I am wondering if there is any literature on general formula of the Moebius function of subgroup lattices of any finite abelian group $G$? What I know is


When $G$ is cyclic, the Moebius function is simply the classical number theoretic one.


When $G=(\mathbb Z/p\mathbb Z)^r$, the formula involves the number of $k$-dimensional linear subspace of $G$.


But is there some formula for any finite abelian group?



Moebius function of finite abelian groups
Since every interval of the subgroup lattice $\mathcal{L}(G)$ of a
finite abelian group $G$ is isomorphic to the subgroup lattice of some
finite abelian group, we can restrict ourselves to
$\mu(\hat{0},\hat{1})$, where $\hat{0}$ is the bottom element (the
trivial subgroup of $G$) and $\hat{1}$ is the top element (the group
$G$ itself) of $\mathcal{L}(G)$. If $G$ and $H$ have relatively prime
orders, then $\mathcal{L}(G\times H)= \mathcal{L}(G) \times
\mathcal{L}(H)$. Thus (EC1, second ed., Proposition 3.8.2)
$\mu\_{G\times H}(\hat{0},\hat{1}) =
\mu\_G(\hat{0},\hat{1})\mu\_H(\hat{0},\hat{1})$, so we can assume that $G$
has prime power order $p^n$. Then $\mathcal{L}(G)$ is atomic (i.e.,
$\hat{1}$ is a join of atoms, or $G$ is generated by the subgroups of
order $p$) if and only if $G$ is elementary abelian (a product of
groups of order $p$). Since for any finite lattice $L$,
$\mu(\hat{0},\hat{1})\neq 0$ implies that $L$ is atomic (EC1,
Corollary 3.9.5), we have $\mu(\hat{0},\hat{1})=0$ unless $G$ is
elementary abelian. Finally, if $G$ is elementary abelian of order
$p^n$, then it is well-known (e.g., EC1, equation (3.34)) that
$\mu(\hat{0},\hat{1})=(-1)^n p^{{n\choose 2}}$.


