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This is a follow-up to my question How many subgroups can there be in a group of order 512? Is it always 2 mod 4?, which in turn was inspired by the question What finite groups have as many elements as subgroups? of Richard Stanley. The suggestion in my question had counterexamples, found by Joachim König. This has caused me to reformulate the question as the following conjecture.

Conjecture. Let $G$ be a group of order $2^n$ with $n\geqslant 1$. Then the number of subgroups $s(G)$ is congruent to $-n-1$ modulo $4$ with the following exceptions. If $n$ is even and $G$ is cyclic, then $s(G)=n+1$. If $n$ is odd and $G$ is dihedral then $s(G)=2^n+n-1$. If $n$ is odd and $G$ is semidihedral then $s(G)=\frac{3}{4}.2^n+n-1$. If $n$ is odd and $G$ is generalised quaternion then $s(G)=\frac{1}{2}.2^n+n-1$.

For example, if $|G|=16$ and $G$ is non-cyclic then $s(G)$ is $3$ mod $4$. If $|G|=32$ and $G$ is not dihedral, semidihedral, or generalised quaternion, then $s(G)$ is $2$ mod $4$. If $|G|=64$ and $G$ is non-cyclic then $s(G)$ is $1$ mod $4$. And so on. For groups of order $512$, this would mean that $s(G)$ is $2$ mod $4$ except for three cases with $s(G)$ equal to $520$, $392$ and $264$ for dihedral, semidihedral, generalised quaternion respectively.

One consequence of this would be that for a group $G$ of order $2^n$ with $|G|=s(G)$, $n$ would have to be congruent to $3$ modulo $4$. This gives one tiny bit of information on the original question of Stanley.

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    $\begingroup$ After I posted this question, Richard Stanley pointed me to the paper of Easterfield, "An extension of a theorem of Kulakoff" (1938). This paper looks as though it ought to contain a proof of the above conjecture, but it will take some untangling of notation to confirm that. At least the list of exceptions is the same. $\endgroup$ Commented Jun 10 at 19:23

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After I posted this question, Richard Stanley pointed me to the paper of Easterfield, "An extension of a theorem of Kulakoff" (1938). This solved the problem.

From this paper, one can deduce that my conjecture is true as follows. Let $G$ be a group of order $2^n$. Let $m_k$ be the number of subgroups of $G$ of order $2^k$ (this is called $n_k$ in the paper, but I'm changing the notation to avoid confusion). One of the main results of the paper is that if $G$ is not cyclic, dihedral, semidihedral, or generalised quaternion, then for $0<k<n$, $m_k$ is congruent to $3$ modulo $4$. This can be found on line 7 of page 317. Thus the total number of subgroups is congruent to $1 + 3(n-1) + 1$ modulo $4$, which is $-n-1$ modulo $4$. One can then easily check that for the cyclic group this congruence is only incorrect when $n$ is even, while for the dihedral, semidihedral and generalised quaternion groups it is only incorrect when $n$ is odd.

It follows that if a group $G$ of order $2^n$ satisfies $|G|=s(G)$ then $n$ is congruent to $3$ modulo $4$. Thus for example $512$ is not a member of OEIS sequence A368538. The next few terms in the list are $560$ (the groups are numbered $117$ and $156$), $576$ (groups $204$, $1430$, $1986$, $5491$), $588$ (group $8$), $624$ (group $167$), $640$ ($23$ different groups), $672$ (group $630$), $704$ ($13$ different groups), $720$ (groups $545$ and $737$).

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  • $\begingroup$ What are the conventions? Should one accept one's own answer when it solves the problem? $\endgroup$ Commented Jun 11 at 7:24
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    $\begingroup$ Yes, after waiting 48 hours: mathoverflow.net/help/self-answer $\endgroup$ Commented Jun 11 at 8:48

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