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Any $z \in \widehat{\mathbb{Z}} = \lim_{n} \mathbb{Z}/n\mathbb{Z}$ defines an operation on all finite groups: if $G$ is a finite group and $g \in G$, say $g^n=1$, then map it to $g^{z_n}$. This defines a map of sets $G \to G$, which is natural with respect to all homomorphisms of finite groups. The same works for torsion groups. And these are all operations, as can be seen from the Yoneda Lemma applied to the ind-representable forgetful functor $\mathbf{FinGrp} \to \mathbf{Set}$.

Of course, we get an actual homomorphism when $z = 0$ or $z = 1$. I believe that these are the only cases (right?).

Question. If $z \in \widehat{\mathbb{Z}} \setminus \{0,1\}$, is there always a finite group $G$ of some order $n$ such that $G \to G$, $g \mapsto g^{z_n}$ is not a homomorphism? How can we prove this?

For $z \in \mathbb{Z} \setminus \{0,1\}$, we can argue as follows: In the free group $F = \langle a,b \rangle$ we have $(ab)^z \neq a^z b^z$. Since $F$ residually finite, there is a finite group $G$ and a homomorphism $\varphi : F \to G$ such that $\varphi((ab)^z) \neq \varphi(a^z b^z)$, and we are done. (Is there a more direct proof which constructs $G$ in this case?)

Equivalently, my question is how to prove $Z(\mathbf{FinGrp}) \cong \{0,1\}$, where $Z(-)$ is the center of a category. Equivalently, $Z(\mathbf{ProFinGrp}) \cong \{0,1\}$.

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    $\begingroup$ There is a concrete construction of a finite group which shows any two words which are different in the free group are different in a finite group. Just take the finite part of the Cayley graph of the free group containing the vertices and edges used reading those words from $1$ and remove all other vertices and edges. Then you get a bunch of partial permutations of a finite set for each generator. You can extend these to permutations of that set and this gives your desired group $G$. $\endgroup$
    – Benjamin Steinberg
    Commented Nov 20, 2022 at 15:06

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It suffices to look at symmetric groups, as these are generated by transpositions. Indeed, let $z \in \widehat{\mathbf Z}$ and suppose it induces a homomorphism $\phi_z$ on $S_n$ for all $n$ given by $g \mapsto g^{z_n}$ if $g$ has order $n$. We will show that $z$ is either $0$ or $1$.

  • If $z_2 = 0 \in \mathbf Z/2\mathbf Z$, then each $\phi_z \colon S_n \to S_n$ is a homomorphism killing all transpositions, therefore $\phi_z$ is the trivial homomorphism. Thus $z_n = 0 \in \mathbf Z/n\mathbf Z$ for all $n$, as $1 = \phi_z(g) = g^{z_n}$ for any element $g \in S_n$ of order $n$.
  • If $z_2 = 1 \in \mathbf Z/2\mathbf Z$, then each $\phi_z \colon S_n \to S_n$ is the identity on all transpositions, hence $\phi_z$ is the identity. Thus $z_n = 1 \in \mathbf Z/n\mathbf Z$ for all $n$, again since $g = \phi_z(g) = g^{z_n}$ for $g \in S_n$ of order $n$. $\square$
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  • $\begingroup$ This is nice. Basically odd and even make sense in the free procyclic group and "non-trivial" even or odd order powers are not homomorphisms on symmetric groups. $\endgroup$
    – Benjamin Steinberg
    Commented Nov 20, 2022 at 16:12
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    $\begingroup$ I really love this proof! The essence is that every finite group can be embedded into a finite group which is generated by involutions, so any natural endomorphism is determined by its behaviour on involutions (and this is determined by $\mathbb{Z}/2$). $\endgroup$
    – Martin Brandenburg
    Commented Nov 20, 2022 at 18:09
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    $\begingroup$ Actually the same argument works with $D_n$, as it is also generated by elements of order $2$. This makes more explicit the elements $a,b$ such that $a^zb^z \neq (ab)^z$ if $z \neq 0,1$. You no longer get an injection $G \hookrightarrow S_n$ for every finite group $G$, but still a roof $D_n \leftarrow C_n \to G$ for every $(G,g)$. (The reason I picked $S_n$ initially is mostly that we know its endomorphisms, but that turned out to be unimportant.) $\endgroup$
    – R. van Dobben de Bruyn
    Commented Nov 22, 2022 at 8:39

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