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Universal coefficient theorem for group homology and cohomology
I'd like to mention that it is Proposition 7.1 in Chapter VI of Brown's book. (Many chapters of this book have section 7 and all items are numbered 7.x there.)
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What is the centralizer of a Coxeter element?
Blokhina relies on an article of Subbotin and Stekolshchik: link.springer.com/article/10.1007%2FBF01077574 where they treat the statement "the Coxeter graph is a tree" in a more restrictive sense, requiring in addition that all labels $m_{st}$ equal 3. This can be seen from the second paragraph of their article, where they notice that the value of the bilinear form on any pair of simple roots is either 0 or -1/2. Thus, Blokhina's result refers only to those Coxeter systems whose graph is a tree AND all labels $m_{st}$ are equal to 3.
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Examples of common false beliefs in mathematics
@ziggurism list all rationals in some order: $\{r_1,r_2,r_3,\dots\}$. Then $$\langle r_1\rangle\subset \langle r_1,r_2\rangle\subset \langle r_1,r_2,r_3\rangle\subset\dots\subset \mathbb Q$$ Each group in this tower, except the last one, is cyclic, hence free of rank 1.
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Jokes in the sense of Littlewood: examples?
$\frac{sin x}{n} = 6$, just cancel n's, you get "six"
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Paradoxical Mathematical Objects Pending for Construction
@SebastianBechtel Well, the standard definition of the derivative, which is well-known to the general mathematical audience, physicists, engineers is essentially integer-indexed. First derivative = speed, second = acceleration, etc. It is mind-blowing to think that something can be 'in between'.
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growth of a free group automorphism is same for finite index subgroups?
Thanks! I didn't realize that G' is f.i. in G. It follows from Svarc-Milnor lemma then that they are q.i.
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Decomposition of Braid Groups
@YCor Sorry, did not notice your comment earlier. What I meant, that for $B_4$ this is a known decomposition which is more interesting than splitting off a copy of $\mathbb Z$.
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growth of a free group automorphism is same for finite index subgroups?
Mark, thank you for the insight! I was wondering why it is obvious that $G$ is quasi-isometric to $G'$?
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growth of a free group automorphism is same for finite index subgroups?
Thank you for your reply, it may turn useful if one would like to avoid dealing with train-tracks.
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growth of a free group automorphism is same for finite index subgroups?
Dear Professor Mosher, Thank you for such extended answer. What would you recommend for a (possibly gentle) introduction into this train-track machinery?
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