bio | website | halgebra.math.msu.su/staff/… |
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location | Moscow | |
age | ||
visits | member for | 1 year, 10 months |
seen | Apr 1 at 17:06 | |
stats | profile views | 1,000 |
Mar 8 |
awarded | Popular Question |
Jan 22 |
awarded | Organizer |
Jan 22 |
revised |
subsets of groups which have to be closed no matter what
arXiv's tags added |
Jan 22 |
suggested | suggested edit on subsets of groups which have to be closed no matter what |
Jan 22 |
awarded | Nice Answer |
Jan 22 |
answered | subsets of groups which have to be closed no matter what |
Jan 21 |
revised |
Fantastic properties of Z/2Z
added 12 characters in body |
Jan 21 |
answered | Fantastic properties of Z/2Z |
Jan 21 |
answered | Examples of cancellative normal semigroups |
Jan 20 |
comment |
Why the axiomatic rank of the variety of groups is equal to three?
Your octonion argument is much simpler and works perfectly. The algebra of all unit octonions or all non-zero octonions (with operations $\cdot$, ${\ }^{-1}$, and $1$) is non-associative but satisfies all two-variable laws that are consequences of the group axioms. |
Jan 19 |
answered | Non finitely based varieties of groups defined by finitely many variables |
Jan 17 |
comment |
Almost uniquely generated groups
Oh yes, thank you! |
Jan 17 |
awarded | Citizen Patrol |
Jan 17 |
comment |
Almost uniquely generated groups
"Any finite $p$-group which is relatively free in some variety has this property by the same argument. In light of Geoff's answer perhaps these are the only ones?" -- No. The quaternion group of order 8 is not relatively free but satisfies the conditions. |
Jan 17 |
comment |
Almost uniquely generated groups
Why $\{a_{1},b_{1},g_{2},g_{3}, \ldots, g_{n} \}$ is irredundant? |
Jan 17 |
comment |
Almost uniquely generated groups
No. $S$ is unique up to automorphisms; so, the complement of $S$ is not necessary the Frattini subgroup.Take a cyclic group of prime order, for example. |
Jan 16 |
comment |
Almost uniquely generated groups
Yes. And probably you mean Question 2 as you use that the orders are finite. |
Jan 16 |
comment |
Almost uniquely generated groups
This condition is indeed weaker. See the (wrong) answer of M. Shahryari. |
Jan 16 |
comment |
Almost uniquely generated groups
No. Even $F_1$ does not satisfy. The question is about inclusion-minimal sets. |
Jan 16 |
comment |
Almost uniquely generated groups
It seems that you use the letter $n$ in two different senses. You are talking about $m$-generated free groups in the varieties of nilpotent groups of exponent $p^n$, right? |