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Some experimentation leads me to suspect that if $G$ is a finite group with $|G|=512$ then the number of subgroups of $G$ is congruent to $2$ modulo $4$. Does anyone see a good reason why this might be true? Is there a more general fact of which this is a special case?

For comparison, if $|G|$ is a power of $4$ then the number of subgroups is congruent to $1$ mod $2$.

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    $\begingroup$ In a 2-group, mod 4 you have to count the number of normal subgroups, and the (even) number of subgroups having normalizer of index 2. You might count both separately and see if there also an overwhelming pattern. $\endgroup$ Commented Jun 10 at 15:56
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    $\begingroup$ For a theorem on the number mod 4 of subgroups of order $2^k$ of a $2$-group, see T. E. Easterfield, An extension of a theorem of Kulakoff, Math. Proc. Camb. Phil. Soc. 34 (1938), 316-320, at cambridge.org/core/journals/…. $\endgroup$ Commented Jun 10 at 19:10
  • $\begingroup$ @RichardStanley I think this paper may confirm the conjecture in my new question, but it'll take a bit of untangling. At least it gives the same list of exceptions. Thanks for the reference! $\endgroup$ Commented Jun 10 at 19:20

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According to Magma, SmallGroup(512,j) for $j=2042,2043,2044$ are some examples (and also the first in database order) with number of subgroups congruent to $0$ mod $4$ (namely, $520$, $392$ and $264$ respectively).

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  • $\begingroup$ Thanks! So I was wrong! Examples seem quite rare. $\endgroup$ Commented Jun 10 at 15:17
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    $\begingroup$ I still think this requires some explanation. These groups have a cyclic subgroup of order 128, so they're quite simple to describe. For example, for groups of order $256$ the number of subgroups is $3$ mod $4$ with one exception: the cyclic group, where it's $1$ mod $4$. I'm sure there's an interesting statement lurking around here somewhere. $\endgroup$ Commented Jun 10 at 15:33
  • $\begingroup$ In the case of groups of order $128$, the number of subgroups is always zero mod four except SmallGroup$(128,j)$ with $j=161,162,163$. So I wonder whether for groups of order $512$, $j=2042,2043,2044$ are the only exceptions. It would take quite a lot of work to find out, without some sort of abstract argument. $\endgroup$ Commented Jun 10 at 18:06
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    $\begingroup$ Okay, I seem to be getting somewhere. It looks to me like the exceptions are dihedral, semidihedral, and generalised quaternion. So now it's still possible that for groups of order $2^{2n-1}$, these are the only exceptions to the statement that the number of subgroups is congruent to $2n$ modulo four. The plot thickens. $\endgroup$ Commented Jun 10 at 18:17
  • $\begingroup$ I've posted a new question giving a precise conjecture on the number of subgroups of a finite $2$-group modulo four. $\endgroup$ Commented Jun 10 at 19:04

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