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Jan
11
comment $G$ faithfully acting on $S$ with $|G|=81$ and $|S|=9$, then the action is transitive
I could have raised the question as follows "why all 3-sylows of S_9 are transitive?" but I preferred to put the original question. Thank you anyway.
Jan
11
accepted $G$ faithfully acting on $S$ with $|G|=81$ and $|S|=9$, then the action is transitive
Jan
11
asked $G$ faithfully acting on $S$ with $|G|=81$ and $|S|=9$, then the action is transitive
Nov
23
reviewed Approve Can I avoid the Abel partial summation technique and instead prove uniform convergence in this way?
Nov
23
reviewed Approve How to find a bijection from [0,1] into (0,1)?
Nov
1
comment Fundamental group of $X = \{(p, q)|p \neq −q\}\subset S^n \times S^n$
Thanks for your multiple answers.
Nov
1
accepted Fundamental group of $X = \{(p, q)|p \neq −q\}\subset S^n \times S^n$
Nov
1
comment Fundamental group of $X = \{(p, q)|p \neq −q\}\subset S^n \times S^n$
I did it! I proved it explicity using your hint. If $\gamma$ is such a geodesic in time $t$ I follow the curve until $\gamma(t)$. The fact that the geodesic is unique gives me easily the continuity of the homotopy, I think if $p=-q$ is when the homotopy I think fails to be continuous.
Nov
1
comment Fundamental group of $X = \{(p, q)|p \neq −q\}\subset S^n \times S^n$
That is a very good idea, thanks.
Nov
1
revised Fundamental group of $X = \{(p, q)|p \neq −q\}\subset S^n \times S^n$
edited body; edited title
Nov
1
comment Fundamental group of $X = \{(p, q)|p \neq −q\}\subset S^n \times S^n$
Youre right, I mean $S^n$. I denoted as the symmetric group.
Nov
1
asked Fundamental group of $X = \{(p, q)|p \neq −q\}\subset S^n \times S^n$
Oct
3
awarded  Popular Question
Sep
1
awarded  Notable Question
Jun
16
reviewed Approve Showing that $\Bbb R^3$ is not homeomorphic to $S^3$.
Jun
12
comment Abelian $p$-group with unique subgroup of index $p$
Why every proper subgroup is contained in $H$?, $H$ is maximal, but is not obvious that $H$ is the unique maximal. For example, a group of index $p^2$ could be also a maximal subgroup.
Jun
10
reviewed Approve A limit question
Jun
9
comment What is the last digit?
Or a smaller basis.
May
30
comment Proof to show function f satisfies Lipschitz condition when derivatives f' exist and are continuous
yes, youre right
May
30
accepted Working with $z$, $\overline{z}$ instead of $\operatorname{Re}(z)$, $\operatorname{Im}(z)$