Reputation
Top tag
Next privilege 50 Rep.
Comment everywhere
Badges
5
Newest
 Commentator
Impact
~455 people reached

  • 0 posts edited
  • 0 helpful flags
  • 2 votes cast
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
Apr
29
accepted Image of subgroup under group automorphism lies in itself
Apr
29
asked Image of subgroup under group automorphism lies in itself
Dec
10
comment Inner automorphism of a group-ring
When $ u = \sum_{g \in G} r_g g $ then $ \text{supp}(u) = \{ g \in G | r_g \neq 0 \} $.
Dec
9
asked Inner automorphism of a group-ring