# 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.

• If the group is commutative, you can pick $a_1=\prod_{g\in G} g$ (as the product does not depend on the order of multiplication). In the case where there is no element of order two, you can choose $a_1=e_G$ (the neutral element) and order the rest in such a way that you group $g$ and $g^{-1}$ next to each other. I have no idea how to do it in the general case. Commented Jun 9, 2018 at 22:03
• If $G$ is non-abelian, fixing an arbitrary $a_1 \in G$ leaves you with $(n - 1)!$ different ways to choose the remaining terms. Some of the resulting products may repeat, but perhaps "enough" of them do not to sweep out the group itself. Commented Jun 9, 2018 at 22:25
• @rwbogl That might work for perfect groups. But the product of all elements of $G$ is well-defined up to $G'$, so if $G$ has a nontrivial abelianization $G/G'$, then there are elements that won't work as $a_1$. Commented Jun 9, 2018 at 23:49
• Note this is equivalent to being able to find a product of all elements is the identity. If $a_1=a_2\cdots a_n$ then $a_1^{-1}=a_n^{-1}\cdots a_2^{-1}$ err nevermind it's all the elements not all but one Commented Jun 10, 2018 at 0:15
• If there are 4 elements of order two, if two of them commute, then their product is another element of order two, and the product of the three elements is trivial.
– san
Commented Aug 12, 2018 at 3:58

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 $$X_0=\{ x\in X\setminus \{x_0\}: \ xx_0=x_0x\}$$ 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.