Reputation
Top tag
Next privilege 75 Rep.
Set bounties
Badges
6
Newest
 Curious
Impact
~583 people reached

  • 0 posts edited
  • 0 helpful flags
  • 2 votes cast
Mar
4
asked Lower central series starts with index 0, but upper central series starts with index 1, why?
Feb
15
awarded  Curious
Feb
14
comment Axiom of choice - Equivalence relation - Representatives
First off, thank you all! But what if I use the power set $ 2^X $ of $ X $. As it contains any possible subset of $ X $, there must be at least on subset of $ X $ that contains one element of every equivalance class. Then I could say, Let R be such a subset? The power set should guarantee me the existence of such a set.
Feb
14
asked Axiom of choice - Equivalence relation - Representatives
Jan
19
accepted $S^2$ covered by 6 sets - Borsuk-Ulam
Jan
19
revised $S^2$ covered by 6 sets - Borsuk-Ulam
added 106 characters in body
Jan
19
awarded  Commentator
Jan
19
comment $S^2$ covered by 6 sets - Borsuk-Ulam
Assume A,B,C cover $S^2$. Create the continuous map $ f:S^2 \rightarrow \mathbb{R}^2 : x \mapsto (d(x,A),d(x,B)) $. Then Borsuk-Ulam gives a point $ x $ with $ f(x) = f(-x) $. If $ f(x) = (0,0) $ then $ x, -x $ lie in A (and B). If not they lie in C.
Jan
19
asked $S^2$ covered by 6 sets - Borsuk-Ulam
Oct
9
comment Prove that $ \text{H}^1(G/P,\text{Z}(P)) = 0 $ for a normal Sylow p-subgroup P
Ah ok, thank you! Do you have any hint for me how to prove it? I mean, are there any non-trivial module homomorphism? Then I would need to show directly, that the cochain of morphism groups is exact for $ n \geq 1 $...
Oct
9
awarded  Editor
Oct
9
revised Prove that $ \text{H}^1(G/P,\text{Z}(P)) = 0 $ for a normal Sylow p-subgroup P
edited title
Oct
9
asked Prove that $ \text{H}^1(G/P,\text{Z}(P)) = 0 $ for a normal Sylow p-subgroup P
Aug
20
comment p-element centralizing a Sylow p-subgroup
Thank you very much =)
Aug
19
comment p-element centralizing a Sylow p-subgroup
@Bungo: What if I have an arbitary set of p-elements. Is the subgroup generated by these p-elements not automatically a p-subgroup?
Aug
19
comment p-element centralizing a Sylow p-subgroup
@ahulpke: I have $P \subseteq \left<P,g\right>$ and $\left<g\right> \subseteq \left<P,g\right>$, right? But that would be true for any $g \in G$. So I always could generate larger $p$-subgroups, for example with 2 different Sylow p-subgroups? So I definitely have something wrong in my understanding. Besides from that I have $\left<P,g\right> = P\left<g\right>$ as $gx=xg$ for all $x \in P$.
Aug
19
asked p-element centralizing a Sylow p-subgroup
Apr
29
comment Image of subgroup under group automorphism lies in itself
Thank you so much! This almost has driven me mad. I also thought of $ \mathbb{Z} $ and $ 2\mathbb{Z} $, but not as subgroups of $ \mathbb{Q} $.
Apr
29
awarded  Scholar
Apr
29
awarded  Supporter