# Is $Z(\Sigma) \cong E$?

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.

• 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. – YCor Mar 2 at 19:44
• 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. – YCor Mar 2 at 19:53
• 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. – YCor Mar 2 at 20:03
• @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. – Yanior Weg Mar 2 at 21:12
• 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. – Derek Holt Mar 7 at 8:24