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\scriptsize \rm\bf Journal of the Iranian Mathematical Society\\
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\scriptsize \rm ISSN (print): ????-????, ISSN (on-line): ????-???? \\
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\scriptsize Vol. {\bf\rm x} No. x {\rm(}201x{\rm)}, pp. xx-xx.\\
\scriptsize $\copyright$ 201x Iranian Mathematical Society\\
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\title{A note on factorizations of finite groups}
\author{George M. Bergman}
\thanks{{\scriptsize
\hskip -0.4 true cm MSC(2010): Primary: 20D60. % ; Secondary: none
\newline Keywords:
factorization of a finite group as a product of subsets.\\
Received: dd mmmm yyyy, Accepted: dd mmmm yyyy.\\
Archived at \url{http://arXiv.org/abs/2003.12866}\,.
After publication, any updates, errata, related references,
etc., found will be recorded at
\url{http://math.berkeley.edu/~gbergman/papers/}.
}}
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Communicated by\;
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\begin{abstract}
In Question 19.35 of the Kourovka Notebook,
M.\,H.\,Hooshmand asks whether, given a finite group $G$
and a factorization %% numerical
$\mathrm{card}(G)= n_1\ldots n_k,$ one can always
find subsets $A_1,\ldots,A_k$ of $G$ with $\mathrm{card}(A_i)=n_i$
such that $G=A_1\ldots A_k;$ equivalently, such that the
group multiplication map $A_1\times\ldots\times\nolinebreak A_k\to G$ is
a bijection.
We show that for $G$ the alternating group on $4$ elements,
$k=3,$ and $(n_1,n_2,n_3) = (2,3,2),$ the answer is negative.
We then generalize some of the tools used in our proof,
and note a related open question.
\end{abstract}
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\section{\bf The example.}\label{S.eg}
In this section we develop the example described in the Abstract.
\begin{defn}[after {\cite[\S1]{MHH_factor}}, cf.\ {\cite[p.\,6]{S+S}}]\label{D.decomp}
Suppose $G$ is a group, $k$ is a positive integer,
and $A_1,\ldots,A_k$ are subsets of $G.$
In this situation, if the multiplication map
$A_1\times\ldots\times A_k\to G$ is
bijective, we shall write $G = A_1\cdot\ldots\cdot A_k,$
and call this a $\!(k\!$-fold\textup{)} {\em factorization} of $G.$
\textup{(}If $G$ is finite, we see that the
above bijectivity condition can
alternatively be expressed, as in \cite[Question 19.35]{Kourovka19},
by the conditions $G=A_1\ldots A_k,$
and $\mathrm{card}(G)=\mathrm{card}(A_1)\ldots\mathrm{card}(A_k).)$
\end{defn}
Observe that if $G=A_1\cdot\ldots\cdot A_k$ is a $\!k\!$-fold
factorization of $G,$ then for $1\leq j< k,$
$G=(A_1\ldots A_j)\cdot\nolinebreak(A_{j+1}\ldots A_k)$
is a $\!2\!$-fold factorization.
Though the example we are working toward is a $\!3\!$-fold
factorization, the key to its proof will be a
property of $\!2\!$-fold factorizations, namely
\begin{lem}[cf. {\cite[Cor.\ 2.14(a)]{MHH_f-rep}}]\label{L.div}
Let $G=A\cdot B$ be a factorization of a finite group.
Then the order of the subgroup of $G$ generated by $A$ is a
multiple of $\mathrm{card}(A),$
and the order of the subgroup generated by $B$ is a
multiple of $\mathrm{card}(B).$
\end{lem}
\begin{proof}
Let $H$ be the subgroup of $G$ generated by $A.$
For each $b\in B,$ the set $A\,b$ is contained in
a single right coset of $H;$ hence every right coset of $H$
is a disjoint union of sets of cardinality $\mathrm{card}(A),$
hence $\mathrm{card}(H)$ is a multiple of $\mathrm{card}(A).$
The statement about the subgroup generated
by $B$ is seen in the same way.
\end{proof}
We will also use the following observation.
\begin{lem}[{after \cite{MHH_factor}}]\label{L.e_in}
If $G=A_1\cdot \ldots\cdot A_k$ is a factorization of a
group $G,$ then for all $g,\,h\in G,$\linebreak
$(g\,A_1)\cdot A_2\cdot\ldots\cdot A_{k-1}\cdot (A_k\,h)$
is also a factorization of $G.$
Hence if for some positive integers
$n_1,\ldots,n_k,$ $G$ has a $\!k\!$-fold factorization with
$\mathrm{card}(A_i)=n_i$ $(i=1,\ldots,k),$ it has such a
factorization in which $A_1$ and $A_k$ both contain the identity
element~$e.$\qed
\end{lem}
We can now prove
\begin{prop}\label{P.2,3,2}
Let $G$ be the alternating group on $4$ elements, a
group of order~$12.$
Then $G$ has no factorization $A_1\cdot A_2\cdot A_3$
with $(\mathrm{card}(A_1),\,\mathrm{card}(A_2),\,\mathrm{card}(A_3)) = (2,3,2).$
\end{prop}
\begin{proof}
Recall that the elements of exponent $2$ in $G$ form a normal
subgroup $N\cong Z_2\times Z_2,$ that $G$ is an extension
of $N$ by a group of order $3$ whose action by
conjugation cyclically permutes the
three proper nontrivial subgroups of $N,$ and that all elements
of $G$ not in $N$ have order~$3.$
Suppose $G=A_1\cdot A_2\cdot A_3$ were a factorization
with $\mathrm{card}(A_1)=2,$ $\mathrm{card}(A_2)=3,$ $\mathrm{card}(A_3)=2.$
By Lemma~\ref{L.e_in} we can assume without loss of generality
that $A_1$ and $A_3$ have the forms
$\{e,\,g\}$ and $\{e,\,h\}$ respectively.
The orders of the groups these sets generate are the
orders of $g$ and $h,$ and by Lemma~\ref{L.div}, are even.
But the only elements of $G$ of even order have order~$2,$ hence $A_1$
and $A_3$ are in fact subgroups (which may or may not be distinct).
Since $A_1$ and $A_3$ are contained in~$N,$
for $G=A_1\,A_2\,A_3$ to hold, $A_2$ must contain
representatives of all three cosets of $N$ in $G.$
Moreover, elements of $G$ act transitively on the set of
$\!2\!$-element subgroups of $N;$ so $A_2$ must contain
an element $g$ that conjugates $A_1$ to $A_3.$
Hence when we multiply out $A_1\,A_2\,A_3,$ the result contains
$A_1\,g\,A_3 = g\,A_3\,A_3.$
But the multiplication map $A_3\times A_3\to A_3$ is not
one-to-one; from which we see that the multiplication map
$A_1\times A_2\times A_3\to G$ cannot be one-to-one,
contradicting the definition of a factorization.
\end{proof}
This completes our negative answer to \cite[Question 19.35]{Kourovka19}
for $k=3.$
What about larger $k$?
If we allow factorizations with some factors equal to $1,$
then for any $k,$ a negative example with
cardinalities $n_1,\dots,n_k$ yields negative
examples for all $k'>k,$ by keeping the same $G$ and $n_1,\ldots,n_k,$
and taking $n_{k+1}=\ldots=n_{k'}=1;$
so the only remaining open case is $k=2.$
However, Hooshmand (personal correspondence) has indicated that he
intended only factorizations of $\mathrm{card}(G)$ into factors $>1.$
With that restriction, the problem remains open for all $k>3;$ I do not
know whether there is an easy way to modify the present example
to cover those cases.
Hooshmand posed the question for $k=2$ in {\cite{MHH_overflow}}, and
refers to that case in \cite{Kourovka19} as of particular interest;
his preprint \cite{MHH_factor} concerns that case, and
also mentions the case $k>3$ at the end, as Problem~IV.
The case $k=2$ is also studied in \cite{BGC}, after a discussion
of the history of the subject of group factorizations.
\section{Strengthening our lemmas}\label{S.stronger}
In the context of Lemma~\ref{L.div}, the order of the subgroup
$H$ of $G$ generated by $A$ can change on left-multiplying
$A$ by an element $g\in G,$ a fact we implicitly used
when we applied Lemma~\ref{L.e_in}
in the proof of Proposition~\ref{P.2,3,2}.
In the next result, modified versions of that subgroup are
noted whose orders are not so affected.
Also, while Lemma~\ref{L.div} is applicable only to the first
and last sets $A_1$ and $A_k$ in a factorization
$G = A_1\cdot \ldots\cdot A_k,$
part~(iii) below obtains a similar, though
weaker, condition on the cardinalities of the other $A_i.$
(This will be slightly improved in Lemma~\ref{L.div3}.)
\begin{lem}\label{L.div2}
Let $A_1\cdot \ldots\cdot A_k$ be a factorization of a finite group $G.$
Then
\textup{(i)} $\mathrm{card}(A_1)$ divides
the order of the subgroup of $G$ generated
by the set $A_1^{-1} A_1 = \{g^{-1} h\mid g,h\in A_1\},$ which
can also be described as generated by any one of
the subsets $g^{-1} A_1$ $(g\in A_1).$
Moreover, that order is also the order of the subgroup generated
by $A_1\,A_1^{-1} = \{g\,h^{-1}\mid g,h\in A_1\},$ equivalently,
by any of the subsets $A_1\,g^{-1}$ $(g\in A_1).$
\textup{(ii)} Similarly, $\mathrm{card}(A_k)$ divides
the order of the subgroup of $G$ generated
by $A_k\,A_k^{-1},$ equivalently,
by any of the subsets $A_k\,g^{-1}$ $(g\in A_k),$
and that order is also the order of the subgroup generated
by $A_k^{-1} A_k,$ equivalently,
by any of the subsets $g^{-1} A_k$ $(g\in A_k).$
\textup{(iii)} For $1