Here is a fun question. Consider the dihedral group $\mathcal{D}_4=\left\langle a,b\mid a^4=b^2=1, bab=a^{-1}\right\rangle$ of order $8$. This group has exactly $8$ genuine subgroups (but not all different up to isomorphism). Are there other finite groups that have as many genuine subgroups as their order? Can there be infinitely many such groups? What would be a reasonable way to tackle this question?

Edit: By genuine subgroup I mean proper subgroup, forgot the terminology there.

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    $\begingroup$ The cyclic group of order 2 has exactly 2 subgroups ... $\endgroup$ – Hagen von Eitzen Jun 21 '16 at 8:08
  • $\begingroup$ Hmm, this is a tricky question. In one "end", the elementary abelian groups have more subgroups, and in the other, cyclic groups have fewer, so one might expect that there could well be infinitely many such groups. $\endgroup$ – Tobias Kildetoft Jun 21 '16 at 8:09
  • $\begingroup$ In general, $D_n$ has $\tau(n)+\sigma(n)-2$ proper subgroups, with number of divisors and sum of divisors function. For which $n$ is this equal to $2n$ ? $\endgroup$ – Dietrich Burde Jun 21 '16 at 8:11
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    $\begingroup$ @DietrichBurde I have a feeling that is a very hard question (closing in on some stuff possibly related to perfect numbers and that sort of thing). $\endgroup$ – Tobias Kildetoft Jun 21 '16 at 8:17
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    $\begingroup$ I have never the expression genuine subgroup. $\endgroup$ – Derek Holt Jun 21 '16 at 8:20

The question is quite general, so that I will content myself with all dihedral groups $D_n$, with $n\ge 3$. Here we do know how many subgroups we have:

Theorem. (Stephan A. Cavior, 1975) If $n \geq 3$, then the number of subgroups of $D_{2n}$ is $\tau(n) + \sigma(n)$.

The theorem says that the number of “all” subgroups, including $\{1\}$ and $D_{2n}$, is $\tau(n) + \sigma(n)$, the number of divisors of $n$ plus the sum of all divisors of $n$. We have to exclude the trivial group and the whole group to obtain the number of proper subgroups. Hence we want to solve $$ \tau(n) + \sigma(n)=2n+2. $$ This has many solutions, e.g., $n=4,10,44, 2336, 8896, 34432, 449295, 549775212544 \cdots$, and it is a nice question to decide whether we have infinitely many solutions or not.

Edit: Thanks to Ivan Neretin for providing the link to OEIS A066229.

  • $\begingroup$ Would you bother to add this as a comment to oeis.org/A066229? $\endgroup$ – Ivan Neretin Jun 21 '16 at 8:47
  • $\begingroup$ Why, adding small pieces of info to OEIS is quite easy. Also, it will stay there forever, and might prove kinda helpful one day, or spark someone's interest. $\endgroup$ – Ivan Neretin Jun 21 '16 at 9:04
  • $\begingroup$ @IvanNeretin It seems that you are the right person to do this, then. What kind of comment do you want to add ? $\endgroup$ – Dietrich Burde Jun 21 '16 at 9:06
  • $\begingroup$ "Also, numbers n such that the number of non-trivial proper subgroups of a dihedral group D_{2n} is the same as its order". $\endgroup$ – Ivan Neretin Jun 21 '16 at 9:11
  • $\begingroup$ OK, great. This is a good idea. You can give a reference to MSE too, if you want. $\endgroup$ – Dietrich Burde Jun 21 '16 at 9:13

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