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.

  • $\begingroup$ You might look at Pálfy, P. P. On faithful irreducible representations of finite groups. Studia Sci. Math. Hungar. 14 (1979), no. 1-3, 95–98 (1982). Mathscinet says he uses the mobius function of the lattice of normal subgroups. I don't have access. $\endgroup$ Mar 10, 2017 at 22:40

1 Answer 1


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?

  • 1
    $\begingroup$ Re "does this theorem exist in the literature?": In Pálfy's paper on p. 97, a chain of equalities is displayed, and from the 1st and the 3rd line, on can extract the equality $$ \sum_{ \substack{ \phi \in \operatorname{Irr}(G) \\ \operatorname{Ker} \phi = 1 } } ( \deg \phi )^2 = \sum_{ N \vartriangleleft G } \mu(N) | G/N|. $$ $\endgroup$ Mar 27, 2017 at 12:47
  • $\begingroup$ @FriederLadisch: you are right, thanks! $\endgroup$ Mar 27, 2017 at 14:03
  • $\begingroup$ @FriederLadisch: do you understand in what sense the (complicated) formula of Pálfy's paper is more interesting than this (easy) formula extracted above? $\endgroup$ Mar 28, 2017 at 10:48
  • $\begingroup$ I think the point is that Palfy's formula depends only on the minimal normal subgroups of $G$ (together with the action of $G$ on the abelian ones), while the "easy" formula depends on all normal subgroups. $\endgroup$ Mar 28, 2017 at 15:18

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