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Let $G = \{ g_i | i = 1, ...,n \}$ be a finite group and denote by $G!$ the multiset consisting of all the products of all different elements of $G$ in any order, that is $$ G! = [ \prod_i g_{\sigma(i)} | \sigma \in S_n] $$.

I'm interested in knowing how $G!$ behaves as a set, and also how often does every element appear (i.e., how it behaves as a multiset).

In the case of an abelian $G$, $G!$ is either 1 or (iff exists) the single element of order 2 in $G$.

We can use the abelian case to get a (seemingly tight) upper bound on $G!$ as a set, by considering modding by $[G,G]$: projecting every such product to the quotient, which is abelian, we get $a^{\#[G,G]} \in G/[G,G]$ where $a$ is the single element of order 2 in $G/[G,G]$ if it exists, and the identity otherwise. Thus, if either $a$ is the identity or $\#[G,G]$ is even, $G!$ is entirely contained in $[G,G]$, and otherwise contained in the coset corresponding to $a$.

The reason this bound seems tight is that it's generally easy to get elements in the commutator group (up to the element of order 2): if $a,b,a^{-1},b^{-1}$ are distinct, $[a,b] \in G!$ by putting every other element of $G$ right next to it's inverse except for $a^{\pm1}, b^{\pm1}$ and likewise for products of commutators and so on.

Is it always the case that $G!$ is a coset of the commutator group? How often does each element appear? It might also be useful to look at the action of $Aut(G)$ on $G!$, but I'm not totally sure what can that tell us.

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  • $\begingroup$ It would be useful to use distinct notation for the multiset and the underlying set. $\endgroup$ – YCor Mar 1 '20 at 8:29
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    $\begingroup$ Yes, your $G!$ (as a set) is always either $[G,G]$ (if the order of $G$ is odd, or its Sylow $2$-subgroup is non-cyclic) or $z[G,G]$ if $G$ has cyclic Sylow $2$-subgroup, where $z$ is the involution in the Sylow $2$-subgroup. This was apparently a conjecture of Golomb, proved by Denes and Hermann. (I think the multiplicity is just the order of the derived subgroup.) $\endgroup$ – James Mar 1 '20 at 8:40
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    $\begingroup$ @James Why not post that as a proper answer to the question? $\endgroup$ – Johannes Hahn Mar 1 '20 at 19:09
  • $\begingroup$ @James: Golomb or Fuchs? $\endgroup$ – user6976 Mar 1 '20 at 20:06
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Yes, your $G!$ (as a set) is always either $[G,G]$ (if the order of $G$ is odd, or its Sylow $2$-subgroup is non-cyclic) or $z[G,G]$ if $G$ has cyclic Sylow $2$-subgroup, where $z$ is the involution in the Sylow $2$-subgroup. This was apparently a question/conjecture of Golomb (see p. 973) and independently of Fuchs (in a 1964 seminar), proved by Dénes and Hermann.

(Addressing Mark Sapir's comment, I could not find a published reference for Fuchs, but did manage to track down this paper by Dénes and Keedwell which contains a discussion of some of the history of the question.)

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