Timeline for Isometry group of an integer as of the corresponding $\Omega(n)$-parallelotope
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 14, 2014 at 10:34 | answer | added | user45639 | timeline score: 1 | |
| Apr 13, 2014 at 12:03 | comment | added | Sylvain JULIEN | You're perfectly right. $G(1)$ is the trivial group and if $a\mid b$ then $G(a)$ is a subgroup of $G(b)$. | |
| Apr 13, 2014 at 11:50 | comment | added | user45639 | @SylvainJULIEN If I really understand your question (I'm not pretty sure...), when $~n~$ is a prime $~p,~$ then the parallelotope is degenerated and become a segment of lenght $~p,~$ so $~G(n)=\{Id,~Half~turn\}=\mathbb{Z} / 2\mathbb{Z},~$ when $~n=p^2~$ with $~p~$ prime, the parallelotope is a square and $~G(n)~$ is the dihedral group $~D_4,~$ and if $~n=pq~$ with $~p,~q~$ distinct primes, $~G(n)~$ is the only non-cyclic group of order $4,$ i.e. the dihedral group $~D_2~$ or $~(\mathbb{Z} / 2\mathbb{Z})\times(\mathbb{Z} / 2\mathbb{Z}).~$ Am I wrong ? Yours truly. | |
| Apr 12, 2014 at 19:18 | comment | added | Sylvain JULIEN | $\Omega(n)$ is the total number of prime factors of $n$ counted with multiplicity, $\omega(n)$ is the number of distinct prime factors of $n$. $\Omega(360)=6$, $\omega(360)=3$. | |
| Apr 12, 2014 at 17:03 | comment | added | Lee Mosher | What do the symbols $\Omega(n)$ and $\omega(n)$ stand for? | |
| Apr 12, 2014 at 10:31 | history | asked | Sylvain JULIEN | CC BY-SA 3.0 |