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.

  • 3
    $\begingroup$ 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. $\endgroup$ Jun 9, 2018 at 22:03
  • $\begingroup$ 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. $\endgroup$
    – Robert D-B
    Jun 9, 2018 at 22:25
  • 4
    $\begingroup$ @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 $a_1$. $\endgroup$ Jun 9, 2018 at 23:49
  • $\begingroup$ 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 $\endgroup$
    – N8tron
    Jun 10, 2018 at 0:15
  • $\begingroup$ 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. $\endgroup$
    – san
    Aug 12, 2018 at 3:58

1 Answer 1


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.


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