A nonempty subset $D$ of a group $G$ is called

$\bullet$ decomposable if $D\subseteq DD$, that is every element $x\in D$ is can be written as the product $x=yz$ of some elements $y,z\in D$;

$\bullet$ product-one if there exists $n\in\mathbb N$ and pairwise distinct elements $x_1,\dots,x_n\in D$ such that $x_1\cdots x_n=1$.

Problem 1. Let $D$ be a finite decomposable subset of a group. Is $D$ product-one?

Remark 1. For commutative groups this problem was posed by Gjergji Zaimi and solved affirmatively by Lev, Nagy, and Pach.

Remarks 2. For some non-commutative groups like generalized dihedral groups the answer to Problem is also affirmative, see my partial answer below. This partial answer suggests the following

Problem 2. Let $G$ be a group containing an Abelian subgroup of index 2. Is every finite decomposable set in $G$ product-one?

  • 2
    $\begingroup$ Suppose $G=\text{Sym}(X)$ for some finite $X$ and $d$ is the metric on $G$ defined by $d(f,g)=|\{x\in X\mid f(x)\neq g(x)\}|$, and $L$ is the loss function where $L(D)=\sum_{g\in D}d(g,DD)$. Then $L$ measures how close $D$ is to being decomposable and $L(D)=0$ iff $D$ is decomposable. I have therefore tried to find decomposable subsets of $\text{Sym}(X)$ simply by minimizing $L(D)$ using evolutionary algorithms and artificial intelligence, but all my examples were trivial in the sense that they always had subsets of the form $\{x,x^{-1}\}$ where $x^{2}\neq e$ or $x=e$. $\endgroup$ Jun 26, 2021 at 21:11
  • 3
    $\begingroup$ Few obvious comments, for finite $G$. For a minimal counterexample (or an inductive proof), we can assume that $1 \not\in D$ and that $D$ is not contained in any maximal subgroup of $G$. With minimality we can also assume that no nonempty proper subset of $D$ is decomposable, so for all $x \in D$, the set $D \setminus \{x\}$ is not decomposable. Then for all $x \in D$, we must have $D \subseteq Dx \cup xD \cup \{x\}$. $\endgroup$
    – spin
    Jun 27, 2021 at 9:23
  • $\begingroup$ @spin We can also assume that $D$ contains no elements of order 2. $\endgroup$ Jun 27, 2021 at 15:24
  • $\begingroup$ @TarasBanakh Can you expand on that last comment? $\endgroup$
    – verret
    Jun 27, 2021 at 22:43
  • 1
    $\begingroup$ Possibly you view it as trivial, but anyway a simple remark is that there exist $0<m\le |D|$ and $x_1,\dots,x_m\in S$ (possibly not distinct) such that $\prod x_i=1$. Indeed, make $D$ an oriented graph with $x\to y$ if $x\in yD$. By assumption for every $x$ there exists $y$ with $x\to y$. So there is an oriented simple loop of size $0<m\le |S|$: $y_0,\dots,y_{m-1}$, with $y_i=y_{i+1}x_i$ for some $x_i\in D$, $i$ modulo $m$. Hence $y_0=y_1x_0=y_2x_1x_0=\dots=y_{m-1}x_{m-2}\dots x_0=y_0x_{m-1}\dots x_0$, so $x_{m-1}\dots x_0=e$. $\endgroup$
    – YCor
    Jul 3, 2021 at 7:27

3 Answers 3


Also an extended comment.

We can consider a (finite) set $D$ with two self-maps $u,v:D\to D$, and consider the group $G_{u,v}$ of presentation $$G_{u,v}=\langle x:x\in D\mid x=u(x)v(x),\forall x\in D\rangle.$$ The question is equivalent to whether there for every nonempty finite set $D$ and $u,v$ there exists an injective nonempty product of elements of $D$ representing $e$ in $G_{u,v}$.

From previous comments by OP, this holds if $G_{u,v}$ is commutative, or if the image of $D$ in $G_{u,v}$ has an element of order $\le 2$.

An observation:

Proposition 1: if $(D,u,v)$ is a counterexample, then $u$ is non-injective on every $v$-cycle, and vice versa (in particular, $u,v$ are non-injective). Also, every cycle of $u$ or $v$ has length $\ge 3$.

Indeed, if $x=u^nx,ux,\dots,u^{n-1}x$ is a $n$-cycle of $u$, then in $G_{u,v}$, $x=u(x)v(x)=u^2x.vux.vx=\dots u^nx.vu^{n-1}x\dots vux.vx$, so $vu^{n-1}x\dots vux.vx$. Since $(D,u,v)$ is a counterexample, it follows that the elements $vu^{n-1}x,\dots ,vux,vx$ are not pairwise distinct.

If there is a $1$-cycle of $u$, say $ux=x$, then $x=ux.vx=x.vx$, so $vx=1$. If there is a $2$-cycle of $u$, say $u^2x=x$, the $x=ux.vx=u^2x.vux.vx$, so $vux=vx$, and in turn $(vx)^2=e$. But the case when $D$ has an element of order $\le 2$ was already excluded.

The other statements hold by symmetry.$\Box$

This reproves that $|D|\le 3$ is excluded, since there should be a cycle, say of length $n$; by non-injectivity $n<|D|$, and by the above, $n\ge 3$, so $|D|\ge 4$. Let's now exclude $D=4$.

Proposition 2 If $(D,u,v)$ is a counterexample then $|D|\ge 5$.

Let me write $i$ instead of $x_i$. So, there is a 3-cycle of $u$, and $v$ is non-injective on it. Up to reindex, $D=\{1,2,3,4\}$ $u:1\mapsto 2\mapsto 3\mapsto 1$ and $v(1)=v(2)$.

  1. suppose $v(1)\neq 4$. Since $v$ has no fixed point, we deduce $v(1)=v(2)=3$: $1=23,2=33,3=1*$. Since $\{1,2,3\}$ is not a counterexample, we get $3=14$. Hence $1,2\in\langle 3\rangle$, and in turn $4\in \langle 1,2,3\rangle=\langle 3\rangle$. So $G_{u,v}$ is cyclic and this case (commutative) is already discarded.
  2. so $v(1)=4$: $1=24,2=34,3=1i$. Then $1=24=344=1i44$, hence $i44=e$. If $i=2$ or $i=3$ this yields $14=e$ or $24=e$; also $i=4$ is impossible since $v$ has a 3-cycle. So $v(3)=1$: $1=24$, $2=34$, $3=11$. Then $1=24=344=1144$, so $144=e$, thus $1\in\langle 4\rangle$, so $3=11\in\langle 4\rangle$, and $2=34\in\langle 4\rangle$. Thus $\langle 1,2,3,4\rangle$ is cyclic, contradiction.
  • $\begingroup$ @TarasBanakh ($e$ neutral element) If there is a 2-cycle, say $a\mapsto b\mapsto a$, so $a=bc$, $b=ad$, for some $c,d\in D$. Hence $a=bc=adc$, so $dc=1$. If $d\neq c$, $dc=e$ contradicts $D$ being non-product-one. So $d=c$, and hence $c^2=e$, and this is excluded since you checked $D$ has no element of order $2$. $\endgroup$
    – YCor
    Jul 15, 2021 at 15:48

GAP shows that the groups SmallGroup(27,3), SmallGroup(27,4), SmallGroup(36,11), SmallGroup(39,1) SmallGroup(48,3) do contain many 5-element decomposable sets, which are not product-one. So, the lower bound 5 for the smallest cardinality of a counterexample, obtained by @YCor in his answer, is the best possible.

Below I write down 5-element decomposable non-product-one sets found by GAP in the groups

SmallGroup(27,3): [ f1, f2, f1 * f2, f1^2 * f2, f1 * f2^2 ]

SmallGroup(27,4): [ f1, f2, f1 * f2 * f3, f1^2 * f2 * f3, f2^2 * f3^2 ]

SmallGroup(36,11): [ f1, f2 * f3, f1^2 * f3, f1 * f2^2 * f3, f1^2 * f2^2 * f4 ]

SmallGroup(39,1): [ f1, f2, f1 * f2, f1^2 * f2, f2^4 ]

SmallGroup(48,3): [ f1, f2, f1 * f2, f2 * f3, f1^2 * f2 ]

These 5 groups are the only groups of order $\le 50$ that contain decomposable non-product-one sets.

  • $\begingroup$ I'd be curious of explicit values of such 5-element subsets in groups of order 27 or 39. $\endgroup$
    – YCor
    Jul 5, 2021 at 7:00
  • 2
    $\begingroup$ Thanks! Indeed I got the (27,3) example. Namely, if in an arbitrary group $x,y$ are non-commuting elements of order 3 such that the abelianization of $\langle x,y\rangle$ has order $9$, then $\{x,y,xy,x^2y,xy^2\}$ works. This applies in the non-abelian group of order 27 and exponent 3. $\endgroup$
    – YCor
    Jul 5, 2021 at 7:41

This is not an answer, but too long for a comment. Below I write down some conditions (on a group or a decomposable set) guaranteeing that a decomposable set in a group is product-one.

Proposition 1. Let $G$ be a group containing an abelian subgroup $A$ of index 2 such that for every $x\in A$ and $y\in G\setminus A$ we have $yx=x^{-1}y$. Every finite decomposable set $D$ in $G$ is product-one.

Proof. First observe that the intersection $D\cap A$ is not empty. Otherwise, $D\subseteq DD=(D\setminus A)\cdot(D\setminus A)\subseteq (G\setminus A)(G\setminus A)=A$ would be empty. If $D\cap A$ contains a decomposable subset of the abelian group $A$, then it is product-one by the result of Lev, Nagy, and Pach. So, we assume that $D\cap A$ contains no decomposable subset. In particular, $D$ does not contain the identity 1 of the group $G$. Since $D\cap A$ is not decomposable, there exists an element $x\in D\cap A$ such that $x=yz$ for some elements $y,z\in D\setminus A$. It follows from $1\notin D$ that $y,z$ differ from $x$.

First we assume that $y\ne z$. Then $yxz=x^{-1}yz=x^{-1}x=1$, witnessing that $D$ is product-one.

Now assume that $y=z$. In this case $x=yz=y^2$. Since $D$ is decomposable, $y=uv$ for some elements $u,v\in D$. Since $1\notin D$, the elements $u,v$ differ from $y$. Assuming that $u=v$, we conclude that $y=uv=u^2\in A$ as the group $A$ has index 2 in $G$. But this contradicts the choice of the points $y=z\notin A$. So, $u\ne v$. If $x\notin\{u,v\}$, then $yxuv=x^{-1}yuv=x^{-1}y^2=x^{-1}x=1$ and we are done. So, we assume that $x\in\{u,v\}$. If $x=u$, then $x=y^2=uvy=xvy$ implies $vy=1$. If $x=v$, then $x=y^2=yuv=yux$ implies $yu=1$. In both cases we have found two elements in $D$ whose product is equal 1, witnessing that $D$ is product-one.

Corollary. Decomposable sets in the dihedral groups $D_{2n}$ and dicyclic groups $Q_{4n}$ are product-one.

Proposition 2. If a finite decomposable subset $D$ of a group contains an element of order 2, then $D$ is product-one.

Proof. If $D$ contains the identity of the group, then $D$ is product-one. So, we assume that $1\notin D$. If $D$ contains a sequence of pairwise distinct points $x_0,\dots,x_n$ such that $x_{k-1}=x_k^2$ for any positive $k<n$ and $x_n=x_0^2$, then $$x_0=x_1^2=x_1x_2^2=\dots =x_1x_2\cdots x_{n-1}x_n^2=x_1x_2\cdots x_nx_0^2$$which implies that $x_1x_2\cdots x_nx_0=1$ and hence $D$ is product-one. So, we assume that $D$ does not contain such a sequence. Fix any element $x\in D$ of order 2 and let $n\in\mathbb N$ be the largest number for which there exists a sequence $x_0,\dots,x_n$ in $D$ such that $x_0=x$ and $x_{k-1}=x_k^2$ for all positive $k\le n$. Our assumption guarantees that $n$ is well-defined and the points $x_0,\dots,x_n$ are pairwise distinct. By the decomposability of $D$, there exist two element $y,z\in D$ such that $x_n=yz$. The maximality of $n$ guarantees that $y\ne z$. If the doubleton $\{y,z\}$ is disjoint with the set $\{x_0,\dots,x_n\}$, then $$1=x^2=x_0x_1\cdots x_nyz$$and hence $D$ is product-one. So we assume that $\{y,z\}\cap\{x_0,\dots,x_n\}\ne\emptyset$ and find the largest number $i$ such that $x_i\in\{y,z\}$. It follows from $x_n=yz$ and $1\notin D$ that $i<n$. If $x_i=z$, then $$x_i=x_{i+1}^2=x_{i+1}\cdots x_nyz=x_{i+1}\cdots x_nyx_i$$implies that $x_{i+1}\cdots x_ny=1$, witnessing that $D$ is product-one. If $x_i=y$, then $$x_i=x_{i+1}^2=yzx_n\cdots x_{i+1}=x_izx_n\cdots x_{i+1}$$ implies that $zx_n\cdots x_{i+1}=1$, which means that $D$ is product-one.

Proposition 3. Every decomposable set $D$ of cardinality $|D|\le 3$ is product-one.

Proof. If $D=\{a\}$ is a singleton, then $a=a^2$ and $a=1$, which means that $D$ is product-one. Now assume that $D=\{a,b\}$ is a doubleton and $1\notin D$. Then $a=b^2$ and $b=a^2=b^4$, which implies that $ab=b^3=1$ and hence $D$ is product-one. Finally, assume that $D=\{a,b,c\}$ and $D$ contains no decomposable subsets of cardinality $<3$.

First we assume that some element of $D$ is the square of some other element of $D$. We lose no generality assuming that $b=a^2$. By the decomposability of $D$, $c\in\{a^2,b^2,ab,ba\}=\{a^2,a^4,a^3\}$ and hence $\{a,b,c\}$ is a decomposable subset of the abelian group $\{a^n:n\in\mathbb Z\}$. By the result of Lev, Nagy and Pach, the decomposable set $D$ is product-one.

If $D$ contains an element of order 2, then $D$ is product-open by Proposition 2.

It remains to consider the case when no element of $D$ is the square of another element of $D$ and no element of $D$ has order 2. By the decomposability of $D$, $a=bc$ or $a=cb$. We lose no generality assuming that $a=bc$ and hence $c=b^{-1}a$. By the decomposability of $D$, $b=ac$ or $b=ca$. If $b=ac$, then $b=ac=ab^{-1}a$ and hence $c^2=(b^{-1}a)^2=1$, which is forbidden by our assumption. So, $b=ca=b^{-1}a^2$ and hence $b^2=a^2$.

By the decomposability of $D$, $c=ab$ or $c=ba$. If $c=ba$, then $ba=c=b^{-1}a$ and hence $b^2=1$, which is forbidden by our assumption. So, $ab=c=b^{-1}a$ and hence $aba^{-1}=b^{-1}$. Now consider the group $H$ generated by the elements $a,b$ and the cyclic subgroup $B$ generated by the element $b$. It follows from $a^2=b^2\in B$ and $aba^{-1}=b^{-1}\ne b$ that the group $B$ has index 2 in $H$ and for every elements $x=b^n\in B$ and $y=ab^m\in H\setminus B$ we have $yx=ab^mb^n=ab^nb^m=b^{-n}ab^m=x^{-1}y$. Since $D=\{a,b,c\}=\{a,b,b^{-1}a\}\subseteq H$, we can apply Proposition 1 and conclude that the decomposable set $D$ is product-one.


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