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Jun 10 at 13:47 comment added Dave Benson Some experimentation suggests that if $G$ is a finite group with $|G|=512$ then the number of subgroups of $G$ is $2$ modulo $4$. This would mean that groups of order $512$ are not candidates. I've asked about this in a separate question.
Jun 10 at 9:58 comment added YCor The next $n$ questions could be when the number of subgroups is the number of elements minus $k$, for $k=1,\dots, n$...
Jun 8 at 19:01 comment added Richard Stanley I found out that OEIS sequence A368538 gives the order of such groups (but not the number of them of a given order) up to order 511.
Jun 8 at 15:50 vote accept Richard Stanley
Jun 7 at 7:29 history became hot network question
Jun 7 at 6:36 answer added Dave Benson timeline score: 21
Jun 6 at 22:41 history edited Richard Stanley CC BY-SA 4.0
removed error about S_2
Jun 6 at 21:01 comment added Peter Kagey More generally, OEIS sequence A083874 lists the dihedral groups that work.
Jun 6 at 20:53 comment added Francesco Polizzi $D_{104}$ is another dihedral group that works
Jun 6 at 20:43 history edited GH from MO
edited tags
Jun 6 at 20:39 comment added Will Sawin To anyone whose suspects that $D_6=S_3, D_{28}$ imply that $D_n$ works for $n$ an even perfect number: The pattern stops after that, since the number of subgroups of $D_n$ is the sum of all divisors (not necessarily proper) of $n/2$ plus the number of divisors (not necessarily proper) of $n/2$. For $n=2^{p-1} (2^p-1)$, with $2^p-1$ prime an even perfect number, the only proper divisor of $n$ that is not a divisor of $n/2$ is $2^{p-1}$, while the number of divisors of $n/2$ is $2(p-1)$, and these agree if and only if $p=2,3$ which gives the two examples found by testaccount.
Jun 6 at 20:27 comment added testaccount Well $S_3$ does have six elements and six subgroups. Some small examples where the number of elements is the number of subgroups: $C_1$, $C_2$, $S_3$, $C_2 \times C_4$, $D_{28}$, $C_6 \times S_3$. With a computer we find that there are $13$ such examples with $|G| \leq 100$, up to isomorphism.
Jun 6 at 20:20 comment added Sam Hopkins You could also ask about conjugacy classes of elements and conjugacy classes of subgroups.
Jun 6 at 20:13 history asked Richard Stanley CC BY-SA 4.0