User salvatore siciliano - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T10:42:53Z http://mathoverflow.net/feeds/user/14653 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/129161/quotient-of-lie-rings-and-quotient-of-lie-groups/129186#129186 Answer by Salvatore Siciliano for Quotient of Lie rings and quotient of Lie groups! Salvatore Siciliano 2013-04-30T08:04:55Z 2013-04-30T08:04:55Z <p>The question is not posed in a clear way but, if I am interpreting correctly, it is enough to recall that the universal enveloping algebra $U(g/I)$ of the Lie algebra quotient $g/I$ is isomorphic to $U(g)/B$, where $B$ is the two-sided ideal of $U(g)$ generated by $I$. (Of course, here $g$ is identified with its isomorphic image in $U(g)$.) This is a rather elementary fact: see e.g. Theorem 1 in Chapter 5 of the book "N. Jacobson: Lie algebras". </p> http://mathoverflow.net/questions/126379/reference-needed/126382#126382 Answer by Salvatore Siciliano for Reference needed Salvatore Siciliano 2013-04-03T12:51:22Z 2013-04-03T13:26:33Z <p>According with MathSciNet and Zentralblatt, it seems that there is no author's corrigendum for this paper. However, a corrected version of his results can be found in the following paper:</p> <p>M. Lewis - J. Riedl: Affine semi-linear groups with three irreducible character degrees, J. Algebra 246 (2001), no. 2, 708–720.</p> http://mathoverflow.net/questions/125921/the-number-of-cyclic-subgroups/125926#125926 Answer by Salvatore Siciliano for The number of cyclic subgroups Salvatore Siciliano 2013-03-29T16:17:24Z 2013-03-29T16:31:14Z <p>A counterexample to your claim: The symmetric group $S_3$ contains a unique cyclic subgroup of order 3 (the alternating group $A_3$), however 1 is not a multiple of 2 (the greatest divisor of $|S_3|=6$ that is relative prime to 3). </p> http://mathoverflow.net/questions/125817/mathematical-paper-that-just-links-two-different-fields-of-sciences/125841#125841 Answer by Salvatore Siciliano for Mathematical Paper That Just Links Two Different Fields of Sciences Salvatore Siciliano 2013-03-28T16:08:11Z 2013-03-28T16:17:28Z <p>Further journals devoted to papers linking mathematics with other scientific areas are the following:</p> <p>Mathematical medicine and biology;</p> <p>Bulletin of mathematical biology;</p> <p>Computers and mathematics with applications;</p> <p>Journal of mathematical imaging and vision;</p> <p>Mathematical geology;</p> <p>Mathematical geosciences.</p> http://mathoverflow.net/questions/124393/name-for-ideal-generated-by-lie-subalgebra/124415#124415 Answer by Salvatore Siciliano for Name for ideal generated by Lie subalgebra Salvatore Siciliano 2013-03-13T13:40:15Z 2013-03-13T13:58:22Z <p>In analogy with the notion of normal closure in group theory, the smallest ideal of ${\mathfrak g}$ containing the subalgebra ${\mathfrak m}$ is indeed called the ideal closure of ${\mathfrak m}$ and denoted by ${\mathfrak m}^{\mathfrak g}$. This terminology is rather diffused in the literature: for instance, you can find it in many papers by Amayo, Stewart, Towers, etc. </p> http://mathoverflow.net/questions/115471/a-certain-theorem-about-finite-dimensional-lie-algebras-over-an-algebraically-clo/115487#115487 Answer by Salvatore Siciliano for A certain theorem about finite-dimensional Lie algebras over an algebraically closed field with zero characteristic. Salvatore Siciliano 2012-12-05T09:38:36Z 2012-12-07T15:49:13Z <p>Condition (i) can be removed, as already observed by Daniel. In positive characteristic you can find a counterexample in the book "J. Humphreys: Introduction to Lie algebras and representation theory" (Chapter 2, Section 4, page 20, Exercise 4), so condition (ii) cannot be relaxed. Finally, let $H={\mathbb F}x+{\mathbb F}y+{\mathbb F}z$ be the Heisenberg algebra with basis $x,y,z$, where $z$ is central in $H$ and $[x,y]=z$. Consider the ring of polynomials ${\mathbb F}[t]$ as a left $H$-module with $x$ acting as $d/dt$, $y$ acting by multiplication by $t$, and $z$ acting as the identity. Now consider the split extension $L=H\ltimes {\mathbb F}[t]$. Then $L$ is solvable of derived length 3, but the derived subalgebra $[L,L]={\mathbb F}z+{\mathbb F}[t]$ is not nilpotent. Thus condition (iii) cannot be removed.</p> http://mathoverflow.net/questions/115170/groups-with-an-automorphism-of-order-two-fixing-only-two-elements/115188#115188 Answer by Salvatore Siciliano for Groups with an automorphism of order two fixing only two elements Salvatore Siciliano 2012-12-02T17:08:50Z 2012-12-02T17:08:50Z <p>MacKay [On the structure of a special class of $p$-groups, Quart. J. Math. Oxford Ser (2) 38, 489-502] and, indipendently, Kiming [Structure and derived length of finite $p$-groups possessing an automorphism of $p$-power order having exactly $p$ fixed points, Math. Scand. 62, 153-172] showed that if a finite $p$-group $G$ admits an automorphism of order $p^n$ with exactly $p$ fixed points, then $G$ contains a subgroup $H$ of index bounded by a function of $p$ and $n$ which is nilpotent of class at most 2 (and $H$ is abelian if $p=2$).</p> http://mathoverflow.net/questions/112715/why-when-classification-of-simple-objects-is-simple-e-g-unknown-classifica/114082#114082 Answer by Salvatore Siciliano for Why/when classification of simple objects is "simple" ? E.g. (unknown) classification of simple Lie algebras in char =2,3... Salvatore Siciliano 2012-11-21T17:34:10Z 2012-11-21T17:34:10Z <p>According with the introduction of Strade's book "Simple Lie algebras over fields of positive characteristic. Structure Theory", it seems that a possible list of known finite-dimensional simple Lie algebras over algebraically closed fields of characteristic 3 could be close to complete. A discussion on this topics can be found in Section 4.4 of that book. On the other hand, in characteristic 2 the situation seems to be more complicated. For example, in the paper [Yu. Kochetov - D. Leites: Simple Lie algebras of characteristic 2 recovered from superalgebras and on the notion of a simple group, in Proceedings of the International Algebraic Conference in the Memory of A.I. Malcev, Novosibirsk, Contemp. Math. 131 (1992), 59-67] the authors have constructed simple Lie algebras in characteristic 2 from superalgebras. Thus one expects that a greater variety of constructions could get many more examples in this exceptional characteristic. </p> http://mathoverflow.net/questions/103522/nilpotent-lie-algebras/103632#103632 Answer by Salvatore Siciliano for Nilpotent Lie Algebras Salvatore Siciliano 2012-07-31T20:17:37Z 2012-08-01T13:28:10Z <p>Of course, the easiest case is when $\mathfrak{g}$ has nilpotence class 2 (that is, $[[\mathfrak{g},\mathfrak{g}],\mathfrak{g}]=0$). Under this assumption one has trivially that $im(ad_{\xi})⊆Z(\mathfrak{g})$, and $im(ad_{\xi}$) contains a 1-dimensional ideal whenever $\xi$ is not in $Z(\mathfrak{g})$.</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99805#99805 Answer by Salvatore Siciliano for Blackbox Theorems Salvatore Siciliano 2012-06-16T20:41:12Z 2012-06-16T20:41:12Z <p>The Feit–Thompson Theorem stating that every finite group of odd order is solvable.</p> http://mathoverflow.net/questions/96850/enveloping-algebras-which-are-solvable-as-lie-algebras/96851#96851 Answer by Salvatore Siciliano for Enveloping algebras which are solvable as Lie algebras Salvatore Siciliano 2012-05-13T20:43:58Z 2012-05-13T20:43:58Z <p>If $F$ has characteristic different from 2 the answer is yes. This follows from Corollary 6.1 in the paper by D. Riley - A.Shalev: The Lie structure of enveloping algebras, J. Algebra 162, 46-61 (1993).</p> <p>On the other hand, in characteristic 2 this conclusion is false. For instance, if $L$ is a 2-dimensional nonabelian Lie algebra or a 3-dimensional Heisenberg algebra, then one can see by explicite calculations that $U(L)$ is Lie solvable of derived length 3. </p> http://mathoverflow.net/questions/90831/maximal-dimension-of-abelian-ideals-of-a-lie-algebra-and-extensions-of-the-ground Maximal dimension of abelian ideals of a Lie algebra and extensions of the ground field Salvatore Siciliano 2012-03-10T16:42:52Z 2012-05-11T21:09:39Z <p>For a Lie algebra $L$ of dimension $n$ over a field ${\mathbb F}$ we denote by $\beta(L)$ the maximal dimension of abelian ideals of $L$. In general, $\beta(L)$ is not preserved under extensions of the ground field (see e.g. Example 2.7 in <a href="http://homepage.univie.ac.at/dietrich.burde/papers/burde_39_max_ab.pdf" rel="nofollow">http://homepage.univie.ac.at/dietrich.burde/papers/burde_39_max_ab.pdf</a>). Do you know any example in which $\beta(L)&lt;\beta(L\otimes_{\mathbb F} \bar{{\mathbb F}})=n-1$, where $\bar{\mathbb F}$ is the algebraic closure of ${\mathbb F}$? (In other words, is it possible that $L\otimes_{\mathbb F} \bar{{\mathbb F}}$ contains an abelian ideal of codimension 1 and $L$ has no abelian ideal of codimension 1?)</p> <p>I am mainly interested in the case where $L$ is a restricted Lie algebra over a field of characteristic $p>0$. </p> http://mathoverflow.net/questions/95865/examples-of-conjectures-that-were-widely-believed-to-be-true-but-later-proved-fal/95934#95934 Answer by Salvatore Siciliano for Examples of conjectures that were widely believed to be true but later proved false Salvatore Siciliano 2012-05-03T23:27:59Z 2012-05-03T23:27:59Z <p>The solution in negative of the isomorphism problem for integral group rings. A counterexample was found by Martin Hertweck: </p> <p><a href="http://www.jstor.org/discover/10.2307/3062112?uid=3738296&amp;uid=2129&amp;uid=2&amp;uid=70&amp;uid=4&amp;sid=56137438593" rel="nofollow">http://www.jstor.org/discover/10.2307/3062112?uid=3738296&amp;uid=2129&amp;uid=2&amp;uid=70&amp;uid=4&amp;sid=56137438593</a></p> http://mathoverflow.net/questions/90045/schur-multipliers-for-lie-algebras/90054#90054 Answer by Salvatore Siciliano for Schur `multipliers' for Lie algebras Salvatore Siciliano 2012-03-02T16:07:28Z 2012-03-02T16:07:28Z <p>Take a look at the Ph.D. Thesis of P.G. Batten:</p> <p><a href="http://www4.ncsu.edu/~stitz/Multipliers%20and%20Covers%20of%20Lie%20Algebras.pdf" rel="nofollow">http://www4.ncsu.edu/~stitz/Multipliers%20and%20Covers%20of%20Lie%20Algebras.pdf</a></p> http://mathoverflow.net/questions/89785/borel-weil-theorem-references/89789#89789 Answer by Salvatore Siciliano for Borel-Weil Theorem-References Salvatore Siciliano 2012-02-28T19:29:34Z 2012-02-28T19:29:34Z <p>J.P. Serre: "Représentations linéaires et espaces homogènes kählériens des groupes de Lie compacts (d'après Armand Borel et André Weil)", Séminaire Bourbaki (Paris: Soc. Math. France) 2 (100), 1995, 447–454. </p> <p>J. Tits: "Sur certaines classes d'espaces homogènes de groupes de Lie, Acad. Roy. Belg. Cl. Sci. Mém. Coll. 29 (1995).</p> <p>M. Sepanski: Compact Lie groups., Graduate Texts in Mathematics, 235, New York, Springer, 1995. (Theorem 7.58).</p> http://mathoverflow.net/questions/88103/comparables-to-journal-of-algebra-journal-of-pure-and-applied-algebra/88109#88109 Answer by Salvatore Siciliano for Comparables to Journal of Algebra, Journal of Pure and Applied Algebra Salvatore Siciliano 2012-02-10T15:18:12Z 2012-02-10T15:18:12Z <p>In my opinion the question is somewhat vague. However, I think that Algebra and Representation Theory or Algebra and Number Theory are examples of specialized journals in algebra of level more or less similar to J. Algebra or Journal of Pure and Applied Algebra. </p> http://mathoverflow.net/questions/79667/when-is-mathbbg-mr-enough-to-determine-r/79701#79701 Answer by Salvatore Siciliano for When is $\mathbb{G}_m(R)$ enough to determine $R$? Salvatore Siciliano 2011-11-01T11:29:19Z 2011-11-01T14:29:19Z <p>It is also worth to recall that if $F$ is a field then a free algebra $F[X]$ has $F^\times$ as group of units for any set $X$...</p> <p>Anyway, there is an extensive literature devoted to study rings with a fixed group if units. A sample of this is the paper:</p> <p>I. Stewart: Finite rings with a specified group of units, Math. Z. 126 (1972), 51-58.</p> http://mathoverflow.net/questions/79601/groups-whose-centralisers-are-finite/79637#79637 Answer by Salvatore Siciliano for Groups whose centralisers are finite Salvatore Siciliano 2011-10-31T14:40:57Z 2011-10-31T14:40:57Z <p>In the opposite direction, B. Hartley and M. Kuzucuoglu proved that in an infinite locally finite simple group the centralizer of every element is infinite. See Theorem A2 of the following paper:</p> <p>B. Hartley - M. Kuzucuoglu: “Centralizers of Elements in Locally Finite Simple Groups”, Proc. London Math. Soc. 62 (1991), 301-324.</p> http://mathoverflow.net/questions/79417/existence-of-cartan-subalgebra/79457#79457 Answer by Salvatore Siciliano for Existence of Cartan subalgebra Salvatore Siciliano 2011-10-29T09:52:50Z 2011-10-30T18:15:23Z <p>Let me just add some remarks. In general, if $L$ is a finite dimensional Lie algebra over an arbitrary field $F$ then a subalgebra $H$ of $L$ is called a Cartan subalgebra if $H$ is nilpotent and self-normalising in $L$. If $L$ is semisimple and $F$ has characteristic zero (as in the case asked by the OP) then the Cartan subalgebras of $L$ are precisely the maximal tori of $L$. (A torus of $L$ is an abelian subalgebra consisting of semisimple elements). Note that the existence of a Cartan subalgebra is always assured whenever the ground field has more than $\dim_F L$ elements. In particular, finite dimensional Lie algebras over infinite field always have Cartan subalgebras. Moreover, the Cartan subalgebras coincides with the minimal Engel subalgebras of $L$. (A subalgebra of $L$ is called an Engel subalgebra if it is the null Fitting component of $L$ with respect to $ad x$ for some $x\in L$.) See the paper </p> <p>R.E. Barnes: On Cartan Subalgebras of Lie Algebras, Math. Z. 101 (1967), 350-355.</p> <p>On the other hand, the existence of Cartan subalgebras of Lie algebras defined over small fields still remains an OPEN problem. </p> <p>It is also worth to mention that solvable Lie algebras always have Cartan subalgebras.</p> <p>Finally, if $L$ is a finite dimensional restricted Lie algebra over a field of characteristic $p>0$, then $H$ is a Cartan subalgebra of $L$ if and only if is the centralizer of a maximal torus of $L$. (Here a torus is an abelian subalgebra consisting of semisimple elements; an element $x$ of $L$ is semisimple it $x$ is in the restricted subalgebra generated by $x^{[p]}$). </p> http://mathoverflow.net/questions/78145/examples-of-finite-dimensional-non-simple-non-abelian-lie-algebras/79070#79070 Answer by Salvatore Siciliano for Examples of finite dimensional non simple non abelian Lie algebras Salvatore Siciliano 2011-10-25T12:35:27Z 2011-10-25T12:35:27Z <p>I think it is also a good idea to take a look at the following papers by Willem de Graaf et al. about nilpotent and solvable Lie algebras of small dimension over arbitrary fields:</p> <p><a href="http://arxiv.org/abs/math/0404071" rel="nofollow">http://arxiv.org/abs/math/0404071</a></p> <p><a href="http://arxiv.org/abs/math/0511668" rel="nofollow">http://arxiv.org/abs/math/0511668</a></p> <p><a href="http://arxiv.org/abs/1011.0361" rel="nofollow">http://arxiv.org/abs/1011.0361</a> </p> http://mathoverflow.net/questions/75192/ring-with-z-as-its-group-of-units/75209#75209 Answer by Salvatore Siciliano for Ring with Z as its group of units? Salvatore Siciliano 2011-09-12T10:15:43Z 2011-09-12T10:20:52Z <p>The example provided by Noam answers the first question. The second question is very old and, indeed, too general. See e.g. the notes to Chapter XVIII (page 324) of the book "László Fuchs: Pure and applied mathematics, Volume 2; Volume 36". In particular, rings with cyclic groups of units have been studied by RW Gilmer [Finite rings having a cyclic multiplicative group of units, Amer. J. Math 85 (1963), 447-452], by K. E. Eldridge, I. Fischer [D.C.C. rings with a cyclic group of units, Duke Math. J. 34 (1967), 243-248] and by KR Pearson and JE Schneider [J. Algebra 16 (1970) 243-251]. </p> http://mathoverflow.net/questions/73856/when-a-group-ring-is-a-local-ring/73859#73859 Answer by Salvatore Siciliano for When a group ring is a local ring Salvatore Siciliano 2011-08-27T18:27:05Z 2011-08-27T19:34:10Z <p>Dear Marco, it is well-known that if $F$ is a field of characteristic $p>0$ and $G$ is a group then the augmentation ideal of the group algebra $FG$ is nilpotent if and only if $G$ is a finite $p$-group (see the book "D. Passman: The algebraic structure of group rings", Lemma 1.6 of Chapter 3, page 70). In that case the Jacobson radical $J$ of $FG$ clearly coincides with the augmentation ideal which has codimension $1$ in $FG$, and you are done. You can also find the description of the Jacobson radical of the group algebra of a finite $p$-group over a field of characteristic $p>0$ in the book "R. Pierce: Associative algebras" (Corollary in Section 4.7).</p> http://mathoverflow.net/questions/73564/when-can-we-lift-an-action-of-lie-algebra/73582#73582 Answer by Salvatore Siciliano for when can we lift an action of Lie algebra? Salvatore Siciliano 2011-08-24T16:14:02Z 2011-08-24T16:14:02Z <p>For a reference about the well-known fact that finite-dimensional representations of a connected and simply connected Lie group are in one-to-one correnspondence with finite-dimensional representations of its Lie algebra, the OP is referred e.g. to "Fulton-Harris: Representation Theory", Section 8.1.</p> http://mathoverflow.net/questions/73261/hopf-algebras-and-quantum-groups/73278#73278 Answer by Salvatore Siciliano for Hopf Algebras and Quantum Groups Salvatore Siciliano 2011-08-20T13:42:28Z 2011-08-20T13:42:28Z <p>I think it is also worth to mention the book "S. Dascalescu: Hopf algebras. An introduction" as a suitable textbook on the algebraic theory of Hopf algebras. However, since it is not dealing with quantum groups, it could be timely to use it together with some books mentioned by MTS on this subject. </p> http://mathoverflow.net/questions/72419/a-good-book-of-functional-analysis/72422#72422 Answer by Salvatore Siciliano for A good book of functional analysis Salvatore Siciliano 2011-08-09T01:47:10Z 2011-08-09T01:47:10Z <p>I am an algebraist and not an analyst, however my favourite book on this area is "Walter Rudin: Functional Analysis".</p> http://mathoverflow.net/questions/72233/restricted-lie-algebras-of-low-dimension Restricted Lie algebras of low dimension Salvatore Siciliano 2011-08-06T11:03:17Z 2011-08-08T21:12:13Z <p>Over the decades there has been a lot of papers devoted to the classification of Lie algebras of low dimension. Do you know any paper dealing with the problem of determining (up to restricted isomorphisms) restricted Lie algebras $(L,[p])$ of low dimension over a field of characteristic $p>0$? </p> http://mathoverflow.net/questions/72098/lie-locally-nilpotent-associative-algebras Lie locally nilpotent associative algebras Salvatore Siciliano 2011-08-04T16:41:07Z 2011-08-07T15:12:23Z <p>Let $A$ be an associative algebra over a field. Then $A$ can be regarded as a Lie algebra via the Lie bracket defined by $[a,b]=ab-ba$ for every $a,b\in A$. The algebra $A$ is called Lie locally nilpotent if it is locally nilpotent as a Lie algebra. Also, $A$ is said to be locally Lie nilpotent if every finitely generated associative subalgebra of $A$ is nilpotent as a Lie algebra. Clearly, if $A$ is locally Lie nilpotent then it is Lie locally nilpotent. Is the converse true?</p> http://mathoverflow.net/questions/64463/when-is-a-restricted-enveloping-algebra-a-domain-a-finitely-generated-domain/68846#68846 Answer by Salvatore Siciliano for When is a restricted enveloping algebra a domain? A finitely generated domain? Salvatore Siciliano 2011-06-26T10:00:53Z 2011-06-26T10:00:53Z <p>Let $g$ be a restricted Lie algebra over a field of positive characteristic $p$. It is not difficult to see that a necessary condition such that the restricted enveloping algebra $u(g)$ of $g$ is a domain is that $g$ has no nonzero $p$-algebraic elements. (An element $x \in g$ is said to be $p$-algebraic if the restricted subalgebra generated by $x$ is finite-dimensional.) The converse of this property is a well-known open question posed by V. Petrogradsky (see "DNIESTER NOTEBOOK: Unsolved Problems in the Theory of Rings and Modules", Problem 3.59). Apparently, this is the Lie theoretical analog of the Kaplansky Problem about zero-divisors of group algebras of torsion-free groups. </p> http://mathoverflow.net/questions/68667/is-there-an-analog-of-clifford-theorem-in-the-setting-of-lie-algebras Is there an analog of Clifford Theorem in the setting of Lie algebras? Salvatore Siciliano 2011-06-23T22:06:05Z 2011-06-24T17:40:20Z <p>A classical theorem of Clifford states that if G is a finite group and K a field, then every irreducible right KG-module is a completely reducible right KN-module, where N is any normal subgroup of G. Is there a Lie theoretic analog of this result? That is, if L is a finite-dimensional Lie algebra, I an ideal of L, and M an irreducible L-module, is M a completely reducible I-module? I expect the answer is negative, but what about a counterexample? </p> http://mathoverflow.net/questions/60447/how-does-the-group-algebra-look-as-a-lie-algebra/62874#62874 Answer by Salvatore Siciliano for How does the group algebra look as a Lie algebra Salvatore Siciliano 2011-04-25T00:02:46Z 2011-04-25T00:02:46Z <p>A full characterization of group algebras which are solvable or nilpotent as Lie algebras can be found in the paper [I.B.S. Passi - D. Passman - S.K. Sehgal: Lie solvable group rings, Can. J. Math. 25 (1973), 748-757].</p> http://mathoverflow.net/questions/130096/linear-algebra-ii Comment by Salvatore Siciliano Salvatore Siciliano 2013-05-08T16:46:05Z 2013-05-08T16:46:05Z This is not a research level question: you are referred to any respectable book of linear algebra... http://mathoverflow.net/questions/124617/logs-and-floors Comment by Salvatore Siciliano Salvatore Siciliano 2013-03-15T15:14:47Z 2013-03-15T15:14:47Z There is something wrong in the mentioned equation...By assuming the base is 2, for n=2 you will have a counterexample! Of course, for other bases the conclusion is analogous... http://mathoverflow.net/questions/116764/reference-finite-p-groups/116783#116783 Comment by Salvatore Siciliano Salvatore Siciliano 2012-12-19T15:54:50Z 2012-12-19T15:54:50Z Dear Professor Mann, welcome on MathOverflow! http://mathoverflow.net/questions/115471/a-certain-theorem-about-finite-dimensional-lie-algebras-over-an-algebraically-clo/115650#115650 Comment by Salvatore Siciliano Salvatore Siciliano 2012-12-08T15:21:41Z 2012-12-08T15:21:41Z @ Leonard: I do not entirely agree with your Update and, indeed, it is rather usual to present Lie algebras by structure contants (provided Jacobi identity is satified). After all, a this stage it would be an easy task to present my counterexample just specyfing the brackets on a basis... http://mathoverflow.net/questions/115471/a-certain-theorem-about-finite-dimensional-lie-algebras-over-an-algebraically-clo Comment by Salvatore Siciliano Salvatore Siciliano 2012-12-08T06:01:54Z 2012-12-08T06:01:54Z @ Leonard. What you wrote is correct. In any case, for the semidirect of a Lie algebra by a module you can look up, for instance, the book &quot;Hilton-Stammbach: A course in homological algebra&quot; (Chapter VII, Section 2)&quot; http://mathoverflow.net/questions/115471/a-certain-theorem-about-finite-dimensional-lie-algebras-over-an-algebraically-clo Comment by Salvatore Siciliano Salvatore Siciliano 2012-12-07T15:49:49Z 2012-12-07T15:49:49Z Dear Leonard, I just modified the example and put it in a &quot;more natural&quot; form (it is realized as a semidirect product of a Heisenberg algebra by an infinite dimensional left module). http://mathoverflow.net/questions/115471/a-certain-theorem-about-finite-dimensional-lie-algebras-over-an-algebraically-clo/115650#115650 Comment by Salvatore Siciliano Salvatore Siciliano 2012-12-07T15:05:19Z 2012-12-07T15:05:19Z I just quoted another counterexample http://mathoverflow.net/questions/110819/finite-subgroup Comment by Salvatore Siciliano Salvatore Siciliano 2012-10-27T22:42:02Z 2012-10-27T22:42:02Z Of course, the group mentioned by Derek is just the Prufer group. http://mathoverflow.net/questions/109234/inner-automorphisms-of-matrix-algebras Comment by Salvatore Siciliano Salvatore Siciliano 2012-10-09T15:58:10Z 2012-10-09T15:58:10Z Why specify that $\Psi$ is inner? By Skolem-Nother Theorem every automorphism of $M_4(\mathbb{R})$ is inner... http://mathoverflow.net/questions/98587/simple-lie-algebras-and-jordan-decomposition/98590#98590 Comment by Salvatore Siciliano Salvatore Siciliano 2012-06-01T18:19:51Z 2012-06-01T18:19:51Z Professor Humphreys: if $L$ is restricted and simple then the adjoint representation $L⟶ad(L)$ is a restricted isomorphism (the p-map on $ad(L)$ is the ordinary p-power of linear transformations). If $F$ is perfect then for any element $x\in L$ we have the Jordan-Chevalley decomposition $x=x_s+x_n$, where $x_s$ is a semisimple element and $x_n$ is a p-nilpotent element of $L$, and $[x_s,x_n]=0$. It is easy to see that $ad(x)=ad(x_s)+ad(x_n)$ is just the Jordan decomposition of the linear transformation $ad(x)$,so $ad(L)$ contains the semisimple and nilpotent parts of its elements in this case. http://mathoverflow.net/questions/98040/refrence-request Comment by Salvatore Siciliano Salvatore Siciliano 2012-05-26T17:35:53Z 2012-05-26T17:35:53Z Gerald Edgar is right: when a group with the required properties exists then it is just the intersection. Obviously, it is always true that $gcd(|H|,|K|)≥|H∩K|$, but equality does not necessary hold, in general, neither when the intersection is not trivial. (For instance, consider the elementary abelian $p$-group of order $p^3$ and take as $H$ and $K$ to subgroup of order $p^2$ having intersection of order $p$). http://mathoverflow.net/questions/96850/enveloping-algebras-which-are-solvable-as-lie-algebras/96851#96851 Comment by Salvatore Siciliano Salvatore Siciliano 2012-05-14T08:19:41Z 2012-05-14T08:19:41Z Polynomial identities of (ordinary and restricted) enveloping algebras have been studied by a lot of people (Latysev, Bahturin, Passman, Petrogradski, Riley, Shalev, etc.) and this topic is one of my interest research. In particular, restricted enveloping algebras which are PI were characterized by Passman and, indipendently, by Petrogradski in 1991, and in the paper mentioned in my answer the authors established when a restricted enveloping algebra is Lie nilpotent, bounded Lie Engel or Lie solvable (in odd characteristic). http://mathoverflow.net/questions/90831/maximal-dimension-of-abelian-ideals-of-a-lie-algebra-and-extensions-of-the-ground/96712#96712 Comment by Salvatore Siciliano Salvatore Siciliano 2012-05-11T22:43:39Z 2012-05-11T22:43:39Z Very nice! Even better, it seems that this argument works in the infinite dimensional case, as well! http://mathoverflow.net/questions/90831/maximal-dimension-of-abelian-ideals-of-a-lie-algebra-and-extensions-of-the-ground/96515#96515 Comment by Salvatore Siciliano Salvatore Siciliano 2012-05-10T14:41:33Z 2012-05-10T14:41:33Z I see: thank you very much for pointing this out to me. http://mathoverflow.net/questions/90831/maximal-dimension-of-abelian-ideals-of-a-lie-algebra-and-extensions-of-the-ground Comment by Salvatore Siciliano Salvatore Siciliano 2012-05-05T12:11:50Z 2012-05-05T12:11:50Z darij: I agree with you.