5
$\begingroup$

Let $G$ be a finite group and $\mathcal{L}(G)$ its subgroup lattice.
Let $s(n):= max\{|\mathcal{L}(G)| \text{ for } |G|=n \}$.

There is an OEIS page for the sequence $s(n)$: A018216
1, 2, 2, 5, 2, 6, 2, 16, 6, 8, 2, 16, 2, 10, 4, 67, 2, 28, 2, 22, 10, 14, 2, 54, 8, 16, 28, 28, 2, 28, 2, 374, 4, 20, 4, 78, 2, 22, 16, 76, 2, 36, 2, 40, 12, 26, 2, 236, 10, 64, 4, 46, 2, 212, 14, 98, 22, 32, 2, 80, 2, 34, 36, 2825, 4, 52, 2, 58, 4, 52, 2, 272

Now let $[H,G]$ be an interval of finite groups and $|[H,G]|$ its cardinal.
Let $t(n):= max\{|[H,G]| \text{ for } |G:H|=n \}$.

I didn't find an OEIS page for the sequence $t(n)$.
This is computable for the indices $<32$, using GAP or MAGMA (I've no access to, currently)

Question: Is there an integer $n$ such that $t(n)>s(n)$?

Remark: If we observe that $\forall n \ t(n)=s(n)$, then a proof of this fact would answer this post.
Else, I'm interested in the smallest counter-example.

Remark: Alexander Hulpke has checked that $t(n)=s(n)$, $\forall n \le 47$ (see comment below).

$\endgroup$

migrated from math.stackexchange.com Jun 15 '16 at 16:59

This question came from our site for people studying math at any level and professionals in related fields.

  • 1
    $\begingroup$ A somewhat related question. Mostly adding this, because Gerry and Derek give more links there. $\endgroup$ – Jyrki Lahtonen Jun 5 '16 at 10:27
  • $\begingroup$ Could you please add a definition for an interval of finite groups? $\endgroup$ – Alexander Konovalov Jun 6 '16 at 20:21
  • $\begingroup$ @AlexanderKonovalov: $[H,G] = \{K \ | \ H \le K \le G\}$. $\endgroup$ – Sebastien Palcoux Jun 6 '16 at 20:25
  • 2
    $\begingroup$ I just verified (using the lists of transitive groups in degrees 32-47 due to Derek Holt) that $t(n)=s(n)$ for $n\le 47$. I would be very surprised if a group could have more block systems than a regular abelian group, but that is gut feeling and no proof. $\endgroup$ – ahulpke Jun 6 '16 at 22:10

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.