Timeline for Generalization of $(HK:H)=(K:H\cap K)$
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Jan 13, 2016 at 3:02 | review | Close votes | |||
| Jan 17, 2016 at 12:45 | |||||
| Jan 7, 2016 at 18:08 | vote | accept | user63850 | ||
| Jan 7, 2016 at 14:36 | answer | added | Geoff Robinson | timeline score: 2 | |
| Jan 4, 2016 at 21:58 | review | Close votes | |||
| Jan 6, 2016 at 15:49 | |||||
| Jan 4, 2016 at 21:47 | comment | added | Russ Woodroofe | As you might have already noticed, you don't need $H$ to normalize $K$ for (*) to hold, only for the two groups to permute setwise. | |
| Jan 4, 2016 at 21:15 | comment | added | Geoff Robinson | I think the question is too vague in its present form. It has to depend on more that just $|H|$ and $|K|$, but when $H$ and $K$ are cyclic of the same prime order $p$ but $H$ and $K$ are not conjugate in $\langle H,K \rangle$, all $(H,K)$ double cosets have size $p^{2}$ ( so you have no chance of even knowing the number of double cosets unless you can already determine $|\langle H,K \rangle |$. | |
| Jan 4, 2016 at 21:02 | comment | added | user63850 | @Geoff Robinson: Maybe one should expect a sum on the RHS ? | |
| Jan 4, 2016 at 21:00 | comment | added | eric | Isn't the generalisation just your comment?? | |
| Jan 4, 2016 at 20:49 | comment | added | Geoff Robinson | I think that for any pair of positive integers $m,n >1$ there is a cyclic group $H$ and $K$ of respective orders $m,n$ with $\langle H,K \rangle$ arbitrarily large. This is clear when $m = n =2$, for example. | |
| Jan 4, 2016 at 20:35 | history | asked | user63850 | CC BY-SA 3.0 |