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Nov
27
comment Powers of finite simple groups
Oh, I see: mathoverflow.net/a/53162/24165
Nov
23
comment Powers of finite simple groups
The automorphism groups of finite simple groups are well known. So, we have to calculate the (non-reduced) Euler function $\phi_n(G)$ (ie. the number of generating n-tuples). In Section 1.1, Collins describes a technique of such calculations that allowed Hall (in 1936) to calculate, e.g., $\phi_2(A_5)=19\cdot 120$ (ie. $h_2(A_5)=19$).
Nov
23
awarded  Custodian
Nov
23
reviewed Approve MMSE estimator expressed through cumulants
Nov
22
revised Is the Amitsur-Levitzki identity essentially unique?
an arXiv tag added
Nov
22
suggested approved edit on Is the Amitsur-Levitzki identity essentially unique?
Nov
22
awarded  Nice Answer
Nov
22
comment bounding from below the cardinality of a set of generators of the $n$-fold cartesian product group of a finite group
See (my answer to) a similar question: mathoverflow.net/q/187736/24165 .
Nov
22
revised Powers of finite simple groups
added 1 character in body
Nov
22
revised Is the Amitsur-Levitzki identity essentially unique?
deleted 2 characters in body
Nov
22
comment Is the Amitsur-Levitzki identity essentially unique?
... and why "combinatorics"?
Nov
22
comment Is the Amitsur-Levitzki identity essentially unique?
Why this is tagged "commutative algebra"?
Nov
22
revised Is the Amitsur-Levitzki identity essentially unique?
added 1 character in body
Nov
22
answered Is the Amitsur-Levitzki identity essentially unique?
Nov
22
revised Powers of finite simple groups
added 1 character in body
Nov
22
answered Powers of finite simple groups
Nov
20
comment Normal Covering of a Finite Group
This Corollary 5.5 follows immediately from the theorem of Brodie, Chamberlain and Kapp, PAMS 1988, see Nick Gill's answer: mathoverflow.net/a/185604/24165 .
Nov
9
comment Is the equational theory of commutative vN regular rings decidable?
Thomas, every finitely generated associative commutative ring is residually finite. (For free rings (= polynomial rings), this is almost obvious.)
Nov
9
comment Possible cardinality and weight of an ordered field
Taras, you may write an answer and accept your own answer to make the question "closed".
Nov
9
comment Explicitly showing that a free group is LERF
Surely, there is an algorithm for finding this finite-index subgroup and the free complement in this subgroup. Just look at the proof of Hall's free-factor theorem. It is quite easy using Stallings graphs..