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 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 $P \subseteq \left$ and $\left \subseteq \left$, 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 $\left = P\left$ 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