the test words, but everyone deserves a second chance, I think ), so it seems to have disappeared. I am not sure I have the energy to do it again right now, but here (in precis, though not precise) are three points I thought worth making/suggesting:

The group determinant has been mentioned: in a 1991 Proc AMS paper, Formanek and Sibley proved that the group determinant determines the group. Perhaps you could use the analogue of the group determinant to tease out the properties that an algebraic structure $G$, whose "formal character theory" satisfies the properties you can get without Schur and Wedderburn, would have. Such a "group determinant" would not a priori have the property that the mutiplicity of an irreducible factor equals its degree.
It is possible to get quite far into the structure of the group algebra just using the symmetric algebra structure of the complex group algebra of $G$ induced by the linear form $t$ with $t(\sum_{g \in G} a_{g}g ) = a_{1}.$ Since $t$ vanishes on nilpotent elements, it follows that no non-zero right ideal of $\mathbb{C}G$ consists of nilpotent elements and that for each minimal (two-sided) ideal $A$ of $\mathbb{C}G$, $Z(A)$ is $1$-dimensional. You might argue that this is using representation theory, since it is not a priori immediately obvious that $t$ vanishes on nilpotent elements until one notes that (up to the multiple $|G|$ ) $t$ is the trace afforded by the regular representation.
Frobenius's theorem, and other normal complement theorems of a similar nature suggest that there might be an analogue (under certain hypotheses) of the transfer homomorphism but when the target group is not necessarily Abelian. That theorem can be proved in teh case that $H$ is solvable by using the usual transfer homomorphism, and what the theorem says in the end (in the general case) is that the identity homomorphism from $H$ to itself extends to a homomorphism from $G$ onto $H.$ If one tries a transfer-type proof, it looks as though it almost would work, except that the order of products matters when the target group is non-Abelian. Nevertheless, the Theorem does in teh end say that the homomorphism you would like to define by "transfer" is well-defined after all.

Answer by Geoff Robinson for On finite groups with same complex-valued character table

2013-05-18T18:02:43Z

This is a complicated question. A pair of non-isomorphic groups with the same character table is sometimes called a "Brauer Pair". There are many such pairs, especially among $p$-groups.

Answer by Geoff Robinson for Why are Schur multipliers of finite simple groups so small?

2013-05-17T19:39:18Z

I would be very surprised if you receive a "conceptual" answer to this problem- though I would be delighted to be proved wrong. Regarding your last comment, there have been examples recently where computational evidence has indicated that human intuition about the size of cohomology groups was probably faulty, being based on limited evidence.

Regarding the comment about the bad general bound for the size of the Schur multiplier of a finite group, it can get quite big for $p$-groups, as you no doubt know. If my memory is correct, an elementary Abelian $p$-group of order $p^{n}$ has Schur multiplier of order $p^{n(n-1)/2}$, as is well-known.

Answer by Geoff Robinson for A catalog of faithful representations of finite groups?

2013-05-16T07:08:46Z

It may be useful to note that if $M$ is a subgroup of the finite group $G$ and $M$ contains no non-identity normal subgroup of $G$ then the permutation character of $G$ afforded by the action on the cosets of $M$ is faithful, so subtracting the trivial character gives a faithful complex character of degree $[G:M]-1.$ This usually won't be the minimal degree of a faithful character, but it is a handy upper bound for the minimal degree.

Answer by Geoff Robinson for Modular reductions of simple characters

2013-05-08T01:27:48Z

I am working from memory, but I believe that there is a proof in Feit's Book "Representations of finite groups" that if the prime $p$ is sufficiently ramified in the dvr then it can even be the case that the reduction (mod $p$) of (a module affording) $\chi$ can be completely reducible. In any case, that is a true statement.

Answer by Geoff Robinson for Relationship between the number of Sylow subgroups with element orders in finite group

2013-05-04T07:46:29Z

One relationship is that the number of $p$-singular elements ( that is, elements whose order is divisible by $p$) is divisible by the number of Sylow $p$-subgroups of $G$. This is a consequence of a theorem Frobenius, together with Sylow's theorem, though I don't recall seeing the fact stated in print. Let $P$ be a Sylow $p$-subgroup of $G.$ Frobenius proved that if $n$ divides the order of finite group $G,$ then the number of solutions of $g^{n}=1$ in $G$ is an integer multiple of $n.$ Hence the number of solutions of $x^{[G:P]} = 1$ in $G$ is divisible by $[G:P].$ This number is also $|G| - \#$ ($p$-singular elements of $G$). Hence the number of $p$-singular elements of $G$ is divisible by $[G:P]$. This is in turn divisible by $[G:N_{G}(P)],$ which is the number of Sylow $p$-subgroups of $G.$

Answer by Geoff Robinson for element of order n such that $\pi(n)=\pi(G)$, where $\pi(n)$ denote the prime divisors of $n$

2013-04-28T18:51:33Z

In the case of solvable groups, this may not say much about the structure of the group. For example, if $G$ is a finite $\{p,q\}$ -group, where $p,q$ are distinct primes (hence $G$ is solvable by Burnside's $p^{a}q^{b}$-theorem), then it is quite unusual (though certianly not impossible) for $G$ not to have an element of order $pq.$ If $G$ contains an elementary Abelian $p$-group of order $p^{2}$ and an elementary Abelian $q$-group of order $q^{2}$, then $G$ will contain an element of order $pq.$ If, for example, $G$ contains no elementary Abelian $q$-subgroup of order $q^{2},$ then the Sylow $q$-subgroups of $G$ are cyclic or generalized quaternion

Answer by Geoff Robinson for Why didn't finite group theorists consider groups where all centralizers of non-identity elements are solvable?

2013-04-27T13:29:56Z

Groups in which the centralizer of every involution is solvable were classified by D. Gorenstein and various co-authors. Also J. G. Thompson classified finite groups such that the normalizer of every non-identity solvable subgroup is solvable. Results of this kind were in some ways more general than the problem you asked about.

Answer by Geoff Robinson for About isomorphism of $PGL(2)$ and $SO(3)$

2013-04-23T02:57:17Z

Put the bilinear form $\langle, \rangle$ on $2 \times 2$ real matrices by setting $\langle A,B \rangle = {\rm tr}(AB).$ The space of matrices breaks with respect to this form as the orthogonal direct sum of the space of scalar matrices and the $3$-dimensional subspace of matrices of trace zero. Now ${\rm GL}(2,\mathbb{R})$ acts by conjugation on the the matrices of trace zero, and preserves this bilinear form in that action. Furthermore, scalar matrices (and nothing more) in ${\rm GL}(2,\mathbb{R})$ are in the kernel of this action, so the action is really one of ${\rm PGL}(2,\mathbb{R}).$ Every matrix in ${\rm GL}(2,\mathbb{R})$ has the eigenvalue $1$ in this action- a scalar matrix certainly does and any non-scalar matrix $A$ fixes the matrices of trace zero in ${\rm span}(I,A).$ Every element of ${\rm PGL}(2,\mathbb{R})$ acts with determinant $1$ in this action, as diagonal elements clearly do. This gives an embedding of ${\rm PGL}(2,\mathbb{R})$ in the special orthogonal group determined by this form,and dimension shows that it is surjection.

Answer by Geoff Robinson for Upper bound on order of finite subgroups of GL_n(Z_p)?

2013-04-20T20:59:31Z

There may be a section in the old Curtis and Reiner "Theorems of Blichfeld, Burnside and Frobenius" which answers this to a large extent, although they don't think about $p$-adics. Thanks to Jordan's theorem, there is an Abelian normal subgroup of (fixed) bounded index. So the question reduces to limiting the size of Abelian (normal) subgroups. But for the rings $R$ you discuss, there are only finitely many roots of unity in any such $R,$ and for any given $R,$ there is an explicit bound on the number of roots of unity in $R$, and an explicit bound on the size of Abelian subgroups (in fact, for a primitive absolutely irreducible group, the largest normal Abelian subgroup consists of scalar matrices).

An alternative approach is to note that the kernel of any reduction (mod $p$) (strictly, reduction (mod the appropriate prime ideal containing $p$) of a finite subgroup is a finite normal $p$-group, and the image group is a subgroup of a finite ${\rm GL}(n,q),$ ($q$ a fixed -in terms of $R$-power of $p$). There are many ways to obtain explicit bounds on the size of the normal $p$-subgroup- over some extension field, it is a monomial group, etc.

In view of questions below, let me expand a little. It is commonplace in modular representation theory to work with a $p$-modular system, which is a triple $(\mathbb{K},R,F)$ such that $R$ is a complete discrete valuation ring of characteristic $0$ such that $R$ has field of fractions $\mathbb{K},$ and $F$ is the residue field $R/J(R)$. This triple is usually taken to be large enough for $\mathbb{K}$ to contain a splitting field for a finite group $G$ and its subgroups ( for example, by assuming that $R$ contains a primitive $|G|$-th roots of unity, which we now do). It is also commonplace to identify $\mathbb{C}$-valued characters of $G$ with characters afforded by finite dimensonal $\mathbb{K}G$-modules. Details are often glossed over, but this is all perfectly permissible.

Any character of a finite dimensional $\mathbb{C}G$-module is afforded by some $\mathbb{Q}[\omega]G$-module, where $\omega$ is a primitive $|G|$-th complex root of unity. This module is determined uniquely up to isomorphism by its character. Since $\mathbb{Q}[\omega]$ is isomorphic to a subfield of $\mathbb{K}$ under current assumptions, every $\mathbb{C}G$-module ``comes from" a $\mathbb{K}G$-module. Conversely, every character afforded by a $\mathbb{K}G$-module may be decomposed uniquely using the standard inner product on the character ring into a sum of complex irreducible characters, each of which may be afforded by a $\mathbb{Q}[\omega]G$-module. So for most purposes, there is little difference between studying $\mathbb{C}G$-modules and $\mathbb{K}G$-modules, and, in particular, the maximum index of an Abelian normal subgroup of a finite subgroup of ${\rm GL}(n,\mathbb{K})$ can be no larger than the corresponding bound for ${\rm GL}(n,\mathbb{C}).$ As mentioned in comments, if one works with primitive irreducible groups, all Abelian normal subgroups are central. But large groups of scalar matrices can't be finessed, though Jordan's theorem in the primitive case shows that this is the only real obstacle to an absolute bound.

Answer by Geoff Robinson for wreath product and matrix presentation

2013-04-13T12:49:54Z

Since it came up in comments, I will give an answer about the group I believe was intended to be asked about. A monomial matrix is a matrix which has one non-zero entry in each row and one non-zero entry in each column. The monomial $n \times n$ matrices who non-zero entries are all $\pm 1$ form a group, which may be thought of as an abstract group as $Z_{2} \wr S_{n}.$ This matrix group has a normal elementary Abelian subgroup of order $2^{n}$ consisting of all its diagonal matrices. This group has many Sylow $2$-subgroups when $n >2,$ but they are all conjugate within it, so in particular are all isomorphic, and they all contain the normal subgroup consisting of all its diagonal matrices. I believe that the group which was intended to be asked about was such a Sylow $2$-subgroup in the case that $n = 2^{r-1}$.

Answer by Geoff Robinson for Finite subgroups of $PGL(3,K)$

2013-04-08T22:04:37Z

Edited in view of Derek Holt's comment on Schur indices: These things are well studied in the literature. You probably want to restrict to irreducible subgroups, and it's probably just as well to work with ${\rm GL}(3,K).$ In such a low dimensions, the Schur index usually will not play much of a role. The Schur index usually arises because it can happen that a complex irreducible character $\chi$ may take values in a field $K,$ but might not be afforded by a representation over $K.$ There is, however, a smallest integer $m_{K}(\chi)$ such that the character $m_{K}(\chi) \chi$ is afforded by a representation over $K$ and $m_{K}(\chi)$ divides $\chi(1).$ If $m_{K}(\chi) =3,$ then representation affording $\chi$ can only be realised over a degree $3$ extension of $K$.Except in degenerate cases, we won't have irreducible representations over $K$ which are not absolutely irreducible, since such a a representation would break up over some extension field of $K$ as a sum of Galois conjugate representations of the same degree. The finite irreducible subgroups of ${\rm GL}(3,\mathbb{C})$ have been known for a century or so. Such an imprimitive group has an Abelian normal subgroup $A$ such that $G/A$ is isomorphic to a subgroup of $S_{3}.$ The primitive ones may be rescaled so that all elements are unimodular, and once this is done, we obtain $G/Z(G)$ isomorphic to $A_{5}, A_{6},{\rm PSL}(2,7)$ or else $G$ is a solvable group with $G/O_{3}(G)$ isomorphic to ${\rm SL}(2,3)$ and $[G:Z(G)] = 216.$

Answer by Geoff Robinson for Conjugate in the symmetric groups

2013-04-05T21:20:43Z

Not when $p >5$. Conjugation by $\rho$ induces an automorphism of order $2n$ of $\langle \pi \rangle.$ Since $\pi$ commutes with nothing other than its powers, this means that $\rho$ must be an element of order $2n.$ This is $p-1$ when $p =3$ or $p =5,$ but not for bigger Fermat primes.

Answer by Geoff Robinson for How many conjugacy classes of subgroups does GL(2,p) have?

2013-03-27T22:17:25Z

Another way to look at these is that there are three kinds of subgroup: those which contain a trivial Sylow $p$-subgroup, those which contain exactly one Sylow subgroup of order $p,$ ad those which contain more than one Sylow $p$-subgroup. The last type is easy to deal with: any such subgroup contains all $p+1$ Sylow $p$-subgroups of ${\rm GL}(2,p,$ so contains ${\rm SL}(2,p)$ (and is, in particular, normal). Every subgroup between ${\rm SL}(2,p)$ and ${\rm GL}(2,p)$ occurs, and the number (of conjugacy classes of) such subgroups is the number of divisors of $p-1.$ The second type is also reasonably straightforward: any such subgroup is conjugate to one and only one subgroup of the group of invertible upper triangular matrices and that group of upper triangular matrices contains all upper unitriangular matrices. The first type consists of subgroups of order prime to $p.$ Since matrix representations over finite fields which are conjugate over an extension field are already conjugate over the field of realizability, the conjugacy class of such a group is uniquely determined by its Brauer character. The isomorphism types which can occur are described already in earlier answers.

Answer by Geoff Robinson for a group with all sylow p subgroups cyclic

2013-03-26T00:06:38Z

If the finite group $G$ has a cyclic Sylow $p$-subgroup $P,$ where $p$ is the smallest prime divisor of $|G|,$ then $G$ always has a normal $p$-complement by (for example) Burnside's transfer theorem, though that normal $p$-complement need not be cyclic. However,if the remaining Sylow subgroups of $G$ are also cyclic and $C_{G}(P) = P,$ the normal $p$-complement will also be cyclic. In the early group-theoretic analysis in the proof of the Feit-Thompson odd order theorem, it is proved that if $G$ is a finite group of odd order and $G$ contains no elementary Abelian subgroup of rank $3$ for any prime, then $G$ has a normal Sylow $q$-group where $q$ is the largest prime divisor of $|G|.$

Answer by Geoff Robinson for Measures of non-abelian-ness

2013-03-25T01:17:33Z

Yes, as Arturo says, you probably want what is known as the "commuting probabilty of $G$", cp(G). Bob Guralnick and I proved (among other things) in a Journal of Algebra paper (circa 2006) (without using the classification of finite simple groups) that $cp(G) \to 0$ as $[G:F(G)] \to \infty,$ where $F(G)$ is the largest nilpotent normal subgroup of a finite group $G,$ though sharper results are possible using the classification.

Answer by Geoff Robinson for Maximal soluble subgroups in a parabolic subgroup of finite classical simple group

2012-11-24T17:29:11Z

I think there are examples when the Borel subgroup is not maximal solvable in a parabolic. One can occur when $q \leq 3,$ so for example when $G = {\rm GL}(3,2)$ both the maximal parabolics are themselves solvable (isomorphic to $S_{4}$). Similarly when $p= 3$, the group ${\rm GL}(3,3)$ has a solvable maximal parabolic $P$ with unipotent radical $U$ such that $P/U \cong {\rm GL}(2,3).$ Such behavior perpetuates itself in higher ranks to give solvable parabolic subgroups strictly containing the Borel in characteristic $2$ or $3$. Later comments: Really, from an inductive point of view, it is probably best not to assume that $G$ is simple as an abstract group, and then one can argue inductively on the rank of the associated $BN$-pair.This seems to reduce us to the rank $1$ case and then it would appear that the only case to worry about is when the whole group is solvable (otherwise the Borel intersects the unique component in a maximal subgroup).

Answer by Geoff Robinson for Finite groups having no dihedral subgroup of order $2p$ for any odd prime $p$

2013-03-23T11:09:31Z

Yes, these are just the finite groups in which all involutions (elements of order $2$) lie in $O_{2}(G),$the largest normal $2$-subgroup of $G,$ using the Baer-Suzuki theorem. Clearly, if every involution of $G$ lies in $O_{2}(G)$, then all dihedral subgroups of $G$ are $2$-groups. On the other hand if $G$ has no dihedral subgroup of order $2p$ for any odd prime $p,$ then whenever $t$ is an involution of $G,$we see that for each $g \in G,$ the dihedral group $\langle t,t^{g} \rangle$ is a $2$-group, so $t \in O_{2}(G)$ by the Baer-Suzuki theorem.

Answer by Geoff Robinson for Characters of p-groups

2013-03-18T00:02:15Z

For every non-trivial irreducible character $\chi$ of a finite $p$-group $P,$ we may choose an element $z in P$ such that $\chi(z) = \chi(1) \omega$ for some primitive $p$-th root of unity $\omega$ (this is an easy consequence of Schur's Lemma and the fact that the image of $P$ in the associated representation has non-trivial center). Now let $\zeta$ be a primitive $p^{e}$-th root of unity, where $P$ has exponent $p^{e}.$ Then ${\rm Gal}(\mathbb{Q}[\zeta]/\mathbb{Q})$ acts on the irreducible characters of $P,$ and it is clear that $\chi$ is in an orbit of length divisible by $p-1.$ Since $\chi$ was an arbitrary non-trivial irreducible character, and since all irreducible characters in the same orbit have the same degree it follows both that the total number of non-trivial irreducible characters, and the total number of irreducible non-linear characters of $P$ are multiples of $p-1$ (in fact, the number of irreducible characters of $p$ of any fixed degree $p^{d} >1$ is a multiple of $p-1.$ And yes, a similar statement holds for conjugacy classes. If $C$ is a non-trivial conjugacy class of $P$, say containing an element $x,$ then the length of the conjugacy class only depends on $\langle x \rangle$. If $x$ has order $p^{e},$ then $\langle x \rangle$ has $p^{e-1}(p-1)$ generators, but if $y$ is another generator, and $y^{p^{e-1}} \neq x^{p^{e-1}},$then $x$ and $y$ are not conjugate within $p.$ Since $\langle x \rangle$ contains $p-1$ elements of order $p,$ it follows that the number of conjugacy classses of non-trivial elements of $P$ which have length $[P:C_{P}(x)]$ is divisible by $p-1.$ In particular, the number of conjugacy classes of length greater than $1$ is divisible by $p-1.$

Answer by Geoff Robinson for Subgroups of the general linear groups

2013-03-16T22:29:30Z

REVISION: In fact, it is a consequence of the (now proved) so-called $k(GV)$-problem that the answer is indeed affirmative when $M$ has order prime to $p,$ and something rather stronger holds. There is a book about the $k(GV)$-problem by Peter Schmid. The $k(GV)$-problem is a special case of a problem of R. Brauer. The $k(GV)$-problem is to prove that if $G$ is a group of order prime to $p$ and $V$ is a faithful finite-dimensional ${\rm GF}(p)G$-module, then the semidirect product $GV$ has at most $|V|$ conjugacy classes. This has now been proved by a combination of authors, the final cases being handled by D.Gluck,K.Magaard, U.Riese and P.Schmid. There are examples when the bound is attained. Its proof does require the classification of finite simple groups. Note that one consequence of the $k(GV)$-problem is that under those hypotheses, $k(G) < |V|,$ where $k(G)$ denotes the number of conjugacy classes of $G.$ In particular, this yields $[G:G^{\prime}] < |V|,$ since $k(G)$ is also the number of complex irreducible characters of $G$, and $[G:G^{\prime}]$ is the number of complex linear characters of $G.$
Hence this does imply that $[M:M^{\prime}] \leq (p^{n}-1)$ when $M$ is a subgroup of order prime to $p$ of ${\rm GL}(n,p)$ (in fact, we can even conclude that $M$ has at most $p^{n}-1$ conjugacy classes).

The answer ( to this MO question) may still be affirmative if you stick to completely reducible subgroups $M$ of ${\rm GL}(n,p),$ though I am nt certain of that. In this situation, these are essentially the groups with no non-identity normal $p$-subgroups- it is necessary to take a little care here, because the underlying module for $M$ may need to be replaced by the direct sum of its composition factors under the action of $M.$ However, if the subgroups $M$ with $O_{p}(M)$ are understood, then that covers all necessary groups at least up to isomorphism. By looking at such subgroups $M,$ you eliminate counterexamples to the question such as those arising in Peter Mueller's answer. However, there are completely reducible subgroups of ${\rm GL}(n,p)$ which are not of order prime to $p,$ such as ${\rm GL}(n,p)$ itself. The completely reducible case reduces almost immediately to the case when $M$ is irreducible. After that, there is work to do: I don't see an immediate elementary argument, but Clifford theory begins to come into play( for example, it may be possibe to reduce to the case where the undelying module for $M^{\prime}$ is a direct sum of isomorphic irreducible modules). This may well be a difficult queston when $M^{\prime}$ is non-Abelian simple, or when $M$ itself is almost simple. The authors I would check out for relevant results here would be people like : M.Aschbacher,R.Guralnick, P.Kleidman, M.Liebeck, P.Tiep. Guralnick and Tiep and others have been trying to prove variants of the $k(GV)$-problem when $G$ acts completely reducibly on $V$. The general bounds are necessarily somewhat weaker than $|V|$. I do not know at present whether they would provide an affirmative answer to this MO question in the case that $M$ is a completely reducible subgroup of ${\rm GL}(n,p)$, or whether this special case has been considered by authors such as Guralnick and Tiep.

Answer by Geoff Robinson for What are the connections between pi and prime numbers?

2013-03-15T21:27:07Z

Here is an example of a way to use $\pi$ to prove the infinitude of primes without calculating its value, or using the relatively deep fact that $\pi$ is irrational, but starting from the knowledge of $\zeta(2)$ and $\zeta(4).$ Suppose that there were only finitely many prime numbers $ 2= p_{1}, 3= p_{2}, \ldots, p_{k-1},p_{k}.$ From the formulae $\sum_{n=1}^{\infty} \frac{1}{n^{2}} = \frac{\pi^{2}}{6}$ and $\sum_{n=1}^{\infty} \frac{1}{n^{4}} = \frac{\pi^{4}}{90}$, we may conclude after the fashion of Euler that (respectively) we have: `$\prod_{j=1}^{k} \frac{p_{j}^{2}}{p_{j}^{2}-1} = \frac{\pi^{2}}{6}$' and $\prod_{j=1}^{k} \frac{p_{j}^{4}}{p_{j}^{4}-1} = \frac{\pi^{4}}{90}.$ Squaring the first equation and dividing by the second leads quickly to $\prod_{j=1}^{k} \frac{p_{j}^{2}+1}{p_{j}^{2}-1} = \frac{5}{2}$, so $5\prod_{j=1}^{k} (p_{j}^{2}-1) = 2 \prod_{j=1}^{k}(p_{j}^{2}+1).$ This is a contradiction, since the product on the left is certainly divisible by $3$, whereas every term in the rightmost product except that for $j = 2$ is congruent to $-1$ (mod 3), so we obtain $0 \equiv (-1)^{k}$ (mod 3), which is absurd. (I would be grateful if anyone knows a reference for a proof like this. I can't believe that I am the first person to think of it).

Answer by Geoff Robinson for Subgroups of $SL_2(F)$ generated by unipotent elements

2013-03-14T20:46:27Z

You should look up Dickson's Theorem. It deals pretty comprehensively with the two dimensional case when $F$ is finite, at least in odd characteristic. The proof may be found in Gorenstein's 1968 book "Finite Groups". I don't remember the statement precisely, but I think it's that $\langle \left( \begin{array}{clcr} 1 & 0\\1 &1 \end{array} \right),\left( \begin{array}{clcr} 1 & \lambda\\0 &1 \end{array} \right) \rangle $ is all of ${\rm SL}(2,K)$, where $K$ is the subfield of $F$ generated by $\lambda \neq 0,$ except when $|F| =9,$ in which case ${\rm SL}(2,5)$ may occur.

There is also the work by J.G. Thompson on quadratic pairs, which may be seen as a massive generalization of this result. Going back further historically than that, there is the Hall-Higman theorem, and more recently there has been extensive work on so-called $2F$-modules, which is relevant to the classification of finite simple groups.

Answer by Geoff Robinson for If e^itπ is algebraic , is $t$ a rational number.

2013-03-09T11:43:51Z

The question as stated is very easy to answer. If the question was really intended to be asked about algebraic integers, then there is still a relatively simple direct example to show that the answer is "no". The question about algebraic integers is well-studied, with a substantial literature and the example that follows is one easy instance:

If we take any real algebraic integer $s$ with $0 < s <1,$ then, $t = s + i \sqrt{1-s^{2}}$ is an algebraic integer which lies on the unit circle. Now apply this with $s = \sqrt{2}-1.$ Note that $t$ generates a degree $4$ extension of the rationals. If $t$ were a primitive $m$-th root of unity, we would have $\phi(m) = 4$ so that $m = 8.$ Hence $t$ would be a primitive $8$-th root of unity, but it is not, as each primitive $8$-th root of unity has real part $\pm \frac{1}{\sqrt{2}}.$

Answer by Geoff Robinson for Can we have many 1-dimensional rep, and very few high dimensional reps in a finite group?

2013-03-02T21:04:43Z

Slightly in the other direction, if $p$ is an odd prime and $P$ is a finite $p$-group which has a non-linear irreducible complex character $\chi,$ then $P$ has at least $p-1$ irreducible characters of $\chi(1),$ namely the algebraic conjugates of $\chi.$ I won't spell out the details, but consider an element outside the kernel of $\chi$ which is represented as a central element of order $p$ in the representation. To complete that picture, for each positive integer $n$ and each prime $p$ (including $p=2$), there do exist $p$-groups of order $p^{2n+1}$ which have $p-1$ irreducible characters of degree $p^{n}$ and $p^{2n}$ linear characters ( these are the extra-special groups).

Answer by Geoff Robinson for $G=\langle a\rangle H$ for subgroup $H$

2013-02-16T12:05:42Z

One way to see many counterxamples for finite groups is to note that if $G$ has even order, then $G$ contains an element $a$ of order $2$. Any proper subgroup $H$ of $G$ with $G = \langle a \rangle H$ would have to be normal, since subgroups of index $2$ are always normal. Hence every non-trivial perfect finite group $G$ (of even order) provides a counterexample. In particular, every non-Abelian finite simple group (of even order) does. The parentheses are because every non-trivial perfect finite group does in fact have even order, but even without knowing that, there are plenty of explicit examples.

Answer by Geoff Robinson for Finite supersolvable groups with trivial Frattini subgroup

2013-01-28T00:51:52Z

In a finite solvable group $G,$ one has $F(G)/\Phi(G) = F(G/\Phi(G)),$ while also $\Phi(F(G)) \leq \Phi(G).$ It follows when $\Phi(G) = 1,$ that $F(G)$ is completely reducible as a module for $G/F(G),$ that is to say, it is a direct product of minimal normal subgroups of $G,$ on which $F(G)$ clearly acts trivially by conjugation. If, in addition, $G$ is supersolvable, all of these minimal normal subgroups must be cyclic of prime order, from which it follows that $G/F(G)$ is Abelian. It also follows that when $p$ is the largest prime divisor of $|G|,$ the Sylow $p$-subgroup of $G$ is normal, hence elementary Abelian. It is possible to bound the rank of the other Sylow subgroups of $G/F(G),$ and the exponents, but the answer is not pretty: If the prime divisors of $|G|$ are $p_{1} >p_{2} > \ldots > p_{k}$ and $|F(G)| = p_{1}^{r_{1}}\ldots p_{k}^{r_{k}},$ then the rank of the Sylow $p_{i}$ subgroup of $G/F(G)$ is at most $\sum_{j=1}^{i-1} r_{j},$ and its exponent is at most the largest power of $p_{i}$ dividing any $p_{j}-1$ with $j < i.$

Answer by Geoff Robinson for An extension of the converse to Hall's theorem.

2013-01-18T09:11:52Z

For the record, I believe that P. Hall proved that if $|G|$ has $n$ prime divisors, then $G$ is solvable if and only if $G$ has $n$ Sylow subgroups $P_{1},P_{2}, \ldots ,P_{n},$ one for each prime divisor, such that $P_{i}P_{j} = P_{j}P_{i}$ for $1 \leq i,j \leq n.$ You are asking whether the pairwise permutability condition can be dropped. The proof of the more difficult direction Hall's Theorem is something like the following, given Burnside's $p^{a}q^{b}$-theorem. I have forgotten Hall's proof, so have had to concoct a proof which is largely the same as Hall's except that I need to invoke Glauberman's $ZJ$-theorem, which Hall did not require. For suppose that $G$ has such a set of permutable Sylow subgroups, and we wish to prove that $G$ is solvable. Then we may suppose that $n \geq 3,$ otherwise Burnside's $p^{a}q^{b}$-theorem yields the desired result. By induction, for $1 \leq i \leq n,$ $G$ has a solvable subgroup $H_{i}$ with $G = H_{i}P_{i} = P_{i}H_{i}$ and $H_{i} \cap P_{i} = 1$ (we may take $H_{i} = \prod_{j \neq i} P_{j}$ which is a group by the permutability condition, and has order $[G:P_{i}]).$ We may also suppose that $p_{1} \geq 5,$ and we do. If $P_{1}$ normalizes a non-trivial subgroup $N_{1}$ of $H_{1},$ then we have $\cap _{g \in G} H_{1}^{g}$ = $\cap_{x \in P_{1}} H_{1}^{x} \geq N_{1},$ so $G$ has a non-trivial solvable normal subgroup $K,$ and an induction argument shows that $G/K$ is solvable. Hence for $2 \leq j \leq n,$ we have $O_{p_{j}}(P_{1}P_{j}) = 1.$ Since $P_{1}P_{j}$ is solvable, and $p_{1} \geq 5,$ we have $ZJ(P_{1}) \lhd P_{1}P_{j}$ for each such $j,$ by Glauberman's $ZJ$-theorem. But then $ZJ(P_{1}) \lhd G,$ since it is normalized by each of $P_{1},P_{2}, \ldots ,P_{n}.$ Again, and induction argument shows that $G/ZJ(P_{1})$ is solvable. But I emphasize that this proof requires pairwise permutable Sylow subgroups, and the hypotheses of this question do not require that.

Answer by Geoff Robinson for Which functions are linear combinations of irreducible characters for a given field $\Bbbk$?

2013-01-16T05:15:23Z

I am interpreting your question as talking of $k$-linear combinations of traces of $k$-representations of $G.$ Note that such a function must not only be constant on conjugacy classes, but should also be constant on $p^{\prime}$-sections, where $k$ has characteristic $p.$ Recall that every element of $G$ may be written uniquely in the form $g = ab = ba,$ where $a$ has order a power of $p$ and $b$ has order prime to $p.$ The element $b$ is called the $p^{\prime}$-part of $g.$ Two elements of $G$ are said to be in the same $p^{\prime}$-section of $G$ if and only if their $p^{\prime}$-parts are conjugate. I think this reduces us to the case where $G$ is a cyclic $p^{\prime}$-group, using Brauer's or Conlon's induction theorem. I think you may find the necessary analysis of that case in the 1962 book of Curtis and Reiner. Expanded edit: The work (of Berman I believe, if my memory is accurate), I am alluding to, deals with the fact that dealing with traces of $k$-valued representations forces some equality of traces at group elements at elements which are not $G$-conjugate. We can assume, as I indicated, that $|G|$ is coprime to char $k$ if char $k \neq 0$. So, for example, if $\theta$ is the trace of a $k$-representation and $k$ has $q$ elements, then we must have $\theta(g) = \theta(g^{q})$ for all $g \in G.$ Similarly, if $k = \mathbb{Q},$ and $\theta$ is the trace of a $\mathbb{Q}$-valued representation, we must have $\theta(g) = \theta(h)$ whenever $\langle g \rangle = \langle h \rangle.$ So, let us say that $g$ and $g^{m}$ are $k$-conjugate if $\theta(g) = \theta(g^{m})$ whenever $\theta$ is the trace of a representation over $k.$ Then a $k$-valued class function $\psi$ of $G$ is said to respect $k$ if $\psi(g) = \psi(g^{m})$ whenever $g$ and $g^{m}$ are $k$-conjugate. Then the best you can hope for is that the traces of $k$-representations span the space of $k$-valued class functions which respect $k,$ and this is indeed the case.

Answer by Geoff Robinson for Decomposition of an induced representation

2013-01-13T18:56:03Z

Not really an answer, but this is already difficult in the complex case, when $C$ is a central subgroup. For example, if $C = Z(G),$ and we induce a faithful irreducible $C$-module to $G,$ the number of distinct irreducible constituents is bounded above by the number of conjugacy classes of $G/C,$ but I don't know many other general statements- the number can be as low as $1$ (when we have a character of so-called ``central type"). To understand the general case of inducing an indecomposable $RC$-module to $G$ for a general commutative ring $R$ when $C$ ne not be central, Mackey decompoosition is certainly a helpful tool ( the induced module restricts freely to any subgroup meeting $C$ only in the identity, and more generally, the restriction to a subgroup $H$ of the induced module can be reasonably described by knowing how the original indecomposable restricts to $C \cap H,$ though precise details may require care. When working over $\mathbb{Z}_{p}$, Green's indecomposability theorem may also be useful.

Answer by Geoff Robinson for Norms agreeing on dense subspace

2013-01-11T10:37:10Z

This is standard, and the answer has been been indicated in the comments. Recall that in a normed space $X,$ the triangle inequality easily yields that $| \|x\| - \| y \| |\leq \|x-y\|$ for all $x,y \in X.$

Now let's turn to your dense subspace $V$ of the Banach space $B.$ Take an element $b \in B.$ There is a sequence $(v_{n})$ of elements of $V$ such that $v_{n} \to b$ (with respect to the norm on $B).$ Then by the above remark, the sequence $( \|v_{n} \|)$ is a Cauchy sequence of real numbers whose limit is the real number $\| b \|$. Hence $\| b \|$ is uniquely specified in terms of $ \| \|_{V} $ since you assume that $\| \|$ and $\| \|_{V}$ agree on $V.$ Also, $\|b\|$ is the same as would be assigned if considering $b$ as an element of the completion of $V.$ Hence $B$ embeds isometrically as a susbpace of the completion of $V.$ However, it is clearly dense in that completion, as $V$ already is, and it is a closed subspace of that completion, since any Cauchy sequence of elements of $V$ has a limit which lies in $B.$ Hence $B$ is indeed isometrically isomorphic to the completion of $V.$

Comment by Geoff Robinson

2013-05-23T22:21:38Z

You might check out the publications of Roger Bryant (sometimes with coauthors)

Comment by Geoff Robinson

2013-05-23T22:17:59Z

Thanks Will. Your comment highlighted a misapprehension on my part. Yes, the basic idea is that when we see a cycle of length $d$, we are getting $(d-2)/2$ pairs non-real roots of unity when $d$ is even and $(d-1)/2$ pairs when $d$ is odd.

Comment by Geoff Robinson

2013-05-15T19:53:03Z

The converse may be true: the representation theory of the symmetric group can be reduced to combinatorics.

Comment by Geoff Robinson

2013-05-15T07:03:13Z

I corrected my post- I was thinking about the exponent $p$ case. The other direction is OK, as I explained, because the whole of ${\rm Sp}(2r,p)$ is induced by the action of ${\rm SL}(V)$ since the representation does extend to the appropriate central extension of $H,$ which has a subgroup $Z(G) \times {\rm Sp}(2r,p).$

Comment by Geoff Robinson

2013-05-14T18:13:30Z

@H A Helfgott: I was intending to be supportive of you in the sense that given that the result was known to be true in all but a finite number (however ridiculously large) number of cases, there would seem to be no reason not to believe that the result had now been proved. Sorry if it sounded otherwise, that was not my intention at all.

Comment by Geoff Robinson

2013-05-14T07:02:21Z

Didn't Vinogradov prove it for sufficiently large odd numbers in something like 1937? So, it seems reasonable to believe that deciding the question one way or the other would be a matter of time after that.

Comment by Geoff Robinson

2013-05-11T09:42:27Z

This is proved in Herstein's book "Topics in Algebra" for example-maybe second or third edition. Once you know the order of ${\rm GL}$,it is a matter of verifying that the index of the subgroup of upper unitriangular matrices is prime to $p,$ which is clear.

Comment by Geoff Robinson

2013-05-08T20:44:05Z

@F.Ladisch: Thanks, that sounds right.

Comment by Geoff Robinson

2013-05-08T19:30:20Z

@JIm; I can't remember where it is, and I don't have the book to hand, but I'm pretty sure it's there somewhere. It's not the easiest book to read, I agree, bjut there is a lot in there which is hard to find elsewhere.

Comment by Geoff Robinson

2013-05-05T15:23:49Z

I am not aware of one. Do you think they would be substantially different from odd order groups with all Sylow subgroups Abelian?

Comment by Geoff Robinson

2013-05-03T17:33:33Z

For the first comment: No, I don't think the module extends in general if the inertia group is $G$ (in fact, I know it doesn't). The second one probably works for a projective indecomposable module, but probably not in general.

Comment by Geoff Robinson

2013-05-02T20:42:30Z

I was also forbidden to access it

Comment by Geoff Robinson

2013-05-02T20:17:41Z

Peter May does appear on here occasionally. Can't you contact the math department at Chicago? They must have a copy, though maybe not in electronic form.

Comment by Geoff Robinson

2013-04-28T17:06:22Z

My reading would be that it allows irreducible representations which are non finite dimensional, but intends them always to be irreducible. The statement is not true in general, even for finite groups, if the representations are not absolutely irreducible.

Comment by Geoff Robinson

2013-04-22T21:13:01Z

@Jim: Well, I suppose the argument enough relies on realisability over $\mathbb{Q}[\omega],$ which does require Brauer's characterization of characters and Brauer's induction theorem ( though I suppose that realisability over the algebraic closure of the rationals, hence over some finite extension of the rationals for a given finite group would do, and that must have be classically known).