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13h

comment 
Is there a nonabelian free group inside a group of positive rank gradient?
@DerekHolt: no, many torsionfree groups are residually$p$ (= residually a finite $p$group). The correct interpretation of Andreas's post is a $p$group which is also residually finite (and hence residually$p$) 
14h

comment 
Is there a nonabelian free group inside a group of positive rank gradient?
Silly remark: "residually finite" (RF) is unnecessary, since if it's true in the RF case, letting $N$ be the intersection of finite index subgroups of $G$, we can apply the result to $G/N$ (noting that RG$(G)=$RG$(G/N)$) and lift $F_2$ to $G$. 
Aug
24 
answered  Is a cocompact CAT(0) periodic? 
Aug
23 
comment 
transitive action on finite abelian subgroups
For $n\ge 2$ the group $Sp(2n,p)\ltimes F_p^{2n}$ acts transitively on elements of order $p$ in $F_p^{2n}$ but not transitively on 2planes (i.e. copies of $F_p^2$) in $F_p^{2n}$. 
Aug
23 
comment 
An approximate version of $g^2 = e$ for all $g \in G$, implies $G$ is Abelian
You can ask the questions you like, but I lost time checking the paper you link at to try to find a meaningful sense to your question... you refer to "your own definition" but you don't provide any. 
Aug
23 
comment 
An approximate version of $g^2 = e$ for all $g \in G$, implies $G$ is Abelian
I don't see any link between your $g^2=o(e)$ and the notion of approximate group in your link. And actually, I have no idea what you mean by $g^2=o(e)$. 
Aug
23 
revised 
Finitness of the Burnside Group
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Aug
22 
comment 
What are your favorite instructional counterexamples?
That the Dehn function is big does not mean that the word problem can be solved only slowly. 
Aug
22 
answered  Applications of Lubotzky's linearity theorem? 
Aug
22 
comment 
Are the BaumslagSolitar groups BS(n,n) and BS(n,n) automata groups?
For a group there's no "standard BassSerre tree". But for BaumslagSolitar groups there's one: indeed they are defined as HNN extension of an infinite cyclic group, and the corresponding BassSerre tree is the one I call "standard", the action is with cyclic edge and vertex stabilizers, and in restriction to $L$ the vertex stabilizers are trivial. 
Aug
21 
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Applications of Lubotzky's linearity theorem?
According to Alex himself, this theorem is practically useless. It does not mean that it can't be applied, for instance when you have a group with assumptions that it has many quotients in some suitable sense, it can be applied. But for explicit examples of groups (e.g., given by a presentation, or as groups of automorphisms of some reasonable structure) as in the examples you provide, you don't have much info about finite quotients so you it's not practical: in examples known to be linear, the easiest way to check Lubotzky's criterion is to show that they are linear... 
Aug
21 
comment 
When is a polynomial ring free over a graded subalgebra?
I changed to a title related to the contents of the question. Please feel free to improve it, but avoid titles with little or no math information. 
Aug
21 
revised 
When is a polynomial ring free over a graded subalgebra?
changed to a useful title. Added tag ac. 
Aug
20 
answered  Finitness of the Burnside Group 
Aug
20 
comment 
Finitness of the Burnside Group
What I said is general about recursive presentations. If you have a recursive group presentation on finitely many generators which turns out to be that of a group of order $\le k$, then for some $n$ you get enough equalities in the $n$ball to ensure it has $\le k$ elements, and enough equalities to ensure that all elements in the $(n+1)$ball are in the $n$ball: the algorithm just consisting in computing consequences of relations. But this is just an algorithm that stops if the group has cardinal $\le k$. 
Aug
20 
comment 
Finitness of the Burnside Group
Btw, as you probably observed, the set of $(n,d)$ such that $B(n,d)$ has cardinal $\le k$ is obviously recursively enumerable (with an explicit algorithm). So whether we can test the other inequality is equivalent to determining whether $(n,d)\mapsto B(n,d)$ is computable. 
Aug
20 
comment 
Finitness of the Burnside Group
If $d$ is fixed, there is in principle an algorithm whose input is $n$ and the output is the cardinal of $B(d,n)$. This is just because this sequence is eventually infinite, hence computable, but this does not say what the sequence (nor the algorithm) is. 
Aug
20 
comment 
Finitness of the Burnside Group
What is the input? the pair $(n,d)$? the triple $(n,d,k)$? 
Aug
19 
comment 
Are the BaumslagSolitar groups BS(n,n) and BS(n,n) automata groups?
As pointed out by Derek, your question is ambiguous. My guess was that you understand that (1) being virtually $\mathbf{Z}\times F_k$ implies being automata group and that you were asking (2) why $BS(n,\pm n)$ has this virtual property. If you're asking why (1) holds, somebody else could answer better than me. 
Aug
19 
revised 
Are the BaumslagSolitar groups BS(n,n) and BS(n,n) automata groups?
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