# Multiplying all the elements in a group

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.

• It would be useful to use distinct notation for the multiset and the underlying set.
– YCor
Mar 1, 2020 at 8:29
• 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.) Mar 1, 2020 at 8:40
• @James Why not post that as a proper answer to the question? Mar 1, 2020 at 19:09
• @James: Golomb or Fuchs?
– user6976
Mar 1, 2020 at 20:06

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.