Suppose $FT$ is the class of all isomorphism classes of finite T-groups (a T-group is a group, where all subnormal subgroups are normal). Define, $\Sigma$ as the set of all functions $f$ from $FT$ to $\mathbb{Z}$, such that $f(E) = 1$, where $E$ is the trivial group. Now, if $f, g \in \Sigma$, define $$(f \ast g) (G)= \Sigma_{H \triangleleft G} f(H)g(\frac{G}{H}).$$
It is not hard to see, that $(\Sigma, \ast)$ is a group:
$f \ast (g \ast h) = \Sigma_{K \triangleleft H \triangleleft G} f(K)g(\frac{H}{K})h(\frac{G}{H}) = (f \ast g) \ast h$
The function $e$, such that $e(E) = 1$ and $e(G) = 0$ for any non-trivial $G$, is the identity element.
The inverse to $f$ is the function $f^{-1}$ satisfying the recurrent relation $\Sigma_{H \triangleleft G} f(H)f^{-1}(\frac{G}{H}) = 0$ and $f^{-1}(E) = 1$.
Moreover, it is quite easy to prove that this group is torsion-free:
Suppose $f \in \Sigma$, $G$ is the nontrivial group of minimal order, such that $f(G) \neq 0$. Then $\forall H \triangleleft G$ if $H \neq G$ and $H \neq E$, then $f(H) = 0$ (as $|H| \leq |G|$). Then $f^n(G) = f(G) + f^{n-1}(G) = nf(G) \neq 0$. That means $\forall n \in \mathbb(N) f^n \neq e$
Personally, I think that this group is also very likely to be centerless, but the only thing I managed to prove in this direction was «If $f \in Z(\Sigma)$ and $G$ is simple, then $f(G) = 0$»:
Define $g_H \in \Sigma$ as a function, such that $g(H) = 1$ and $G(K) = 0$ for all non-trivial groups $K$ non-isomorphic to $H$. If $G \cong C_2$, then $0 = g_{C_3} \ast f (S_3) - f \ast g_{C_3} (S_3) = f(S_3) + f(C_2) - f(S_3) = f(C_2)$. If $G$ is simple, but not isomorphic to $C_2$, then it has an automorphism $a$ of order $2$. So $ 0 = f \ast g_{C_2} (G \rtimes \langle a \rangle) - g_{C_2} \ast f (G \rtimes \langle a \rangle) = f(G \rtimes \langle a \rangle) + f(G) - f(G \rtimes \langle a \rangle) = f(G)$.
However, it is not enough to prove that $Z(\Sigma) = E$ and here I am stuck.
