To extend A018216 Maximal number of subgroups in a group with n elements

[Jonathan Post]

Dear Max,
You're right.
For n = 4:
Z_4 has Z_2,  Z_2 X Z_2 has 3 of Z_2
I had somehow miscomputed this way:
Z_4 has Z_2 [which is one subgroup],  Z_2 X Z_2 has 3 of Z_2 [which is3 more subgroups], so 1 + 3 = 4, then add trivial subgroup and entiregroup to get 4+2 = 6.  Yo are correct that I should have said:
Z_4 has Z_2 [which is one subgroup, plus the whole group and thetrivial group makes 3],  Z_2 X Z_2 has 3 of Z_2 [plus the whole groupand the trivial group makes 5] so 3 + 5 = 8.
I made the same error thereafter, which explains why, for instance, Igave a(8) = 24 while you gave it correctly as 28.
Notationally, I used:
Z_n: the cyclic group of order n (often the notation C_n is used, or Z / n Z).
Dih_n: the dihedral group of order 2n (often the notation D_n or D_2n is used )
S_n: the symmetric group of degree n, containing the n! permutationsof n elements.
A_n: the alternating group of degree n, containing the n!/2 evenpermutations of n elements.
Dicn: the dicyclic group of order 4n.
When I got to n=12 for the nonabelians, I'd neglected to mention thatDic_3 = Z_3 ⋊ Z_4 where ⋊ is semidirect product.
At this point, it's clear that I merely stumbled into a question ofadding one or more seqs at the core of Group Theory enumeration ofsubgroups, but that Max, Christian, et al did the heavy lifting, andare due co-authorship.
One seq for the abelian, one for the nonabelian, one for the sum ofthose two? One seq to rule them all, and in the darkness bind them?
Best,
Jonathan Vos Post

On 12/4/07, Max Alekseyev <maxale at gmail.com> wrote:> On Dec 1, 2007 7:00 PM, Jonathan Post <jvospost3 at gmail.com> wrote:>> > We can get another sequence out of this process of seeking the maxima.> >> > a(n) = Sum of number of subgroups of all nonisomorphic groups of order n.> >> > Comment: with multiplicity, i.e. Z_2 X Z_2 has 3 isomorphic subgroups> > Z_2, so contributes 3 to the sum for a(4).  Every group has itself and> > the trivial group as subgroups, so a(1) = 1 where these are the same,> > and n>1 has a(n) => 2.  a(p) = 2 for prime p.> >> >  1, 2, 2, 6, 2, 4, 2, 24, 7, 4, 2, 12, 2, 4, 4> >> > Offset 1,2> >> > examples:> > n  a(n)  explanation (not list itself and trivial group)> > 1  1> > 2  2> > 3  2> > 4  6     Z_4 has Z_2,  Z_2 X Z_2 has 3 of Z_2>> In addition to four Z_2 mentioned above, you have one Z_4, one Z_2 X> Z_2, and two trivial subgroups (of Z_4 and of Z_2 X Z_2). So, a(4)> must be equal 8. Don't you agree?>> This sequence goes as>> 1, 2, 2, 8, 2, 4, 2, 28, 9, 4, 2, 16, 2, 4, 4, 125, 2, 18, 2, 16, 4,> 4, 2, 56, 11, 4, 42, 16, 2, 8, 2, 626, 4, 4, 4, 72, 2, 4, 4, 56, 2, 8,> 2, 16, 18, 4, 2, 250, 13, 22, 4, 16, 2, 84, 4, 56, 4, 4, 2, 32, 2, 4,> 18, 4273, 4, 8, 2, 16, 4, 8, 2, 252, 2, 4, 22, 16, 4, 8, 2, 250, 304,> 4, 2, 32, 4, 4, 4, 56, 2, 36, 4, 16, 4, 4, 4, 1252, 2, 26, 18, 88>> Regards,> Max>