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.

  • $\begingroup$ If you want $\Sigma$ to be a set, you probably want $\mathrm{Fin}$ to be the class of isomorphism classes of groups, or suppose that $f$ is constant on isomorphism classes. $\endgroup$ – YCor Mar 2 at 19:44
  • $\begingroup$ The claim that $\ast$ is associative is false. Indeed, both $f\ast (g\ast h)(G)$ and $(f\ast g)\ast h(G)$ can be described as $\sum_{K,H}f(K)g(H/K)h(G/H)$, with $K\le H\le G$ with $H$ normal, but in the first case, $K$ has to be normal in $G$, while in the second case it only has to be normal in $H$. Finding an explicit triple for which associativity fails is then an easy exercise. $\endgroup$ – YCor Mar 2 at 19:53
  • $\begingroup$ Bonus: check that there exists $f$ such that $(f\ast f)\ast f\neq f\ast (f\ast f)$. Thus $f^n$ is not well-defined. $\endgroup$ – YCor Mar 2 at 20:03
  • $\begingroup$ @YCor, thank you for noticing the mistakes in my definition, that render my question meaningless. I corrected them by replacing the class of all finite groups with the class of all finite T-groups, as T-groups are both normal subgroup-closed and quotient-closed and my argument still seems to work in case, when normality is transitive. $\endgroup$ – Yanior Weg Mar 2 at 21:12
  • $\begingroup$ I suspect that a large proportion of people who work in group theory would not know immediately what $T$-group is. Of course it is not difficult to find out using a search, but you really should define it in the post. $\endgroup$ – Derek Holt Mar 7 at 8:24

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