TITLE: Existence of a faithful irreducible representation using Möbius function
QUESTION [16 upvotes]: Let $G$ be a finite group, $L(G)$ its subgroup lattice and $\mu$ the Möbius function.
Consider the Euler totient of $G$ defined as follows:
$$ \varphi(G) = \sum_{H \le G}\mu(H,G) |H| $$
Let $X=\{M_1, \dots, M_n \}$ be the set of maximal subgroups of $G$. By applying the Crosscut Theorem with $X$ (see this comment of Richard Stanley) and next the inclusion–exclusion principle, we get that:
$$ \varphi(G) = |G \setminus \bigcup_{i=1}^n M_i| $$ In other words, $\varphi(G)$ is the number of elements $g \in G$ such that $\langle g \rangle = G$. It follows that 
$$ \varphi(G) \neq 0 \Leftrightarrow G \text{ cyclic}$$
Note that $\varphi(\mathbb{Z}/n) = \varphi(n)$ the usual Euler's totient function.
Now, consider the dual Euler totient of $G$ defined as follows: 
$$ \hat{\varphi}(G) = \sum_{H \le G}\mu(1,H) |G:H| $$  
Question: $ \hat{\varphi}(G) \neq 0 \Leftrightarrow G$ has a faithful irreducible complex representation?
Remark: We will see below that $(\Rightarrow)$ is true. So the question reduces to $(\Leftarrow)$.
It is true for the finite simple group $G$ of order $<10000$:  
$$ \begin{array}{c|c|c|c|c|c}
G & |G| & \hat{\varphi}(G) \newline 
 \hline  
  A_5 & 60     &  8   &  \newline 
 \hline
  PSL(2,7) & 168    &  228    &  \newline 
 \hline
  A_6 & 360     &   8748   &  \newline 
 \hline
  PSL(2,8) & 504     &   19056   &  \newline 
 \hline
PSL(2,11) & 660     &  24932    &  \newline 
 \hline
PSL(2,13) & 1092     &  105684    &  \newline 
 \hline
  PSL(2,17) & 2448     &   389496  &  \newline 
 \hline
  A_7 & 2520     &  188136    &  \newline 
 \hline
  PSL(2,19)& 3420     &   1148028   &  \newline 
 \hline
  PSL(2,16)& 4080     &  1935584    &  \newline 
 \hline
  PSL(3,3)& 5616      &  395496    &  \newline
 \hline
   PSU(3,3)& 6048     &  507168    &  \newline
 \hline
  PSL(2,23)& 6072     &  2234784    &  \newline
 \hline
   PSL(2,25)& 7800    &  5391800    &  \newline
 \hline
   M_{11} & 7920         &   1044192   &  \newline
 \hline
  PSL(2,27)& 9828          &   7778916   &  \newline
\end{array}$$  
Any idea about the meaning of these numbers?

Proof of $(\Rightarrow)$
Consider the relative version $$ \hat{\varphi}(H,G) = \sum_{K \in [H,G]}\mu(H,K) |G:K|.$$ Warning: $-\hat{\varphi}(H,G)$ is not the Möbius invariant of the bounded coset poset $\hat{C}(H,G)$ because $$-\mu(\hat{C}(H,G)) = \sum_{K \in [H,G]}\mu(K,G) |G:K| $$ and $\mu(K,G) \neq \mu(H,K)$ in general. 
Now if $[H,G]$ is boolean of rank $n$ then $\mu(K,G) = (-1)^n \mu(H,K)$; moreover (independently) by Theorem 3.21 of this paper, if $ \hat{\varphi}(H,G) \neq 0$ then there is an irreducible complex representation $V$ of $G$ such that $G_{(V^H)} = H$. Next, using a dual reformulation of the Crosscut Theorem with $X$ the set of atoms, we can extend the proof of Theorem 3.21 to any interval $[H,G]$ (i.e. without assuming it to be boolean). Finally by taking $H = 1$, we get that for any finite group $G$, if $\hat{\varphi}(G) \neq 0$ then there is an irreducible complex representation $V$ such that $G_{(V)} = 1$, which means that $V$ is faithful.

REPLY [6 votes]: No, the modular maximal-cyclic group $M_4(2)$, of order $16$, has a faithful irreducible complex representation (f.i.c.r.) of dimension $2$, whereas $\hat{\varphi}(M_4(2)) = 0$.
Let $B$ be the subgroup generated by the minimal subgroups of $G$. By the Crosscut Theorem, if $H  \in [1,G] \setminus [1,B]$ then $\mu(1,H) = 0$. Then $$\hat{\varphi}(G) = |G:B|  \hat{\varphi}(B)$$ So that $\hat{\varphi}(G) \neq 0 $ if and only if $\hat{\varphi}(B) \neq 0$. But we can't expect that the existence of a f.i.c.r. for $G$ implies the existence of a f.i.c.r. for $B$ (whereas the converse is true).
 In fact, $G = M_2(4)$ has a f.i.c.r. but not its $B = C_2^2$  
Improved question:  Is $(\Leftarrow)$ true if $B=G$?
[note that $B$ is normal (see here), so $B=G$ for $G$ simple]  
Answer: No, there are exactly two counter-examples of order $\le 100$: $D_8 \rtimes C_2^2$ and $D_8 \rtimes S_3$. They are of order $64$ and $96$ resp., and both check $B=G$, $\hat{\varphi}(G)=0$ and have a f.i.c.r. of dimension $4$.  
We can still specialized the question to $G$ simple.

For answering the comment of Benjamin Steinberg, we will prove an alternative result involving the normal subgroup lattice, and giving a formula shorter than in Pálfy's paper (available here).
Let $G$ be a finite group and $\mathcal{N}(G)$ its normal subgroup lattice. Let $\mu_{\mathcal{N}}$ be the Möbius function for $\mathcal{N}(G)$. Consider the dual normal Euler totient:
$$ \hat{\varphi}_{\mathcal{N}}(G) = \sum_{H \in \mathcal{N} (G)}\mu_{\mathcal{N}}(1,H) |G:H| $$ Let $V_1, \dots, V_r$ be equivalent class representatives of the irreducible complex representations of $G$.
Theorem: $\hat{\varphi}_{\mathcal{N}}(G) = \sum_{V_i \text{ faithful}} \dim(V_i)^2$.
Corollary:  $G$ has a faithful irreducible complex representation iff $\hat{\varphi}_{\mathcal{N}}(G) \neq 0$.

Proof of the theorem: 
Let the fixed-point subspace $V_i^H=\{v \in V_i \ | \ hv=v ,\ \forall h \in H  \}$ 
Claim 1: $|G:H| = \sum_i\dim(V_i) \dim(V_i^H)$
Claim 2: $H$ is a normal subgroup iff $\forall i \ V_i^H = \{0\}$ or $V_i$.
Claim 3: $V_i$ is not faithful iff there is an atom $H$ of $\mathcal{N}(G)$ such that $V_i^H = V_i$.
Consider the set $L=\{ \bigoplus_i V_i^H \  | \ H \in \mathcal{N}(G) \}$ whose lattice structure is the reverse of $\mathcal{N}(G)$.
By applying the Crosscut Theorem and the inclusion-exclusion principle, together with the claims above, the result follows. $\square$  
Reference request: does this theorem exist in the literature?