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Let $L$ be a finite lattice with minimum $\hat{0}$ and maximum $\hat{1}$. The Möbius function $\mu$ for $L$ is defined recursively by: for $\forall a,b \in L$ with $a<b$, $\mu(b,b) = 1$ and $\mu(a,b) = -\sum_{a<c\le b}\mu(c,b)$.
The Möbius number of $L$ is defined by $\mu(\hat{0},\hat{1})$.

Define the Möbius number $\mu(G)$ of a finite group $G$ to be the Möbius number of its subgroup lattice. For the nonabelian finite simple groups of small order, we observe that $$\mu(G) \in |G| \mathbb{Z}$$

Question: Is it true for any nonabelian finite simple group?

Remark: It is true if $|G| < 100000$ (see the table below).

It's false for nonsimple groups because for $p$ (odd) prime, $\mu(S_p) = p!/2$ ([S]) and $\mu(C^2_p) = p$.

Table for the nonabelian finite simple groups of order $< 100000$: $$ \begin{array}{c|c|c|c|c|c} G & |G| & \mu(G) & \mu(G)/|G| & |Out(G)| & \newline \hline A_5 & 60 & -60 & -1 & 2 & \newline \hline PSL(2,7) & 168 & 0 & 0 & 2& \newline \hline A_6 & 360 & 720 & 2 & 4 & \newline \hline PSL(2,8) & 504 & -504 & -1 & 3 \newline \hline PSL(2,11) & 660 & 660 & 1 & 2 \newline \hline PSL(2,13) & 1092 & -1092 & -1 & 2 \newline \hline PSL(2,17) & 2448 & 0 & 0 & 2 \newline \hline A_7 & 2520 & 2520 & 1 & 2 \newline \hline PSL(2,19)& 3420 & 3420 & 1 & 2 \newline \hline PSL(2,16)& 4080 & 0 & 0 & 4 \newline \hline PSL(3,3)& 5616 & 0 & 0 & 2 \newline \hline PSU(3,3)& 6048 & 0 & 0 & 2 \newline \hline PSL(2,23)& 6072 & 0 & 0 & 2 \newline \hline PSL(2,25)& 7800 & 0 & 0 & 4 \newline \hline M_{11} & 7920 & -7920 & -1 & 1 \newline \hline PSL(2,27)& 9828 & 9828 & 1 & 6 \newline \hline PSL(2,29) & 12180 & 12180 & 1 & 2 \newline \hline PSL(2,27)& 14880 & 29760 & 2 & 2 \newline \hline A_8& 20160 & 20160 & 1 & 2 \newline \hline PSL(3,4)& 20160 & -120960 & -6 & 12 \newline \hline PSL(2,37)& 25308& -25308& -1 & 2 \newline \hline PSp(4,3)& 25920& -25920& -1 & 2 \newline \hline Sz(8)& 29120& -29120& -1 & 3 \newline \hline PSL(2,32)& 32736& -32736& -1 & 5 \newline \hline PSL(2,41)& 34440& 68880& 2 & 2 \newline \hline PSL(2,43)& 39732& -39732& -1 & 2 \newline \hline PSL(2,47)& 51888& 0& 0 & 2 \newline \hline PSL(2,49)& 58800& 117600& 2 & 4 \newline \hline PSU(3,4)& 62400& 0& 0 & 4 \newline \hline PSL(2,53)& 74412& -74412& -1 & 2 \newline \hline M_{12} & 95040 & 95040 & 1 & 2 \newline \end{array}$$

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    $\begingroup$ I believe the fact that $\mu(1,S_p)=\frac{p!}{2}$ when $p$ is prime was known well before I published the paper you link to above. I think it appears in a paper of Pahlings. $\endgroup$
    – John Shareshian
    Commented Mar 26, 2017 at 14:42

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The answer to your question is ``yes". Something more general was proved by Hawkes, Isaacs and \"Ozaydin in a 1989 paper in the Rocky Mountain Journal of Mathematics.

CORRECTION: As Sebastien Palcoux notes below, this result is due to Kratzer and Th\'evenaz

More precise results for the groups $PSL_2(p)$ were known to Philip Hall, and $PSL_2(q)$ was handled by Martin Downs, as were the Suzuki groups.

Some results on symmetric groups appear in a paper of mine, and other classes of ``small" almost simple groups are addressed in my thesis,

http://www.math.wustl.edu/~shareshi/thmsri.ps

(Sorry, it is a postscript file and some non-mathematical pages are repeated.)

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    $\begingroup$ I don't know how to deduce it from the paper Hawkes, Isaacs and Özaydin (available here), but by browsing your thesis, I've found Theorem 1.5 refering to Théorème 3.1. of the paper "Fonction de Möbius d'un groupe fini et anneau de Burnside" (1984) by Kratzer and Thévenaz (available here) and stating that $$\mu(G) \in \frac{|G|}{|G:G'|_0} \mathbb{Z}.$$ The result follows immediately for any perfect group. $\endgroup$
    – Sebastien Palcoux
    Commented Mar 26, 2017 at 16:48
  • $\begingroup$ You have the right reference, my memory was incorrect. $\endgroup$
    – John Shareshian
    Commented Mar 26, 2017 at 16:55
  • $\begingroup$ I should mention also that Emilio Pierro has handled some Ree groups, see arxiv.org/pdf/1410.8702.pdf $\endgroup$
    – John Shareshian
    Commented Mar 26, 2017 at 17:40
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    $\begingroup$ @SebastienPalcoux: Theorem 4.5 in the Hawkes-Isaacs-Özaydin paper says that $ |N_G(H):H | $ divides $ m \mu(H,G)$, where $m$ is the square-free part of $ |G:G'H | $. Of course, the case $ H=1$ is the result of Kratzer and Thévenaz. $\endgroup$
    – Frieder Ladisch
    Commented Mar 27, 2017 at 11:44
  • $\begingroup$ @FriederLadisch: Nice! In others words: $$\mu(H,G) \in \frac{|N_G(H):H |}{|G:G'H |_0} \mathbb{Z}.$$ $\endgroup$
    – Sebastien Palcoux
    Commented Mar 27, 2017 at 11:57

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