can one order the elements of a finite group such that their product is equal to the first element in the list? 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 $a_1\cdot\dots\cdot a_n=a_1$. 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 $x\in X$ the centralizer of $x$ has at most $\frac{|G|}{2}$ elements, hence we find a $g\in G\setminus X$ with $y:= gxg^{-1}\neq x$; 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.
 A: 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:


*

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

*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 $$X_0=\{ x\in X\setminus \{x_0\}: \ xx_0=x_0x\}$$ and $$X_1=\{ x\in X: \ xx_0\ne x_0x\}.$$
Then we have the disjoint union
$$ X=\{x_0\}\cup X_0\cup X_1$$
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
$$ x\cdot (x_0 x)\cdot y \cdot (x_0 y)= x_0 \cdot x_0=1,$$
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
$$ g \cdot x \cdot g^{-1}\cdot (x_0 x x_0)= x_0 \cdot x_0=1.$$
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 $G\setminus X$ 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.
