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: $$ \varphi(G) = \left\lvert G \setminus \bigcup_{i=1}^n M_i \right\rvert $$ 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.
