Timeline for Molien for modular representations?
Current License: CC BY-SA 2.5
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 23, 2011 at 18:25 | comment | added | Kevin Ventullo | The union is disjoint by uniqueness of the decomposition $g=h_1h_2$. Anyway, Mariano's argument is cleaner. | |
| Feb 23, 2011 at 16:08 | comment | added | darij grinberg | Oh. Wait. In $G=\bigcup_g P_g\cdot g$, why is the union disjoint? | |
| Feb 23, 2011 at 15:58 | vote | accept | darij grinberg | ||
| Feb 23, 2011 at 16:08 | |||||
| Feb 23, 2011 at 15:58 | comment | added | darij grinberg | Ah, thanks! I was stupid. I confused the "well known lemma" with a lemma I wanted to use in my proof and couldn't show. They only see what they want to see. | |
| Feb 23, 2011 at 15:16 | comment | added | Johannes Hahn | @darij: The reference is "folklore". Write $ord(g)=ab$ with relative prime $a,b$ and choose $s,t$ such that $sa+bt=1$. Then $g=g^1=(g^a)^s \cdot (g^b)^t$. Obviously the two factors commute and $ord((g^a)^s) | b$ and $ord((g^b)^t) | a$. In fact the orders equal those terms because $ab=ord(g)=ord((g^a)^s)ord((g^b)^t) | ba$. One the other hand: If $g=xy$ with $ord(x)=b, ord(y)=a$ and $xy=yx$ then $g^{sa} = (x^{sa})(y^{sa})=x^{sa}=x^{sa \mod b}=x$ and $g^{tb}=(x^{tb})(y^{tb})=y^{tb}=y^{tb \mod a}=y$. | |
| Feb 23, 2011 at 12:55 | comment | added | darij grinberg | Thanks. I didn't even read your post to the end; after the "well known lemma" it was immediately clear to me. I am going to accept your solution as soon as you add a reference to this lemma. | |
| Feb 23, 2011 at 9:09 | history | answered | Kevin Ventullo | CC BY-SA 2.5 |