User maurizio monge - MathOverflow

Is a profinite group with a finite number of simple quotients and Jordan-Hölder factors finitely generated?

2011-06-29T20:58:20Z

Assume $G$ is a profinite group such that the Jordan-Hölder factors appearing in the finite quotients vary in a finite number of isomorphism classes of simple groups. Assume also $G$ to have a finite number of subgroups whose corresponding quotient is simple. Does this imply that $G$ is (topologically) finitely generated?

I'm asking here after some attempt to make work a modification of the principle the for a $p$-group $P$ each set of elements generating $P/\Phi(P)$ is a generating set for $P$. For $P$ groups the question is clearly much simpler, and i have been thinking that elements generating each simple quotient had to be be enough (this is not true, as shown by the simple example $Sym_n$. But the different symmetric groups have bigger and bigger Jordan-Holder factors). However the issue is not totally trivial, because maximal (non-normal) subgroups are in generally not contained in a proper normal subgroup, so it is not possible to replicate a similar proof smoothly. Note that the hypothesis of having a finite number of factors rules out silly couterexamples like $\prod_{i=4}^\infty Alt_i$.

Is anything know about this question? Thanks for the attention!

Irreducible mod-p representation of a semidirect product with trivial p-core

2011-10-31T10:47:19Z

Consider the group $G=H\rtimes{}C$, where $H$ has order prime with $p$ and $C$ is cyclic of order $p^k$, and $C\rightarrow{}\mathrm{Aut}(H)$ is faithful (or equivalently $G$ has trivial $p$-core). Assume $V$ to be an irreducible (not just indecomposable) faithful representation of $G$ over $\mathbb{F}_p$.

I need a reference for the following (supposedly well known?) fact:

Proposition: as $\mathbb{F}_p[C]$-module $V$ is free (that is isomorphic to $\mathbb{F}_p[C]^n$ for some n).

This can be proved going to the algebraic closure $\overline{\mathbb{F}_p}$, and decomposing $V$ into irreducible $H$-representations. Then it's possible to see (I omit the details) that $C$ should act with orbits of cardinality $p$ on the irreducible $H$-factors of $V$, and hence that $V$ is induced from an irreducible $H$-representation.

In particular I would like to have an easy way to say that every extension of $G$ by $V$ is split, this is an easy consequence of the proposition above and Schur-Zassenhaus theorem. And I would be surprised if this was not well-known as well.

Symplectic groups Sp_{2m}(2) as 2-transitive permutation (i.e. Galois) groups

2012-01-06T17:04:36Z

Hello, I am looking for information about the symplectic groups $Sp_{2m}(2)$ as permutation group acting on quadratic forms.

Consider the block matrices \[e=\begin{pmatrix}0&1\\0&0\end{pmatrix}, \qquad f=\begin{pmatrix}0&1\\-1&0\end{pmatrix}=e-e^T\] on the vector space $(\mathbb{F}_2)^{2d}$ equipped with the standard basis. Consider the symmetric bilinear form $\phi(u,v)=ufv^T$, and let $\Omega$ be the set of all quadratic forms $\theta(u)$ such that \[ \phi(u,v)=\theta(u+v)-\theta(u)-\theta(v). \] In particular the quadratic form $\theta_0(u)=ueu^T$ is $\in\Omega$, and any other element of $\Omega$ can be shown to be of the form \[ \theta_a = \theta_0(u)+\phi(u,a). \] Now $Sp_{2m}(2)$ acts on $\Omega$, and it turns out that the action splits in two distinct orbits \[ \Omega^+=\{\theta_a|\theta_0(a)=0\},\qquad \Omega^-=\{\theta_a|\theta_0(a)=1\},\] of size respectively $2^{m-1}(2^m+1)$ and $2^{m-1}(2^m-1)$. The group $Sp_{2m}(2)$ acts 2-transitively on each of these orbits, see Chap. 7 of Permutation Groups (Dixon, Mortimer) for more details.

Question: what can be said of the action of these two sets? Here are two more specific questions: is the stabilizer of one point acting imprimitevely, for some block structure? What are the orbits of a 2-point stabilizer?

Motivation: I am studing the Galois groups of polynomials (trinomials) over function fields in characteristic $p$, which can be proven to be $2$-transitive. This was done by Abhyankar, Galois theory on the line in nonzero characteristic (1996), which computed the Galois group of many trinomials, and I think that his results can be extended to cover more cases. If I'm wrong I will happen to have learned something about 2-transitive groups.

The $2$-transitive permutations groups are classified (affine groups, alternating/symmetric, projective, symplectic $Sp_{2m}(2)$, unitary $PGU_3(q^3)$, Suzuki $Sz(q)$ and Ree $R(q)$, plus a few sporadic groups). Computing the local Galois group at a ramified place it is possible to describe the action of a subgroup, the inertia subgroup, as a permutation group on the roots. This allows to rule out certain familes of $2$-transitive groups, and sometimes it is possible to determine completely the Galois group. And the symplectic group at the moment is the family that I find more difficult to understand.

Answer by Maurizio Monge for 'Important' applications of p-adic numbers outside of algebra (and number theory).

2011-12-26T17:55:19Z

The (unsolved) Hilbert-Smith conjecture states that any locally compact group acting faithfully on a manifold has to be a Lie group: http://en.wikipedia.org/wiki/Hilbert%E2%80%93Smith_conjecture

However, it turns out that it is enough to prove this for $\mathbb{Z}_p$, and the conjecture follows proving that $\mathbb{Z}_p$ has no continuous faithful action on a manifold.

Answer by Maurizio Monge for Produce an irreducible polynomial that can't be proved irreducible by using Eisenstein

2011-11-08T12:28:11Z

It is possible to produce a polynomial that cannot (provably) be proved to be irreducible considering the valuations of the roots, or even any polynomial function in the roots (which can be much more general than a linear substitution), for every possible valuation over the base field.

Let $L/K$ be an unramifed extension of numeber fields (for instance the Hilbert class field of a $K$ with non-trivial class group), generated by $\alpha$ say. Then the minimal polynomial for $\alpha$ over $K$ will do.

Answer by Maurizio Monge for Bijective function on a dense set

2011-10-31T12:39:04Z

Sorry but I could not resist:

The map is defined on $[0,1]\times[0,1]$, and can be written as $$ f(x,y) = \begin{cases} (x, (2-x)y) & \text{ if }y\leq 1/2, \\ (x, xy+1-x) & \text{ if }y> 1/2. \end{cases} $$

Just take $D$ to be the set of $(x,y)$ with rational coordinates in $(0,1)\times(0,1)$, then $f$ is bijective on $D$ (because it can be easily inverted), but it is clearly not injective on $[0,1]\times[0,1]$.

Things to keep in mind while looking for a Postdoc overseas

2011-09-14T16:31:21Z

Hello, I would like to receive some suggestions about what you think to be the best important things that should be kept in mind while looking for a postdoc position. I'm not considering (in this question) the opportunities one may have from direct knowledge, e.g. because they were pointed out from your advisor or in any case about positions in some nearby university where you have some contact.

For instance while it is often not written explicitly, for most positions the candidate is expected to have taken personal contacts with one professor that may be interested, and set up a research project together. But this is certainly not the only matter, and I would like to hear some story from who has some experience in this kind of issues, also for what regards things which can vary a lot depending on the country (i did my Master and I am doing Ph.D. in Italy, with a short period in France).

To make the question more focused, i'm mostly interested in the scenario of someone working in pure mathematics (e.g. number theory) making application from Europe to some university in North America, or possibly also in Asia.

(EDITED as suggested by quid, making the question CW, and more focused and about data rather than about advice).

Reference for decomposition in invariants and derived subgroup in a semidirect product of abelian groups

2011-07-15T09:24:08Z

Let $A$ and $B$ be finite abelian groups with coprime order, and let $G=A\rtimes{}B$ be a semidirect product, via any action. Let $C\subseteq{}A$ be the subgroup of the elements of $A$ which are fixed by the action of $B$, so that $C=Z(G)\cap{}A$. Then we have $$ A = C \oplus G'. $$

Is there a quick reference for this fact? Please note that i'm NOT asking for a proof of this simple (and well known i guess?) fact, i just need a reference to quickly point to in a note, to avoid making it cumbersome. Unless there is a one-line proof that i missed. Thanks for the attention!

Extensions obtained adding torsion points of an elliptic curve

2011-07-04T08:37:55Z

When adding to the rational the $p$-torsion points $E[p]$ of an elliptic curve we obtain an extension containing the $p$-th roots of the unity, and whose Galois group can be embedded in $GL(2, \mathbb{F}_p)$. To what extent are such extensions coming from elliptic curves?

I mean, assume $K/\mathbb{Q}$ to be an extension whose Galois group can be embedded in $GL(2, \mathbb{F}_p)$ and containing the $p$-th roots of the unity (which is required to expect a positive answer), is $K$ obtained adding to $\mathbb{Q}$ the torsion points of some elliptic curve defined over the rationals?

Note that I'm not considering a particular Galois representation, but just Galois groups that can be embedded in some way into $GL(2, \mathbb{F}_p)$. Thanks!

Answer by Maurizio Monge for P-adic Arithmetic Software

2011-06-30T12:43:17Z

This is just a very partial answer that is based on my experience trying to do some work with extension of p-adic numbers.

There is p-adic arithmetic in the free software programs SAGE, PARI and GAP, but their main limitation (i don't know if things have changed recently) was their inability to with relative extension, that is extensions of another field which is itself a proper extension of $\mathbb{Q}_p$. However if you have limited need for extensions they (mostly SAGE and PARI, because GAP is more group-theory oriented) have a very good interface to work with p-adic numbers.

For my thesis work i absolutely needed relative extensions, so i had to use Magma which is not free, but for small computation it can be used online: http://magma.maths.usyd.edu.au/calc/. It has a very good library, which requires some time to learn, like the magma language which requires some learning too, but however the functionality provided is worth the effort.

Sum of products of p-th powers of roots of 1 and monomial symmetric functions

2011-06-12T20:16:15Z

Hello mathematicians, i'm looking for explicit computations of expressions like $$ \sum_{\substack{0\leq i,j,k<n\\i\neq j\neq k \neq i}}\zeta_n^{ip^{k_1}+jp^{k_2}+kp^{k_3}} $$ and its generalizations, where $p$ is a prime, $n$ an integer (not assumed prime with $p$) and $\zeta_n$ is a primitive $n$-th root of the unity, and the sum is over all distinct $i,j,k$ from $0$ to $n-1$. The above expression is also, up to a factor, equal to the Schur polynomial $s_\lambda$ (of algebraic combinatoric fame) with partition $\lambda=(p^{k_1},p^{k_2},p^{k_3})$ evaluated in $1,\zeta_n,\zeta_n^2,\dots,\zeta_n^{n-1}$ .(this is not true, but as was pointed out by Emmanuel the above sum is rather a monomial symmetric function $m_\lambda$)

While i just need the expression for small sums (up to $3$ or $4$ terms) which can also be easily worked out by hand (writing the sum without the condition $i\neq j\neq k\neq i$ and using inclusion-exclusion on the set of indices such that $i=j$, $i=k$, etc), i'm almost sure that this computation has been done before and that it i may just refer to some (more or less) well known result. Thanks for your attention!

Maximal (non-abelian) extensions of number fields unramified everywhere

2011-01-27T19:42:34Z

Hello!

Let $K$ be a number field. All abelian unramified extensions are contained in the Hilbert class field which is a finite extension 'maximal' with respect to this property. For general unramified extensions, is there a bound (depending on $K$) on the degree of an unramified extension over $K$? If so, does the compositum of all unramified extensions also have finite degree over $K$ in general?

Thanks for your attention!

ADDENDUM: as Hunter noticed the answer can be no even just even for solvable groups, when the field admits an infinite class field tower. But perhaps it is still interesting to study the question for extension having simple Galois group, and possibly their compositum. Is there anything known about this case?

Equidistribution in the unit interval of numbers in a real field with bounded Mahler measure

2011-02-23T11:15:04Z

Let $K$ be a real number field, together with a fixed immersion in $\mathbb{R}$, and for each positive real number $M$ consider the set $S_M(K)$ of elements in $K \cap [0,1]$ having Mahler measure smaller than $M$.

When $K = \mathbb{Q}$, the $S_M(K)$ are a Farey sequence, and for $M \rightarrow \infty$ they have been proved to become equidistribuited by Neville ("The Structure of Farey Series", Proceedings of the London Mathematical Society, 1949), with a geometrical method with does not seem to generalize easily. On the other hand, when $K$ is a real quadratic extension of $\mathbb{Q}$ come computational experiment seems to show that the $S_M(K)$ are still uniformly distributed in the unit interval.

Is any there anything known about this problem, or possibly about connected issues, other than Neville's result? Thanks!

Answer by Maurizio Monge for p-subgroups of automorphisms of $p$ groups

2011-02-11T14:23:51Z

Elaborating on Someone's comment, we have that the answer to the question is the following:

Let's consider the semidirect product $H := G\rtimes P$, having $G$ as normal subgroup and where the elements $u \in P$ act on $G$ via the given action, i.e. in the semidirect product we have as $u \cdot x \cdot u^{-1} = u(x)$. Consider now the filtration defined by $$ H_0 := G,\qquad H_{i+1} = [H_i,H]\quad\text{ for }i>0. $$ The $H_i$ are normal in $H$, $H_{i+1} \subsetneq H_i$ and $H_i/H_{i+1}$ is precisely the biggest quotient of $H_i$ on which $H$ acts trivially by conjugation. Since in particular also $G$ acts trivially all the intermediate subgroups are normal in $G$. Consequently refining the chain of the $H_0 \supset H_1 \supset \dots$ so that all the factors are cyclic we obtain the requested composition series.

Best way to introduce the Chinese Remainder Theorem (to a high school student)

2010-10-20T16:25:23Z

What do you think to be the most effective way to teach the Chinese remainder theorem to a smart high school student, which is supposed to only have a soft idea about how modular arithmetic works, and where by effective i mean "putting someone in the best position to use a mathematical result as a tool for solving problems"? Please note that the answer "it should come with a better understanding of modular arithmetic" is not a valid answer (even if i support this point of view actually), because that is something that is going come with undergraduate studies (or a really smart high school student). I'm not asking about a magic wand, just some suggestion about what you consider the most insightful examples or effective ideas. Please note that while being an obvious result for those that are used to it, this theorem requires some examples and exercises to be grasped by a beginner. Thanks!

Answer by Maurizio Monge for Generalizing a problem to make it easier

2010-09-26T15:13:44Z

A very simple example of this phenomenon is also given by the following problem: prove that the maximum determinant of $n\times n$ matrices that have entries in the set $\{-1,+1\}$ is divisible by $2^{n-1}$. In fact, it is much easer proving by induction that all matrices with coefficients in $\{-1,+1\}$ have determinant divisible by $2^{n-1}$, because you can use induction and just say that changing a $-1$ to a $+1$ will change the determinant by the amount $2 * 2^{n-2}k$, by row expansion, and you can do so till when you get a matrix will all entries $1$, which has determinant $0$ when $n>1$. Note that it is not clear how you could use induction to prove the weaker statement about the maximum determinant.

Actually when a statement can be proved by induction it happens quite often that the correct statement that "makes induction work" is a somewhat generalized version of the result to be proved.

Answer by Maurizio Monge for Jokes in the sense of Littlewood: examples?

2010-09-25T16:56:26Z

For consecutive Farey fractions $\frac{a}{b}, \frac{c}{d}$ the mediant is obtained via a "simple's man addition": $$ \frac{p}{q} = \frac{a+c}{b+d} $$ which since $\frac{a}{b},\frac{c}{d}$ are consecutive if and only if $det\begin{pmatrix}a & c\\ b & d\end{pmatrix} = 1$ also turns out to be the rule of invariance of the determinant when you add a column to another column.

Classification of $p$-groups of order $p^n$ with rank $n-1$

2010-09-24T18:00:09Z

Hello, i've been looking for a way to classify the non-trivial $p$-groups $G$ that live in an exact sequence of the form $ 0 \rightarrow \mathbb{Z}/p\mathbb{Z} \rightarrow G \rightarrow (\mathbb{Z}/p\mathbb{Z})^{n-1} \rightarrow 0 $. Was this question settled before? Or is there any explicit computation of $H^2((\mathbb{Z}/p\mathbb{Z})^{n-1}, \mathbb{Z}/p\mathbb{Z})$? Thanks!

Comment by Maurizio Monge

2012-05-16T21:45:25Z

Another way to see immediately that the equation has a solution is via Newton's polygon: it is formed by two sides with slopes 0 and 1, and each corresponds to a non-trivial factor, having degree 1.

Comment by Maurizio Monge

2012-03-27T13:14:39Z

Looks like homework?

Comment by Maurizio Monge

2012-01-09T22:20:23Z

Perhaps their presentation is not the best way to understand them, in any case.

Comment by Maurizio Monge

2012-01-09T19:36:29Z

Groups such that every elements has roots of all orders and are f.g.? I'd really like to see one.

Comment by Maurizio Monge

2012-01-09T14:57:32Z

Thanks for the details, I would not ask you to translate into GAP something you already know how to do in Magma, which could be just as good. It's interesting to know how to do this in GAP, nevertheless.

Comment by Maurizio Monge

2012-01-06T23:43:26Z

Dear Derek, thanks a lot for your reply. Do you know for any reference that I may use to better understand these permutation groups? And did you do your computation by hand or by some computer algebra program? (I didn't find an obvious way to view Sp_2m(2) as permutation group on quadratic forms in GAP)

Comment by Maurizio Monge

2012-01-06T17:13:16Z

I don't want to sound like the (annoying) open-source guy, but I guess you also know GAP, right? I happened to use both actually, and for group theory I found GAP to be danm good, so in any case you may keep in mind that it exists, in case things shouldn't work with magma :)

Comment by Maurizio Monge

2012-01-05T19:43:25Z

I think that I'm missing something... in $S\wr{}G=S^{|G|}\rtimes{}G$, isn't the subgroup of $S^{|G|}$ formed by vectors with equal components a normal subgroup? In the case of $S$ elementary abelian the normal subgroups are exactly the sub-representations, why in the case that $S$ is non-abelian there should be no invariant subgroup? If $S$ is replaced by $S^{m_n}$, furthermore, all elements of $(S^{m_n})^{|G|}$ tuples with the first component equal (of the $m_n$) appears to be invariant under $G$.

Comment by Maurizio Monge

2011-12-19T00:16:52Z

Nice answer. And I should have known, as (if I remember correctly) the counterexamples pointed out occur as solvable subgroups of $GL(n,\mathbb{F}_q)$.

Comment by Maurizio Monge

2011-12-17T14:40:25Z

I can just tell you how, sometimes, you can feel while having to do with people who has a deep understanding of mathematics: youtube.com/watch?v=5blbv4WFriM

Comment by Maurizio Monge

2011-12-12T23:45:37Z

so either the systems collapses, either we have proved that the authors are not model theorists.

Comment by Maurizio Monge

2011-11-19T15:48:29Z

Perhaps Hensel lemma can provide a quite elementary application. You can explaing to anybody what Newton's root-finding method is, and even display Newton's fractal created from the convergence data to a root over the complex numbers. So while p-adic numbers give you a system of numbers which is in some sense analogous to real numbers, the existence of a root can be ensured quite easily.

Comment by Maurizio Monge

2011-11-01T16:08:36Z

By the way, I was noticing that in Bartel's proof $C$ is not required to be cyclic.

Comment by Maurizio Monge

2011-10-31T21:04:53Z

One last question: is it possible to see with the same methods that $G$ in $V\rtimes{}G$ is the unique complement, up to conjugacy?

Comment by Maurizio Monge

2011-10-31T15:18:46Z

In any case thank you very much for illustrating how group cohomology can be used to prove this.