Vtyjkyuk

This question is inspired by this question.

Given a finite group $G$, is there an ordering $G=a_1, dots, a_n$ of its elements such that the product of all group elements in that specified order equals the first element, i.e $a_1cdotdotscdot a_n=a_1$. Equivalently, can we multiply all but one element of the group in such a way to obtain the unit $1$.

Trivial cases (probably not very helpful):

• $G$ is abelian (just define $a_1$ to be the product of all elements).




• $G$ has no element of order $2$ ($a_1=1$ and pair the elements with their inverses in the list).




• $G$ has a unique element of order $2$ (set $a_1$ to be that element and pair the others with their inverses).

Slightly less trivial cases (still probably not that useful)

• $G=S_3$ (symmetric group) if $a=(12), b=(23)$ are the standard generators, then $(aba)=(aba)(a)(ba)(b)(ab)1$




• $G$ with $|G|>6$ has exactly two or three elements of order $2$: Consider the conjugation action of $G$ on the set $X$ of elements of order $2$. If this action is non-trivial, then for at least one $xin X$ the centralizer of $x$ has at most $frac2$ elements, hence we find a $gin Gsetminus X$ with $y:= gxg^-1neq x$; then the product $ygxg^-1$ followed by all the elements of order $geq3$ paired with their inverses is trivial. If the action is trivial, then $X$ lies in the center and is thus an abelian subgroup (because the product of commuting elements of order $2$ has order $leq 2$); hence we can first multiply all elements of $X$, followed by all other elements paired with their inverses.

We will prove a bit stronger result: Given an element $x_0$ of order two, we can order the elements in the group such that one of the following cases occur:

1. The product is trivial, and you can set $a_1=1$.




2. The product is equal to $x_0$ and the first element is $x_0$.

Proof: Let $X$ be the set of elements of order two and set $$X_0= xin Xsetminus x_0: xx_0=x_0x$$ and $$X_1= xin X: xx_0ne x_0x.$$

Then we have the disjoint union
$$ X=x_0cup X_0cup X_1$$
Consider the orbits $x,x_0x$ in $X_0$ corresponding to (left) multiplication by $x_0$, and the orbits $x, x_0 x x_0$ in $X_1$ corresponding to the adjunction with $x_0$.
Clearly all orbits have cardinality two.
For each pair of orbits $x,x_0 x, y, x_0 y$ in $X_0$ consider the product
$$ xcdot (x_0 x)cdot y cdot (x_0 y)= x_0 cdot x_0=1,$$

and for each orbit $ x, x_0 x x_0$ in $X_1$ set $g= x_0 x$ (so $g^-1=x x_0=x_0(x_0 x x_0)$) and consider the product
$$ g cdot x cdot g^-1cdot (x_0 x x_0)= x_0 cdot x_0=1.$$

Note that the sets $ g, g^-1$ are disjoint, since the orbits are disjoint.

So, if there is an odd number of orbits in $X_0$, set $a_1=1$, then take one orbit $ x,x_0 x$ in $X_0$, and consider the product $$ x_0 cdot x cdot (x_0 x)=1.$$
Then form the product of all elements in $G$ multiplying by the products corresponding to pairs of orbits in $X_0$, then by the products corresponding to orbits in $X_1$, and finally multiplying by pairs $ g, g^-1$ of elements in $Gsetminus X$ that have not been used in any of the previous products. Then the product of all elements is trivial.

If there is an even number of orbits in $X_0$, then set $a_1=x_0$ and multiply as before by the remaining elements. Then the product of all the elements is $x_0$, which is the first element.