Timeline for A particular example of solvable group
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Jul 15, 2014 at 20:18 | comment | added | Derek Holt | I think $N$ should be a direct product of copies of $C_{l^d}$ rather than just $C_{l^d}$. I checked that the construction works for $d=2$ with $A=C_7$, $B=C_9$ and $N = C_4^6$, so it seems plausible that it works for higher $d$. Assuming that to be the case, I would say that the question is interesting! | |
| Jul 15, 2014 at 13:24 | comment | added | Peter Mueller | Rather than asking a new question in the comments, you should either edit your question, or raise a new question. In addition, it could be useful to know why you are interested in these things. | |
| Jul 15, 2014 at 13:12 | comment | added | user55910 | Can we prove that $f(exp(G)) \leq k f(G)$ with $k$ constant( maybe $k=3$)? | |
| Jul 13, 2014 at 12:38 | comment | added | Peter Mueller | I doubt that this construction works: The cyclic group $N$ has an abelian automorphism group, so the non-abelian group $H$ cannot act faithfully on $N$. In particular, $H$ cannot act fixed-point-freely, as it should for $N$ being the Frobenius kernel. | |
| Jul 13, 2014 at 10:18 | comment | added | user55910 | $H$ the semidirect product of a cyclic group $A=C_{p^{m}}$ by a cyclic group $B=C_{q^{n+1}}$ which induces an automorphism of order $q^{n}$ on $A$ ($p$ and $q$ primes appropriate). $N$ the nilpotent group $C_{l^{d}}$ with $l \neq p,q$. Let $G=NH$ of Frobenius. If $n=m=d-1$, then $f(exp(G))=3d-1$ and $f(G)=d$ | |
| Jul 12, 2014 at 17:46 | comment | added | Peter Mueller | Which Frobenius groups bring the quotient close to $3$? | |
| S Jul 12, 2014 at 6:00 | history | suggested | Samuel Lelièvre | CC BY-SA 3.0 |
Improve typesetting of math.
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| Jul 12, 2014 at 5:24 | review | Suggested edits | |||
| S Jul 12, 2014 at 6:00 | |||||
| Jul 11, 2014 at 19:46 | comment | added | user55910 | With Frobenuis group, we can construct a solvable group with $ f(exp(G))/f(G) \rightarrow 3$. But I don't know examples with $f(exp(G))=3f(G)$. | |
| Jul 11, 2014 at 19:07 | review | Close votes | |||
| Jul 11, 2014 at 22:03 | |||||
| Jul 11, 2014 at 18:53 | comment | added | Derek Holt | Why $3f(G)$? Do you know examples with $f({\rm exp}(G)) = 3f(G)$? | |
| Jul 11, 2014 at 18:47 | review | First posts | |||
| Jul 11, 2014 at 18:53 | |||||
| Jul 11, 2014 at 18:41 | history | edited | Jeremy Rouse | CC BY-SA 3.0 |
Minor grammatical fixes.
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| Jul 11, 2014 at 18:32 | history | asked | user55910 | CC BY-SA 3.0 |