Let $n \in \mathbb{N}$. Is it true that for every $k \in \{1, \dots, n!\}$ there are some group $G$ and pairwise distinct elements $g_1, \dots, g_n \in G$ such that the set $\{g_{\sigma(1)} \cdot \ \dots \ \cdot g_{\sigma(n)} \ | \ \sigma \in {\rm S}_n\}$ of all products of the $g_i$ obtained by permuting the factors has cardinality $k$?

**Added on Mar 21, 2013:** On Mar 19, 2013, Benjamin Young gave a presentation of
a group in which the above set has cardinality $\leq k$ (see below). As such this is
trivial, since this is the case in particular in every abelian group. The thing that
is missing to turn this into an answer to the question is to prove that the given
relations do not force equality among any other products.

The assertion is true at least for $n \leq 4$. In case $n = 4$, for all $k \in \{1, \dots, 4!\}$ suitable 4-tuples of group elements can be taken from ${\rm S}_5$:

```
k = 1: (), (1,2), (3,4), (1,2)(3,4) [ 215]
k = 2: (), (1,2)(3,4), (1,2,3,4), (2,4) [ 1845]
k = 3: (), (2,3), (2,3,4), (3,4) [ 10230]
k = 4: (), (1,2), (2,3), (3,4) [ 38870]
k = 5: (), (1,2), (2,3), (2,3,4) [ 85350]
k = 6: (), (1,2), (2,3), (2,4) [ 186220]
k = 7: (1,2), (1,2,3,4), (2,3,4), (2,4,3) [ 7920]
k = 8: (1,2), (2,3), (2,3,4), (2,4,3) [ 40560]
k = 9: (1,2), (2,3), (2,3,4), (3,4) [ 39535]
k = 10: (1,2), (1,2)(3,4), (2,3), (2,3,4), [ 96240]
k = 11: (1,2), (1,2)(3,4), (2,3), (2,4) [ 116715]
k = 12: (1,2), (1,2,3), (2,4,3), (2,4) [ 264360]
k = 13: (1,2), (1,2,3), (1,5), (2,3,4) [ 284020]
k = 14: (1,2), (1,2,3), (1,5), (2,4) [ 449940]
k = 15: (1,2,3), (1,2,3,4), (1,5), (2,3,4) [ 525420]
k = 16: (1,2), (1,2,3), (1,5), (2,4,3) [ 814070]
k = 17: (1,2,3), (1,2,3,4), (1,5), (2,3) [1034670]
k = 18: (1,2)(3,4), (1,2,3,4), (1,2,4,3), (1,5) [1208650]
k = 19: (1,2), (1,2,3), (1,2,3,4), (1,5) [1199930]
k = 20: (1,2,3,4), (1,2,4,3), (1,3,2,4), (1,5) [ 968760]
k = 21: (1,2)(3,4), (1,2,3,4), (1,5)(3,4), (2,3) [ 527160]
k = 22: (1,2)(3,4), (1,3), (1,4), (1,5) [ 242340]
k = 23: (1,2,3), (1,2,3,4), (1,5)(3,4), (2,4,3) [ 63240]
k = 24: (1,2), (1,3), (1,4), (1,5) [ 8310]
```

In brackets (added on Sep 18, 2013): the numbers of such 4-tuples of elements of ${\rm S}_5$.

**Added on Sep 18, 2013:** In case $n = 5$, at least for all
$k \in \{1, \dots, 5!\} \setminus \{117, 119\}$, suitable
5-tuples of group elements can be taken from ${\rm S}_6$ -- see 5tuples.txt.

**Update (December 2013):** This question will appear as
Problem 18.50 in:

Kourovka Notebook: *Unsolved Problems in Group Theory*. Editors V. D.
Mazurov, E. I. Khukhro. 18th Edition, Novosibirsk 2014.