most recent 30 from http://mathoverflow.net 2013-05-22T16:55:00Z http://mathoverflow.net/feeds/user/29488 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/115516/a-conjecture-on-solvablity-of-finite-groups M Shahryari 2012-12-05T16:57:16Z 2012-12-13T20:15:59Z

<p>Suppose $G$ is a finite group and $A$ an abelian subgroup. Suppose for some natural number $n\geq 2$, elements of $\gamma_n(G)$ have the form $[a, x]$ where $a\in A$ and $x\in G$. Then $G$ is solvable. </p>

http://mathoverflow.net/questions/116176/groups-whose-all-normal-subgroups-are-principal M Shahryari 2012-12-12T14:19:01Z 2012-12-13T13:29:33Z

<p>My motivation for this question is from Universal Algebra: A congruence of an arbitrary algebra $A$ is said to be principal, if it is generated by a single element. In the case of rings, this is just the notion of principal ideal and for groups it is a normal subgroup which is the normal closure of a single element, more precisely:</p> <p>A normal subgroup of the form $\langle x^G\rangle$ is called a principal subgroup of the group $G$. We say that $G$ is a principal group, if every normal subgroup of $G$ is principal.</p> <p>Is there any classification of principal groups? Is there at least a classification of nilpotent (solvable) principal groups? </p> <p>The same notion can be defined for Lie algebras and also the same questions for Lie algebras arise.</p>

http://mathoverflow.net/questions/115574/ascending-chain-condition-on-ideals-of-free-products M Shahryari 2012-12-06T05:36:59Z 2012-12-11T15:14:39Z

<p>In my previous question: M Shahryari (mathoverflow.net/users/29488), Normal Subgroups of Free Products, <a href="http://mathoverflow.net/questions/114801" rel="nofollow">http://mathoverflow.net/questions/114801</a> (version: 2012-11-28), I asked if a group $A$ has max-n property, is it true that the free product $A\ast \mathbb{Z}$ has also max-n? The answer was NO in that case. Now suppose $F$ is free group of finite rank and $A$ is a group having max-n (maximal condition on normal subgroups). A normal subgroup $N$ of $A\ast F$ is called an ideal if $N\cap A=1$. Is it true that $A\ast F$ has maximal property of ideals?</p>

http://mathoverflow.net/questions/115273/sum-of-commuting-semisimple-operators M Shahryari 2012-12-03T11:24:42Z 2012-12-03T17:14:02Z

<p>Let $V$ be a finite dimensional vector space over a field $K$. An operator $T:V\to V$ is called semi-simple if every $T$-invariant subspace of $V$ has a complement(for algebraically closed fields these are exactly diagonalizable operators). Is it true that the sum (the product) of two commuting semi-simple operators is semi-simple?</p>

http://mathoverflow.net/questions/110373/sylow-theorems-for-infinite-groups/114874#114874 M Shahryari 2012-11-29T11:32:23Z 2012-11-29T11:32:23Z

<p>The best reference for this subject is the book of martyn Dixon: Locally finite groups and Sylow theory.</p>

http://mathoverflow.net/questions/114801/normal-subgroups-of-free-products M Shahryari 2012-11-28T18:27:00Z 2012-11-29T11:05:33Z

<p>Let $G=A\ast \mathbb{Z}$ be the free product of a group $A$ and the cyclic group $\mathbb{Z}$ and suppose $K$ is a subgroup of $G$. By Kurosh Subgroup Theorem we know that $K=F\ast (\ast_{i\in I}(K\cap A^{u_i}))$, where $F$ is free group and $u_i$ are some representatives of double cosets $KxA$ in $G$. Now suppose further that $A$ has ACC on normal subgroups and $K$ is normal. Is it true that $K$ is finitely generated? (this will be true if we can show that $|I|$ and $rank\ F$ are finite). </p>

http://mathoverflow.net/questions/115516/a-conjecture-on-solvablity-of-finite-groups/116319#116319 M Shahryari 2012-12-13T20:46:43Z 2012-12-13T20:46:43Z

@Andreas Caranti: yes and this point was my motivation in fact: it was proved by F. Ladish may be 4 years ago in Comm. Algebra using CFSG. I proved a version of it for Lie algebras using Cartan criterion (above mentioned paper). The above conjecture was one of my favorite problems during the past 3 years.

http://mathoverflow.net/questions/116176/groups-whose-all-normal-subgroups-are-principal/116260#116260 M Shahryari 2012-12-13T11:52:00Z 2012-12-13T11:52:00Z

@Andreas Carant: this is a nice point. Now we can conclude that if $G$ is a finite nilpotent group with all proper normal subgroups principal, then $G$ is cyclic or a $p$-group of maximal class.

http://mathoverflow.net/questions/116176/groups-whose-all-normal-subgroups-are-principal/116187#116187 M Shahryari 2012-12-13T11:45:49Z 2012-12-13T11:45:49Z

@Nick Gill: thank you about the paper.

http://mathoverflow.net/questions/116176/groups-whose-all-normal-subgroups-are-principal/116187#116187 M Shahryari 2012-12-12T19:59:24Z 2012-12-12T19:59:24Z

At least, we know that, the nilpotent and solvable cases are the most easy cases of the problem. There are many other principal groups: Simple groups, Semisimple groups and much more.

http://mathoverflow.net/questions/116176/groups-whose-all-normal-subgroups-are-principal/116187#116187 M Shahryari 2012-12-12T19:58:42Z 2012-12-12T19:58:42Z

@JSpecter and Geoff Robinson: By your comment, if $G$ is a finite nilpotent principal group, then it must be cyclic. For proof, let $G=P_1\times \cdots\times P_n$, where $P_i$'s are Sylow subgroups of $G$. Then every $P_i$ is principal (since it is characteristic) and so $P_i$ is cyclic. So $G$ is cyclic. I don't know is the same true for infinite nilpotent principal groups or not. Also for finite solvable groups, can we obtain a similar result?

http://mathoverflow.net/questions/116176/groups-whose-all-normal-subgroups-are-principal/116187#116187 M Shahryari 2012-12-12T19:50:11Z 2012-12-12T19:50:11Z

@Nick Gill: much more is true: If $G$ is principle, then the factors of its upper central series, lower central series and derived series are cyclic. There are many other immediate consequences of definition: If $G$ is principal, then so is its quotients and its characteristic subgroups. Such a group has max-$n$ property.

http://mathoverflow.net/questions/116176/groups-whose-all-normal-subgroups-are-principal/116187#116187 M Shahryari 2012-12-12T17:59:37Z 2012-12-12T17:59:37Z

@Nick Gill: please send a copy for me mshahryari@tabrizu.ac.ir thank you so much for comment.

http://mathoverflow.net/questions/115574/ascending-chain-condition-on-ideals-of-free-products/115609#115609 M Shahryari 2012-12-11T20:51:45Z 2012-12-11T20:51:45Z

@HW: Thank you so much, now I understand your argument. I will send my results to your email in next days.

http://mathoverflow.net/questions/115574/ascending-chain-condition-on-ideals-of-free-products M Shahryari 2012-12-08T06:31:21Z 2012-12-08T06:31:21Z

@HW: I know examples of equationally noetherian groups, as well as groups which are not. I just asked some details for your answer below. How I should construct the group Gn (what do mean by &quot;long small cancellation relator&quot;?) If I can understand your construction, then I can prove that for any group G, there exists a G-group which is not G-equationally noetherian.

http://mathoverflow.net/questions/115471/a-certain-theorem-about-finite-dimensional-lie-algebras-over-an-algebraically-clo/115650#115650 M Shahryari 2012-12-07T15:07:08Z 2012-12-07T15:07:08Z

You can search the key word &quot;Strongly solvable Lie algebra&quot; +&quot;David Tower&quot;.

http://mathoverflow.net/questions/115574/ascending-chain-condition-on-ideals-of-free-products/115609#115609 M Shahryari 2012-12-06T20:24:57Z 2012-12-06T20:24:57Z

@HW: I have few knowledge of small cancellation theory ( in the level of last chapter of Lyndon-Schupp). So, may I ask you to write some details?

http://mathoverflow.net/questions/115574/ascending-chain-condition-on-ideals-of-free-products M Shahryari 2012-12-06T20:19:10Z 2012-12-06T20:19:10Z

I hope the answer for my question will be negative.

http://mathoverflow.net/questions/115574/ascending-chain-condition-on-ideals-of-free-products M Shahryari 2012-12-06T20:15:32Z 2012-12-06T20:15:32Z

@HW:Thank you for comment. You are right, I'm interested in versions of Hibert's basis theorem for groups. So, I'm trying to recognize groups $A$ having max-n such that A\ast F$ has maximal property on ideals. I need time to read your question and and D. Osin's answer. Thank you again.

http://mathoverflow.net/questions/115516/a-conjecture-on-solvablity-of-finite-groups M Shahryari 2012-12-05T18:34:39Z 2012-12-05T18:34:39Z

@Dima: it is the $n$-th term of the lower central series.

http://mathoverflow.net/questions/115516/a-conjecture-on-solvablity-of-finite-groups M Shahryari 2012-12-05T17:03:30Z 2012-12-05T17:03:30Z

A similar statement is true for finite dimensional Lie algebras of characteristic zero. See my short note in Bull. Australian Math. Soc.(2011): A note on derivations of Lie algebras.