User jp - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T02:28:17Z http://mathoverflow.net/feeds/user/970 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/91151/looking-for-criterion-for-mathbbzg-modules-to-be-projective Looking for criterion for $\mathbb{Z}G$-modules to be projective jp 2012-03-14T08:05:06Z 2012-03-14T22:14:36Z <p>Given a finite group $G$ and a (finitely generated) $\mathbb{Z}G$-module $M$, assume that for each prime $p$ dividing the order $|G|$ of $G$ the $\mathbb{Z}_pG$-module $M^{\mathbb{Z}_p} = M\otimes\mathbb{Z}_p$ is projective.</p> <blockquote> <p>How can I prove that $M$ is projective?</p> </blockquote> http://mathoverflow.net/questions/70371/greatest-common-divisor-of-a2n-1-and-b2n-1/70420#70420 Answer by jp for Greatest common divisor of a^{2^n}-1 and b^{2^n}-1 jp 2011-07-15T10:51:30Z 2011-07-16T13:15:19Z <p>One can rewrite your problem as follows:</p> <p>For $p$ prime, $p\mid a^{2^n}-1$ for some $n$ is equivalent to $\mathrm{ord}_{\mathbb{F}_p^\times}(a)$ being a power of $2$.</p> <p>The probability for a random element of the multiplicative group $\mathbb{F}_p^\times$ to have order a power of $2$ is $\frac{2^n}{p-1}$ where $n$ is chosen maximal among the natural numbers $m$ with $2^m \mid p-1$.</p> <p>A naive (hopefully not too naive) heuristic for the expected number of primes dividing both $a^{2^n}-1$ and $b^{2^n}-1$ for some $n$ is $-$ assuming that both conditions are independent:</p> <p>$$\sum_{n\in\mathbb N}\sum_{\mbox{p\in\mathbb{P} : n maximal w.r.t. p = 1 \bmod 2^n}} \left(\frac{2^n}{p-1}\right)^2 \approx \sum_{n\in\mathbb N} \sum_{q\in\mathbb N} \frac{1}{\log(q\cdot 2^n+1)\cdot q^2}$$</p> <p>For the approximation the heuristics is used that the probability for a number $m$ to be prime is about $\frac{1}{\log m}$. As the latter sum diverges one would expect that infinitely many primes divide your greatest common divisor for some $n$.</p> http://mathoverflow.net/questions/49866/applications-of-finite-continued-fractions/49879#49879 Answer by jp for Applications of finite continued fractions jp 2010-12-19T13:52:56Z 2010-12-19T13:52:56Z <p>The first attacks (discovered by Michael J. Wiener) against using small private exponents in the <a href="http://en.wikipedia.org/wiki/RSA" rel="nofollow">RSA public key crypto system</a> were based on continued fractions. Better attacks are now obtained with the help of the <a href="http://en.wikipedia.org/wiki/Lenstra%E2%80%93Lenstra%E2%80%93Lov%C3%A1sz_lattice_basis_reduction_algorithm" rel="nofollow">LLL</a>-algorithm.</p> http://mathoverflow.net/questions/42986/computing-homology-of-very-large-posets/43235#43235 Answer by jp for Computing homology of very large posets jp 2010-10-22T21:13:55Z 2010-10-22T21:13:55Z <p>You wrote that upper intervals are very nice (i.e., shellable, hence spherical) in your poset, you can try to apply this slightly stronger version of <em>Quillen's Fiber Lemma</em> (compare [Q78, Proposition 7.6] and [AS92, (4.3)]):</p> <p>Let $f: X \to Y$ be a (monotone) map of posets such that the fibers $|\{f \le y\}|$ are $n$-connected for all $y \in Y$. Then $f$ is $(n+1)$-connected, i.e., for all $x \in X$ the induced maps</p> <p>$$f_{x, i}: \pi_i(|X|, x) \to \pi_i(|Y|, f(x))$$</p> <p>of homotopy groups are isomorphisms for $i \le n$ and epimorphisms for $i = n+1$.</p> <p>For $Y$ you take the dual of your poset and for $X$ the dual of the poset without the atoms. If you understand the homology of $|X|$ very well, maybe you are able to show that $f$ is null-homotopic, which implies that $|X|$ is $n$-connected and $|Y|$ is $(n+1)$-connected.</p> <p>[AS92] M. Aschbacher, Y. Segev: Locally connected simplicial maps, Israel Journal of Mathematics 77 (1992), 283-303.</p> <p>[Q78] D. Quillen: Homotopy properties of the poset of nontrivial p-subgroups of a group, Advances in Mathematics 28 (1978), 102-28.</p> http://mathoverflow.net/questions/31699/strengthening-the-induction-hypothesis/40035#40035 Answer by jp for Strengthening the Induction Hypothesis jp 2010-09-26T16:53:43Z 2010-09-26T16:53:43Z <p>A classical example is the proof of <a href="http://en.wikipedia.org/wiki/Sperner%27s_lemma" rel="nofollow">Sperner's Lemma</a> where one replaces the weaker condition of the existence of a simplex with all colors by showing that the number of such simplexes is odd.</p> http://mathoverflow.net/questions/33304/mareys-problem-generating-all-prime-numbers-in-n-1-n-2/33308#33308 Answer by jp for Marey's problem: Generating all prime numbers in $[n_1,n_2]$ jp 2010-07-25T18:20:22Z 2010-07-25T20:34:03Z <p>The fastest approach should be first to sieve the numbers by marking the numbers divisible by small primes (for this you should use only one long division per short prime), then to use a <a href="http://en.wikipedia.org/wiki/Fermat_primality_test" rel="nofollow">Fermat test</a> to the base 2 (as it is more efficient since multiplication with 2 is a left shift) on all unmarked numbers. Finally apply a certain number of <a href="http://en.wikipedia.org/wiki/Miller%E2%80%93Rabin_primality_test" rel="nofollow">Miller-Rabin tests</a> to all candidates passing the Fermat test to reduce your error probability to a level you can tolerate (e.g., $2^{-100}).</p> http://mathoverflow.net/questions/32351/doubly-transitive-groups/32384#32384 Answer by jp for Doubly-transitive groups jp 2010-07-18T18:19:17Z 2010-07-18T18:19:17Z <p>In Section 7.7 "The Finite 2-transitive Groups" of the book <em>Permutation groups</em> by John D. Dixon and Brian Mortimer, the authors describe the complete list of finite 2-transitive groups without proofs but with references.</p> <p>They list eight infinite families: the alternating, symmetric, affine and projective groups in their natural actions, as well as the less known groups of Lie type: the symplectic groups, the Suzuki groups, the unitary groups and the Ree groups. The symplectic groups have two distinct 2-transitive actions, the last three classes are 2-transitive on the sets of points in their action on appropriate Steiner systems. Additional there are 10 sporadic examples of 2-transitive groups.</p> http://mathoverflow.net/questions/18716/sylow-subgroups/19543#19543 Answer by jp for Sylow Subgroups jp 2010-03-27T20:29:37Z 2010-03-27T20:29:37Z <p>An extension of the Vipul's ideas can be found in the article (couldn't find a link to the pdf with google)</p> <p><strong>Subgroup complexes</strong> by <em>Peter Webb</em>, pp. 349-365 in: ed. P. Fong, The Arcata Conference on Representations of Finite Groups, AMS Proceedings of Symposia in Pure Mathematics 47 (1987).</p> <p>But as Mariano already commented, the analogy to the maximal unipotent subgroups of the general linear group was probably not Sylow's motivation. As commented before, he was maybe looking for maximal$p$-subgroups (i.e., maximal with respect to be a$p$-subgroup).</p> <p>This is also the leitmotif of my favorite proof of the Sylow theorems given by Michael Aschbacher in his book <strong>Finite Group Theory</strong>. It is based on Cauchy's theorem (best proved using J.H.McKay's trick to let$Z_p$act on the set of all$(x_1, \dots, x_p) \in G^p$whose product is$1$by rotating the entries) and goes essentially like this:</p> <p>The group$G$acts on the set$\mathrm{Syl}_p(G)$of its maximal$p$-subgroups by conjugation. Let$\Omega$be a (nontrivial) orbit with$S\in\Omega$. If$P$is a fixed point of the action restricted to$S$then$S$normalizes$P$and$PS=SP$is a$p$-group. Hence$P=S$by maximality of both$P$and$S$, and$S$has a <strong>unique</strong> fixed point. As$S$is a$p$-group, all its orbits have order$1$or a multiple of$p$, in particular$|\mathrm{Syl}_p(G)| = 1 \bmod p$. All orbits of$G$are disjoint unions of orbits of$S$proving$\Omega = 1 \bmod p$and$\Omega' = 0 \bmod p$for all other orbits$\Omega'$of$G$. This implies that$\Omega = \mathrm{Syl}_p(G)$, as$\Omega$was an arbitrary nontrivial orbit of$G$, showing that the action of$G$is transitive. The stabilizer of$S$in$G$is its normalizer$N_G(S)$, and as the action is transitive$|G:N_G(S)| = |\mathrm{Syl}_p(G)| = 1 \bmod p$. It remains to show that$p$does not divide$|N_G(S):S|=|N_G(S)/S|$. Otherwise, by Cauchy's theorem there exists a nontrivial$p$-subgroup of$N_G(S)/S$whose preimage under the projection$N_G(S) \to N_G(S)/S$is$p$-subgroup properly containing$S$contradicting the maximality of$S$.</p> http://mathoverflow.net/questions/1083/do-good-math-jokes-exist/8914#8914 Answer by jp for Do good math jokes exist? jp 2009-12-14T21:42:21Z 2009-12-14T21:42:21Z <p>If I remember correctly someone told me that this really happened:</p> <p>A famous mathematician gave a talk (maybe about mathematical physics), after which an as famous physicist sitting in the first row got up, and loudly declared: "That's all nice, but without mathematics, research in physics would be maybe a week behind the state it is now!"</p> <p>The famous mathematician responded: "Yes, the week god needed to create the world."</p> http://mathoverflow.net/questions/8182/is-a-polynomial-with-1-very-large-coefficient-irreducible/8211#8211 Answer by jp for Is a polynomial with 1 very large coefficient irreducible? jp 2009-12-08T17:02:51Z 2009-12-08T17:02:51Z <p>To your 2nd question:</p> <p>Taking n+1 different primes$p_0, p_1, \dots, p_n$you can define$a_i := \prod_{j \ne i} p_j$. By a theorem of Eisenstein ("Eisenstein's irreducibility criterion"), you get that any permutation yields an irreducible polynomial.</p> http://mathoverflow.net/questions/129185/nilpotent-fixed-point-free-groups-of-automorphisms Comment by jp jp 2013-05-01T07:29:22Z 2013-05-01T07:29:22Z @MarkSapir: I don't understand the last sentence of your first comment. Why do you ask about$G$abelian? http://mathoverflow.net/questions/129069/intution-behind-conditional-expectation-when-sigma-algebra-isnt-generated-by-a-p Comment by jp jp 2013-04-29T09:45:30Z 2013-04-29T09:45:30Z I'm not sure, if this is the right place to ask this question as it isn't really research level (see FAQ), please ask at math.stackexchange.com instead. [Did you read <a href="http://en.wikipedia.org/wiki/Conditional_expectation?]" rel="nofollow">en.wikipedia.org/wiki/Conditional_expectation?]</a> http://mathoverflow.net/questions/127447/wreath-product-and-matrix-presentation Comment by jp jp 2013-04-15T13:08:18Z 2013-04-15T13:08:18Z @Geoff: Yes,$r-1$not$r$. http://mathoverflow.net/questions/127447/wreath-product-and-matrix-presentation Comment by jp jp 2013-04-13T14:24:24Z 2013-04-13T14:24:24Z @Geoff: Do you mean with your first description just the$2$-Sylow of the symmetric group$S_{2^r}$? It is for finite fields$\mathbb{F}_q$with$q=3 \bmod 4$also the$2$-Sylow of$GL_{2^r}(\mathbb{F}_q)$. http://mathoverflow.net/questions/121162/math-behind-databases-management-and-sql Comment by jp jp 2013-02-18T16:07:21Z 2013-02-18T16:07:21Z You could take a look at &quot;Applied Mathematics for Database Professionals&quot; by Lex de Haan and Toon Koppelaars. I have to confess that I skipped the &quot;math part&quot; at the beginning as I was interested only in its later chapters, but it is worth checking if you can find it in a library. Do not expect interesting theorems. [Lex de Haan was according to the other author an expert on 3-valued logic, but died before finishing writing the chapter about this topic. If I recall correctly, this unfinished chapter was added as an appendix.] http://mathoverflow.net/questions/119933/a-b-cong-a-c-rightarrow-b-cong-c Comment by jp jp 2013-01-26T13:56:52Z 2013-01-26T13:56:52Z Finite hint: look at subgroups of order$4$in the dihedral group of order$8$. http://mathoverflow.net/questions/119045/is-there-a-characterization-of-groups-in-which-at-least-one-subgroup-is-not-an-en Comment by jp jp 2013-01-16T12:54:07Z 2013-01-16T12:54:07Z Can you say anything about groups that are <i>non-boring</i> for an <b>abelian</b> quotient? (like$Q_8$?) http://mathoverflow.net/questions/119045/is-there-a-characterization-of-groups-in-which-at-least-one-subgroup-is-not-an-en Comment by jp jp 2013-01-16T12:51:31Z 2013-01-16T12:51:31Z Suggestion: How about calling a group$G$<i>boring</i> if all its quotients are isomorphic to some subgroup of$G$? So you are looking for the characterization of <i>non-boring</i> groups. http://mathoverflow.net/questions/119056/homotopy-type-of-a-finite-simplicial-complex Comment by jp jp 2013-01-16T12:26:30Z 2013-01-16T12:26:30Z Adding to HW's 2nd comment: The proper place to ask question like yours is <a href="http://math.stackexchange.com/" rel="nofollow">math.stackexchange.com</a> http://mathoverflow.net/questions/118788/is-sl2-5-irreducible Comment by jp jp 2013-01-13T10:48:18Z 2013-01-13T10:48:18Z I doubt that this a research level question (see FAQ - the link is at the top). The proper place to ask it is math.stackexchange.com. Anyway, if$SL(2, 5)$is reducible, what dimension does a proper invariant subspace have? How does the stabilizer in$SL(2, q)$of the invariant subgroup look like? http://mathoverflow.net/questions/110727/on-a-theorem-of-galois/110740#110740 Comment by jp jp 2012-10-27T12:35:56Z 2012-10-27T12:35:56Z @GH: Another elementary proof of Zassenhaus' theorem (by Ulrich Meierfrankenfeld) you can find in <a href="http://www.math.msu.edu/~meier/Preprints/Frobenius/frobenius.pdf" rel="nofollow">math.msu.edu/~meier/Preprints/Frobenius/&hellip;</a> http://mathoverflow.net/questions/110609/a-simple-question-about-the-center-of-a-finite-group Comment by jp jp 2012-10-25T12:14:29Z 2012-10-25T12:14:29Z @Nick: As I totally agree with Derek that this question is not research level at all, I left it to the reader to figure out what is what. Hint:$\mathrm{GL}_{\mathbb F}(V)$has non-trivial center (and too small fields and vector spaces don't work). http://mathoverflow.net/questions/110609/a-simple-question-about-the-center-of-a-finite-group Comment by jp jp 2012-10-25T07:49:44Z 2012-10-25T07:49:44Z For a finite vector space$V$over the finite field$\mathbb F$take the semi-direct product$V \rtimes \mathop{GL}_{\mathbb F}(V)$. http://mathoverflow.net/questions/103368/automorphisms-of-quotient-groups Comment by jp jp 2012-07-29T17:32:12Z 2012-07-29T17:32:12Z Exercise 11.4 in *Derek Robinson*'s book &quot;A Course in the Theory of Groups&quot; (2nd edition) determines the kernel$K$of the map$Aut(G)\to Aut(N)\times Aut(G/N)$in case that$C:=C_G(N)$equals the center$Z(N)$of$N$: This kernel$K$is an abelian group isomorphic to the group of derivations$\mathop{Der}(G/N, C)$and$K/K\cap\mathop{Inn}(G)$is isomorphic to$H_1(G/N,C)$. http://mathoverflow.net/questions/102532/normalizers-in-symmetric-groups Comment by jp jp 2012-07-24T06:22:42Z 2012-07-24T06:22:42Z Hi Peter, welcome to MO! Do you happen to know, how to get the alternating groups for$n&gt;6\$?