Pseudocompact paratopological groups that are topological
Alex Ravsky
We obtain necessary and sufficient conditions when a pseudocompact pa\-ra\-to\-po\-lo\-gi\-cal group is topological. (2-)pseudocompact and countably compact paratopological groups that are not topological are constructed. It is proved that each 2-pse\-u\-do\-com\-pact paratopological group is pseudocompact and that each Hausdorff $\sigma$-compact pseudocompact paratopological group is a compact topological group. Our particular attention is devoted to periodic and topologically periodic pseudocompact paratopological groups.
1406.2001.tex
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\begin{document}
\title[Pseudocompact paratopological groups that are topological]
{Pseudocompact paratopological groups that are topological}
\author{Alex Ravsky}
\email{oravsky@mail.ru}
\address{Department of Functional Analysis, Pidstryhach Institute for Applied Problems of Mechanics and Mathematics
National Academy of Sciences of Ukraine,
Naukova 2-b, Lviv, 79060, Ukraine}
\keywords{paratopological group,
continuity of the inverse,
totally countably compact paratopological group,
countably compact paratopological group,
2-pseudocompact paratopological group,
saturated paratopological group,
topologically periodic paratopological group,
periodic paratopological group,
pseudocompact topological group,
countably compact topological group,
compact space,
countably pracompact space}
\subjclass{22A15, 54H99,54H11}%MSC 2000
\begin{abstract}
We obtain necessary and sufficient conditions when
a pseudocompact pa\-ra\-to\-po\-lo\-gi\-cal group is topological.
(2-)pseudocompact and countably compact paratopological groups that are not topological are constructed.
It is proved that each 2-pse\-u\-do\-com\-pact paratopological group is pseudocompact and that
each Hausdorff $\sigma$-compact pseudocompact paratopological group is a compact topological group.
Our particular attention is devoted to periodic and topologically periodic
pseudocompact paratopological groups.
\end{abstract}
\maketitle \baselineskip15pt
\section{Introduction}
Let $G$ be a group endowed with a topology $\tau$. A pair $(G,\tau)$
is called a {\it semitopological group} provided the multiplication
$\cdot:G\times G\to G$ is separately continuous. Moreover, if
the multiplication is continuous then $(G,\tau)$ is called a {\it paratopological group}
and the topology $\tau$ is called a {\it semigroup topology} on $G$.
Moreover, if the inversion $(\cdot)^{-1}:G\to G$ is continuous with
respect to the topology $\tau$, then $(G,\tau)$ is a {\it topological group}.
Properties of paratopological groups, their
similarity and difference to properties of topological groups,
are described in book \cite{ArhTka} by Alexander V. Arhangel'skii
and Mikhail G. Tkachenko, in author's PhD thesis~\cite{Rav3} and
papers~\cite{Rav}, ~\cite{Rav2}. New Mikhail Tkachenko's
survey~\cite{Tka2} exposes recent advances in this area.
A standard example of a paratopological group failing to be a
topological group is the Sorgenfrey line, that is the real
line endowed with the Sorgenfrey topology (generated by the
base consisting of half-intervals $[a,b)$, $a<b$).
But it turns out that the inversion on a paratopological group automatically is continuous
under some topological conditions. The search and the investigation of these conditions
is one of main branches in the theory of paratopological groups and it
has a long history.
In 1936 Montgomery~\cite{Mon} showed that every completely metrizable
paratopological group is a topological group. In 1953 Wallace
[Wal] asked: {\it is every locally compact regular semitopological
group a topological group?} In 1957 Ellis obtained a positive
answer of the Wallace question. In 1960 Zelazko used Montgomery's
result and showed that each completely metrizable semitopological
group is a topological group. Since both locally compact and
completely metrizable topological spaces are \u{C}ech-complete (I recall that
\u{C}ech-complete spaces are $G_\delta$-subspaces of Hausdorff compact spaces),
this suggested Pfister ~\cite{Pfi} in 1985 to ask: {\it is every
\u{C}ech-complete semitopological group  a topological group?} In
1996 Bouziad~\cite{Bou} and Reznichenko~\cite{Rez2} independently
answered affirmatively to the Pfister's question. To do this, it
was sufficient for to show that each \u{C}ech-complete
semitopological group is a paratopological group since earlier, N.
Brand~\cite{Bra} had proved that every \u{C}ech-complete
paratopological group is a topological group.
But a problem when a paratopological group is a topological group
has many branches which still are growing by efforts of people throughout the world.
An interested reader can find known results on this subject
in \cite[Introduction]{AlaSan}, in the survey \cite[Section 5.1]{Rav3},
in Section 3 of the survey \cite{Tka2} and farther in the present paper
in which we continue this investigation.
We obtain necessary and sufficient conditions when
a pseudocompact pa\-ra\-to\-po\-lo\-gi\-cal group is topological,
construct (2-)pseudocompact and countably compact paratopological groups that are not topological,
prove that each 2-pse\-u\-do\-com\-pact paratopological group is pseudocompact and that
each Hausdorff $\sigma$-compact pseudocompact paratopological group is a compact topological group.
Our particular attention will be devoted to periodic and topologically periodic
pseudocompact paratopological groups.
The present paper is a first, slightly updated, part of the authors preprint~\cite{Rav7}
which grew too large and will be split into several articles.

\section{Definitions} Now it is the time and the place to recall the definitions of the used
notions.

In this paper the word "space" means "topological space".

{\it Conditions which are close to compactness}. In general topology an essential role play compact spaces which have nice properties.
These properties are so nice that even spaces satisfying slightly weaker conditions
are still useful. We recall that a space $X$ is called
\begin{itemize}
\item {\it sequentially compact} if each sequence of $X$ contains
a convergent subsequence,
\item {\it $\omega$-bounded} if each countable subset of $X$ has the compact closure,
\item {\it totally countably compact} if each sequence of $X$ contains
a subsequence with the compact closure,
\item {\it countably compact at a subset} $A$ of $X$ if each infinite
subset $B$ of $A$ has an accumulation point $x$ in the space $X$
(the latter means that each neighborhood of $x$ contains infinitely many points of the set $B$),
\item {\it countably compact} if $X$ is countably compact at itself,
\item {\it countably pracompact} if $X$ is countably compact at a dense subset of $X$,
\item {\it pseudocompact} if each locally finite family of nonempty open subsets of the space $X$ is
finite,
\item {\it finally compact} if each open cover of $X$ has a countable subcover.
\end{itemize}

The following inclusions hold.

\begin{itemize}
\item Each compact space is $\omega$-bounded.
\item Each $\omega$-bounded space is totally countably compact.
\item Each totally countably compact space is countably compact.
\item Each sequentially compact space is countably compact.
\item Each countably compact space is countably pracompact.
\item Each countably pracompact space is pseudocompact.
\end{itemize}

In these terms, a space $X$ is compact if and only if $X$ is countably compact and finally compact.
A Tychonoff space $X$ is pseudocompact if and only if each continuous real-valued function on $X$ is bounded.
A paratopological group $G$ is {\it left (resp. right) precompact } if for each neighborhood $U$ of
the unit of $G$ there exists a finite subset $F$ of $G$ such that $FU=G$ (resp. $UF=G$).
A paratopological group is left precompact if and only if it is right precompact.
So we shall call left precompact paratopological groups {\it precompact}.
A sequence $\{U_n:n\in\omega\}$ of subsets of a space $X$ is {\it non-increasing} if
$U_n\supset U_{n+1}$ for each $n\in\omega$. A paratopological group $G$ is {\it 2-pseudocompact}
if $\bigcap \ol{U_n^{-1}}\not=\0$ for each non-increasing sequence $\{U_n:n\in\omega\}$ of nonempty open subsets of $G$.
Clearly, each countably compact paratopological group is 2-pseudocompact.
A paratopological group $G$ is {\it left $\omega$-precompact} if for each neighborhood $U$ of
the unit of $G$ there exists a countable subset $F$ of $G$ such that $FU=G$.
A paratopological group $G$ is {\em saturated} if for each nonempty open subset $U$ of $G$
there exists a nonempty open subset $V$ of $G$ such that $V^{-1}\subset U$.
Each precompact paratopological group is saturated and each saturated paratopological group is
quasiregular.

{\it Separation axioms}.
All topological spaces considered in the present paper are {\it not} supposed to satisfy any of the
separation axioms, if otherwise is not stated.
The definitions and relations between separations axioms both for topological spaces and
paratopological groups can be found in the first section of the paper~\cite{Rav2}.
Each $T_0$ topological group is $T_{3\frac 12}$, but for paratopological groups
neither of the implications
$$T_0\Rightarrow T_1 \Rightarrow T_2 \Rightarrow T_3$$
holds (independently of the algebraic structure of the group),
Mikhail Tkachenko's survey~\cite{Tka2} contains a section devoted to
relations between separations axioms for paratopological groups.
Some problems about these relations are still open.
Paper~\cite{Tka2} is devoted to axioms of separation in semitopological groups and related
functors.
Therefore the separation axioms play essential role in the theory of paratopological groups.
In particular, these axioms essentially affect results about automatic continuity of the inversion.
We recall some not very well known separation axioms which we shall use in our paper.
A space $X$ is {\it $T_1$} if for every distinct points $x,y\in X$ there exists an open set
$x\in U\subset X\bs\{y\}$. A space $X$ is {\it $T_3$} if
each closed set $F\subset X$ and every point $x\in X\bs F$ have
disjoint neighborhoods. A space $X$ is {\it regular} if it is $T_1$ and $T_3$.
A space $X$ is {\it quasiregular} if each nonempty open subset $A$ of $X$ contains the closure of some nonempty open
subset $B$ of $X$. A space $X$ is a {\it CB}, if $X$ has a base consisting of
canonically open sets, that is such sets $U$ such that $U=\int\ol U$.
Each $T_3$ regular space is quasiregular and CB.

Denote by $\mathcal B^G$ the open base at the unit of a paratopological group $G$.

Suppose $A$ is a subset of a group $G$. Denote by $\langle A\rangle\subset G$
the subgroup generated by the set $A$.

Now we are ready to go to the next section and to formulate our

\section{Results}

\begin{proposition}\label{TCCTG} Each totally countably compact paratopological group is a topological group.
\end{proposition}
\begin{proof}
Let $G$ be such a group. Put $B=\bigcap\mathcal B^G$. From Pontrjagin conditions \cite[1.1]{Rav}
it follows that $B$ is a semigroup. Put $B'=\ol{\{e\}}=\{x\in G:(\forall U\in\mathcal B^G)(xU\ni e)\}=\{x\in G:xB\ni e\}=B^{-1}$.
Hence $B'$ is a closed subsemigroup of the group $G$. Since $B'$ is a compact
semigroup, there is a minimal closed right ideal $H\subset
B'$. Let $x$ be an arbitrary element of $H$. Then $x^2H=H\ni x$.
Hence $x^{-1}\in H$ and $e\in H$. Consequently $H=B'$ and $xB'=B'$ for
each element $x\in B'$. Therefore $B'$ is a group and $B'=B$.
Since $B=\bigcap\mathcal B^G$, we see that $g^{-1}Bg\subset B$ for each $g\in
G$. Hence $B$ is a normal subgroup of the group $G$.
Since $B$ has the antidiscrete topology, $B$ is a topological group.

Since the set $B$ is closed, we see that the quotient group $G/B$ is a $T_1$-space.
Now let $\{x_n:n\in\omega\}$ be an arbitrary sequence of the group $G/B$ and let $\pi:G\to G/B$ be the
quotient map. Choose a sequence $\{x'_n:n\in\omega\}$ of the group $G$ such that $\pi(x'_n)=x_n$
for each $n\in\omega$. Since $G$ is totally countable compact, we see that there is a subsequence $A$ of
$\{x'_n:n\in\omega\}$ such that the closure $\ol A$ is compact. Since the set $\ol A$ is closed, $\ol A=\ol AB$.
Hence the closed compact set $\pi(\ol A)$ contains a subsequence of the sequence $\{x_n:n\in\omega\}$.
Therefore $G/B$ is a $T_1$ totally countably compact paratopological group. Consequently, from \cite[Cor 2.3]{AlaSan}
it follows that $G/B$ is a topological group. Hence both $B$ and $G/B$ are topological groups.
Then by \cite[1.3]{Rav4} or \cite[Pr. 5.3]{Rav3}, $G$ is a topological group too.
\end{proof}

Proposition~\ref{TCCTG} generalizes Corollary 2.3 in \cite{AlaSan} and Lemma 5.4 in \cite{Rav3}.

\begin{lemma}\label{SatQua} \cite [Pr. 1.7]{Rav2} Each saturated paratopological group is
quasiregular.
\end{lemma}

Given a topological space $(X,\tau)$ Stone \cite{Sto} and Katetov
\cite{Kat} consider the topology $\tau_r$ on $X$ generated by
the base consisting of all canonically open sets of the space
$(X,\tau)$. This topology is called the {\it regularization} of
the topology $\tau$. If $(X,\tau)$ is a paratopological group then $(X,\tau_r)$
is a $T_3$ paratopological group \cite[Ex. 1.9]{Rav2}, \cite[p. 31]{Rav3}, and \cite[p. 28]{Rav3}.
We remind that a paratopological group $G$ is {\it topologically
periodic} if for each $x\in G$ and a neighborhood $U\subset G$ of
the unit there is a number $n\ge 1$ such that $x^n\in U$, see
\cite{BokGur}.

\begin{lemma}\label{RegTG} \cite{RavRez} Suppose $(G,\tau)$ is a quasiregular paratopological group such that $(G,\tau_r)$
is a topological group; then $(G,\tau)$ is a topological group.
\end{lemma}

The following lemma is quite easy and probably is known.

\begin{lemma}\label{RegPCompact} Let $(X,\tau)$ be a topological space. Then $(X,\tau)$ is
pseudocompact if and only if the regularization $(X,\tau_r)$ is
pseudocompact.\qed
\end{lemma}

\begin{lemma}\label{TopPerSat} Each topologically periodic Baire paratopological group is saturated.
\end{lemma}
\begin{proof}
In \cite{AlaSan}, Alas and Sanchis proved that each $T_0$ topologically periodic Baire paratopological
group is saturated. This result answers an author's Question 2 from \cite{Rav5}.
Moreover, from their proof follows that each topologically periodic Baire paratopological group is
saturated.
\end{proof}

\begin{lemma}\label{2PseBai} Each 2-pseudocompact paratopological group is a Baire space.
\end{lemma}

\begin{proof} In \cite[Th. 2.2]{AlaSan}, Alas and Sanchis proved that
each $T_0$ 2-pseudocompact paratopological group is a Baire space.
Nevertheless their proof also proves this lemma.
\end{proof}

\begin{lemma}\label{PseudoTGPre} Each pseudocompact topological group is precompact.\qed
\end{lemma}

\begin{lemma}\label{PreSat} \cite[Pr.3.1]{Rav3} Each precompact paratopological group is saturated.\qed
\end{lemma}

\begin{lemma}\label{BaiSat} (also, see \cite[Th. 2.5]{AlaSan})
Each Baire left $\omega$-precompact paratopological group is saturated.\qed
\end{lemma}

\begin{lemma}\label{ArhRezT3}
Each paratopological group that is a dense $G_\delta$-subset of a $T_3$ pseudocompact space is a topological group.
\end{lemma}
\begin{proof} In \cite[Th. 1.7]{ArhRez}, Arhangel'skii and Reznichenko proved that for each
paratopological group $G$ such that $G$ is a dense $G_\delta$-subset of a {\it regular}
pseudocompact space, it follows that $G$ is a topological group.
Their proof with respective changes also proves this lemma.
\end{proof}

\begin{lemma}\label{2PseT3}
Each $T_3$ 2-pseudocompact paratopological group is a topological group.
\end{lemma}
\begin{proof} Let $G$ be such a group.
Suppose $G$ is not a topological group; then there exists a neighborhood $U\in\mathcal B^G$ such that
$V\not\subset\ol{U^{-1}}\subset (U^{-1})^2$ for each $V\in\mathcal B^G$.
By induction we can build a sequence
$\{U_i:i\in\omega\}$ of open neighborhoods of the unit $e$ of $G$ such that
$\ol{U_0}\subset U$ and $U_{i+1}^2\subset U_i$ for each $i\in\omega$. Put $F=\bigcap\{U_i:i\in\omega\}$.
Then $F$ is a subsemigroup of $G$. Since the group $G$ is 2-pseudocompact, there is a point $x\in G$ such that
$x\in\bigcap\ol{(U_i\bs\ol{U^{-1}})^{-1}}\subset
\bigcap\ol{U_i^{-1}}\subset\bigcap (U_i^{-1})^2\subset F^{-1}$. Moreover,
$x\in\ol{(U_0\bs\ol{U^{-1}})^{-1}}\subset
\ol{(U_0\bs U^{-1})^{-1}}\subset \ol{U_0^{-1}\bs U}\subset G\bs U$.
Since $x\not\in\ol {U_0}$ there is a neighborhood $W\in\mathcal B^G$ such that
$W\subset U_1$ and $Wx\cap U_0=\0$.
Let $m,n\in\omega$ and $n>m\ge 1$. If $\0\not=Wx^n\cap Wx^m$ then
$\0\not=Wx \cap Wx^{m+1-n}\subset Wx\cap U_1F\subset Wx\cap U_1^2\subset Wx\cap U_0=\0$.
Therefore $\{Wx^i:i\ge 1\}$ is an infinite family of disjoint sets.
Choose a neighborhood $V\in\mathcal B^G$ such that $V^2\subset W$.
Since $G$ is 2-pseudocompact, there is
a point $y\in\bigcap_{n\ge 1}\ol{(\bigcup_{i\ge n} Vx^i)^{-1}}$.
There exist numbers $m,n\in\omega$ such that $n>m\ge 1$ and $yV\cap (Vx^m)^{-1}\not=\0$ and
$yV\cap (Vx^n)^{-1}\not=\0$. Then $y^{-1}\in V^2x^m\cap V^2x^n\subset Wx^m\cap Wx^n$.
This contradiction proves that $G$ is a topological group.
\end{proof}

Each countably compact paratopological group and each pseudocompact topological group are
both 2-pseudocompact.

\begin{proposition}\label{2PseP}
Each 2-pseudocompact paratopological group is pseudocompact.
\end{proposition}
\begin{proof} Let $(G,\tau)$ be such a group.
The regularization  $(G,\tau_r)$ of the group $(G,\tau)$ is a $T_3$ 2-pseudocompact
paratopological group. By Lemma~\ref{2PseT3}, $(G,\tau_r)$ is a topological group.
By Lemma~\ref{RegPCompact}, $(G,\tau)$ is pseudocompact.
\end{proof}

An opposite inclusion does not hold. Manuel Sanchis and Mikhail Tkachenko
constructed a Hausdorff pseudocompact Baire not 2-pseudocompact
paratopological group \cite[Th.2]{SanTka}, answering author's
questions from a previous version of the manuscript.

\begin{problem}\label{PCto2PC} Is each countably pracompact Baire paratopological group
2-pseudocompact?
\end{problem}

The answer to Problem~\ref{PCto2PC} is positive provided the group is quasiregular or saturated
or topologically periodic or left $\omega$-precompact.

\begin{proposition}\label{PseTG} Suppose $G$ is a pseudocompact paratopological group; then the following conditions
are equivalent:

1. The group $G$ is quasiregular;

2. The group $G$ is saturated;

3. The group $G$ is topologically periodic Baire;

4. The group $G$ is left $\omega$-precompact Baire;

5. The group $G$ is precompact;

6. The group $G$ is a topological group.
\end{proposition}

\begin{proof} It is easy to prove that Condition 6 implies all other conditions.

($1\Rightarrow 6$) The regularization  $(G,\tau_r)$ of the group $(G,\tau)$ is a $T_3$ pseudocompact
paratopological group. By Lemma~\ref{ArhRezT3}, $(G,\tau_r)$ is a topological group.
Then by Lemma~\ref{RegTG}, $(G,\tau)$ is a topological group too.

$(2\Rightarrow 1)$ By Lemma~\ref{SatQua}, the group $G$ is quasiregular.

$(3\Rightarrow 1)$ By Lemma~\ref{TopPerSat}, the group $G$ is saturated.
Then by Lemma~\ref{SatQua}, the group $G$ is quasiregular.

$(4\Rightarrow 2)$ By Lemma~\ref{BaiSat}, the group $G$ is saturated.

$(5\Rightarrow 2)$ By Lemma~\ref{PreSat}, the group $G$ is saturated.
%($1b\Rightarrow 2$) The proof is very similar to that from \cite{BokGur}. Let $U$ be an arbitrary
%neighborhood of the unit and $\{V_i:i\in\omega\}$ be a sequence of open neighborhoods of
%unit such that $V_0=U$ and $V_{i+1}^2\subset V_i$ for each $i\in\omega$. It is easy to check
%that $(\ol{V_{i+1}^{-1}})^{-1}\subset V_i$. We show that $F=\bigcap \ol{V_i^{-1}}
%\subset U$.
%Let $x\in F$, $x\in\ol{V_i^{-1}}$ for each $i\in\omega$. Then $x^{-1}\in
%(\ol{V_{i+1}^{-1}})^{-1}\subset V_i$ for each $i\in\omega$. Choose $n\in\omega$ such that $x^n\in
%V_1$, then $(x^{-1})^{n-1}\in ((\ol{V_{i+1}^{-1}})^{-1})^{n-1}\subset V_i^{n-1}$ for each
%$i\in\omega$. Now choose $i_0\in\omega$ such that $V_{i_0}^{n-1}\subset V_1$. Then $x=x^n(x^{-1})^{n-1}\in V_1V_1\subset
%U$. Therefore $F\subset U$ and $\bigcap V_i\subset\ol{U^{-1}}$. If there exists a number $i$
%such that $V_i\subset\ol{U^{-1}}$ then $V_i\subset (U^{-1})^2$ and we are done. In the opposite
%case the 2-pseudocompactness
%implies that there is a point $x\in G$ such that for each $W\in\mathcal B$ and each $i$ holds
%$xW^{-1}\cap (V_i\bs\ol{U^{-1}})\not=\0$. Then $xW^{-1}\cap (V_i\bs U^{-1})\not=\0$ and so
%$Wx^{-1}\cap (V_i^{-1}\bs U)\not=\0$. Thus $x^{-1}\in\ol{V^{-1}_i}\subset V_{i-1}^{-1}$. Hence
%$x^{-1}\in F\subset U$, which is impossible since $U$ is open.
\end{proof}

Implication ($1\Rightarrow 6$) generalizes Proposition 2 from \cite{RavRez}.
Implication $(4\Rightarrow 6)$
generalizes Proposition 2.6 from \cite{AlaSan}. Implication ($3\Rightarrow 6$)
answers Question C from \cite{AlaSan} and generalizes Theorem 3 from \cite{BokGur}.

%In \cite{BokGur}, Bokalo and Guran proved that each countably
%compact topologically periodic paratopological group is a topological group.
%Since each countably compact paratopological group
%is 2-pseudocompact, this implication also generalizes the result of Bokalo and Guran.

%\begin{lemma}\label{PseQuaTop} \cite[Pr. 2]{RavRez} Every Hausdorff pseudocompact quasiregular paratopological group is a
%topological group.
%\end{lemma}
%\begin{lemma}\label{Le1} [??] Every Hausdorff pseudocompact periodic solvable topological group
%has a finite period.\qed
%\end{lemma}
%\begin{proposition} Every Hausdoff pseudocompact periodic solvable paratopological
%group has finite period and hence is a topological group.
%\end{proposition}
%\begin{proof} Let $(G,\tau)$ be such a group and $(G,\tau_r)$ be the
%regularization of the group $(G,\tau)$. Then $(G,\tau_r)$ is a regular
%periodic pseudocompact topological group. Lemma \ref{Le1} implies that the group
%has a finite period.
%\end{proof}
%\begin{lemma} \label{PerT0} \cite[Th. 2.7]{AlaSan} Every topologically periodic
%$T_0$ paratopological group is $T_1$.
%\end{lemma}

A group has two basic operations: the multiplication and the inversion.
The first operation is continuous on a paratopological group, but the second may be not continuous.
But there are continuous operations on a paratopological group besides the multiplication.
For instance, a power. Now suppose that there exists a non-empty open subset $U$ of a paratopological group $G$
such that the inversion on the set $U$ coincides with a continuous operation.
Then, since the inversion on the set $U$ is continuous, it follows that the group $G$ is topological.
Let us consider a trivial application of this idea. Let $G$ be a paratopological group
of bounded exponent. There is a number $n$ such that the inversion on the group $G$ coincides with the
$n$-th power. Thus the group $G$ is topological. Another applications are formulated in two next propositions.

\begin{proposition} Each Baire periodic paratopological group is a topological group.
\end{proposition}
\begin{proof} Let $G$ be such a group. Put $B'=\bigcap\{U^{-1}:U\in\mathcal B^G\}$. Then $B'$ is closed subset of $G$.
Since $B'$ is a periodic semigroup, we see that $B'$ is a group.
For each $n>1$ we put $G_n=\{x\in G:x^n\in B'\}$. Since the group $G$ is Baire, there is a
number $n>1$ such that the set $G_n$ has the nonempty interior.

Therefore there are a point $x\in G_n$ and a neighborhood $U$ of the unit of $G$ such that
$(xu)^n\in B'$ for each $u\in U$. Then $V^{-1}\subset B'(xV)^{n-1}x$ for each subset $V$ of $U$.
Let $U_1$ be an arbitrary neighborhood of the unit of $G$. Then $x^n\in B'U_1$.
By the continuity of the multiplication there is a neighborhood $W\subset U$ of the unit such that
$(xW)^{n-1}x\subset B'U_1$. Since $B'\subset U_1$, we see that $W^{-1} \subset B'U_1\subset U_1^2$.
\end{proof}

A Hausdorff topological group $G$ is called {\it Raikov-complete\/}, provided
it is complete with respect to the
upper uniformity which is defined as the least upper bound $\mathcal
L\vee\mathcal R$ of the left and the right uniformities on $G$. Recall
that the sets $\{(x,y): x^{-1}y\in U\}$, where  $U$  runs  over  a
base  at  unit  of $G$, constitute a base of entourages for the
left uniformity $\mathcal L$ on $G$.  In  the  case  of  the  right
uniformity  $\mathcal R$,  the condition $x^{-1}y\in U$ is replaced by
$yx^{-1}\in U$. The {\it Raikov completion $\hat G$\/} of a Hausdorff
topological group $G$ is a completion of $G$ with respect to the
upper uniformity $\mathcal L\vee\mathcal R$. For every Hausdorff topological group
$G$ the space $\hat G$ has a natural structure of a topological
group. The group $\hat G$ can be defined as a unique (up to an
isomorphism) Raikov complete group containing a Hausdorff topological group $G$ as a dense
subspace.

\begin{proposition}~\label{T2PPTG} Each Hausdorff pseudocompact periodic paratopological
group is a topological group.
\end{proposition}
\begin{proof} Let $(G,\tau)$ be such a group and let $(G,\tau_r)$ be the
regularization of the group $(G,\tau)$. Then $(G,\tau_r)$ is a regular
pseudocompact periodic topological group. Let $\hat G$ be the Raikov completion
of the group $(G,\tau_r)$.

We claim that the group $\hat G$ is periodic. Assume the converse. Suppose there exists a non-periodic
element $x\in\hat G$. Then for each positive integer $n$ there exists a neighborhood
$V_n\ni x$ such that $V_n^n\not\ni e$. Since the group $G$ is $G_\delta$-dense
in the group $\hat G$, we see that there exists a point $y\in G\cap \bigcap V_n$.
Then there is a positive integer $n$ such that $y^n=e$. This contradiction proves the
periodicity of the group $\hat G$.

For each $n>1$ we put $\hat G_n=\{x\in\hat G:x^n=e\}$. Since the group $\hat G$ is Baire, there is a
number $n>1$ such that the set $\hat G_n$ has the nonempty interior. Since the group
$G$ is dense in $\hat G$, we see that there are a point $x\in G$ and a neighborhood $U$ of the unit of $G$
such that $(xu)^n=e$ for each $u\in U$.

Let $y$ be an arbitrary point of $G$ and $V$ be an arbitrary neighborhood of the point $y^{-1}$.
Then $x^ny^{-1}=y^{-1}\in V$. By the continuity of the multiplication there is a neighborhood
$W\in\tau$ of the unit such that $W\subset U$ and $(xW)^{n-1}xy^{-1}\in V$. Let $z\in yW$ be an
arbitrary point. Then $(xy^{-1}z)^n=e$. Therefore $z^{-1}=(xy^{-1}z)^{n-1}xy^{-1}\in (xW)^{n-1}xy^{-1}\subset V$.
Thus the inversion on the group $G$ is continuous.
\end{proof}

%\begin{proposition} Every $T_0$ Baire periodic paratopological group is
%a topological group.
%\end{proposition}
%\begin{proof} Let $G$ be such the group. Lemma \ref{PerT0} implies that $G$ is $T_1$.
%For every $n>1$ put $G_n=\{x\in G:x^n=e\}$. Since the group $G$ is Baire there exists a
%number $n>1$ such that the set $G_n$ has nonempty interior. Thus there
%are a point $x\in G$ and a neighborhood $U$ of the unit such that
%$(xu)^n=e$ for each $u\in U$. Then $V^{-1}\subset (xV)^{n-1}x$ for
%every set $V\subset U$. It follows that $G$ is a topological group.
%\end{proof}

T.~Banakh build the next example. This result answers some author's questions and shows that
in general, Proposition~\ref{T2PPTG} cannot be generalized for $T_1$ groups (even for
countably pracompact and periodic groups).
Moreover, in \cite{AlaSan}, Alas and Sanchis proved that each topologically periodic
$T_0$ paratopological group is $T_1$. Example~\ref{BanakhEx} shows that there is a
$T_1$ periodic paratopological group $G$ such that $G$ is not Hausdorff.

\begin{example}\label{BanakhEx} There exists a $T_1$ periodic paratopological group $G$ such that each power of
$G$ is countably pracompact but $G$ is not a topological group.
\end{example}
\begin{proof}
For each positive integer $n$ let $C_n$ be the set $\{0,\dots,n-1\}$ endowed with the discrete
topology and the binary operation "$+'$" such that $x+'y\equiv x+y (\operatorname{mod}n)$ for each $x,y\in C_n$.
Let $G=\bigoplus\limits_{n=1}^\infty C_n$ be the direct sum.
Let $\mathcal F$ be the family of all non-decreasing unbounded functions from $\omega\bs\{0\}$ to
$\omega$. For each $f\in\mathcal F$ put $$O_f=\{0\}\cup\{(x_n)\in G:(\exists m\ge 1)( (\forall n>m)
(x_n=0)\&(0<x_m<f(m)))\}.$$ It is easy to check that the family $\{O_f:f\in\mathcal F\}$ is a base
at the zero of a $T_1$ semigroup topology on $G$.

For each positive integer $m\ge 2$ we put $a^m=(a^m_n)\in G$, where $a^m_n=1$ if $n=m$; $a^m_n=0$ in the opposite case.
Let $f,g\in\mathcal F$ be arbitrary functions and let $x\in G$ be an arbitrary element.
There exists a number $m$ such that $f(m)\ge 2$, $g(m)\ge 2$, and $x_n=0$ for each $n\ge m$.
Then $a^m\in O_g$ and $x+a^m\in O_f$. Therefore $x\in\overline{O_f}$. Hence $\overline{O_f}=G$ for each $f\in\mathcal
F$. Thus $G$ is not Hausdorff. Since each two nonempty open subsets of $G$ intersects, we see that
each two nonempty open subsets of each power of $G$ in the box topology intersects too.
Therefore each power of $G$ in the box topology is pseudocompact.
Put $A=\{0\}\cup \{a^m:m\ge 2\}$. Then $A$ is compact and dense in $G$. Let $\kappa$ be an arbitrary cardinal.
Then $A^\kappa$ is compact and dense in $G^\kappa$. Therefore the space $G^\kappa$ is countably compact at a
dense subset of $G^\kappa$. Thus the space $G^\kappa$ is countably pracompact.
\end{proof}

\begin{problem}\label{prob2} Suppose a paratopological group $G$ satisfies one of the following
conditions:
\begin{itemize}
\item $G$ is Hausdorff countably pracompact topologically periodic left $\omega$-precompact,
\item $G$ is Hausdorff countably pracompact topologically periodic,
\item $G$ is Hausdorff pseudocompact topologically periodic left $\omega$-precompact,
\item $G$ is Hausdorff pseudocompact topologically periodic,
%\item $G$ is regular 2-pseudocompact.
\end{itemize}
Is $G$ a topological group?
\end{problem}

If a paratopological group $G$ satisfies one of Problem~\ref{prob2} Conditions 1-4 and
the group $G$ is quasiregular or saturated or Baire then $G$ is a topological group.

Denote by TT the following axiomatic assumption: there is an infinite torsion-free abelian countably
compact topological group without non-trivial convergent sequences. The first example of such a group
constructed by M.~Tkachenko
under the Continuum Hypothesis \cite{Tka}. Later, the Continuum Hypothesis weakened  to the Martin Axiom
for $\sigma$-centered posets by Tomita in \cite{Tom2}, for countable posets in \cite{KosTomWat},
and finally to the existence continuum many incomparable selective ultrafilters in \cite{MadTom}.
Yet, the problem of the existence of a countably compact
group without convergent sequences in ZFC seems to be open,  see \cite{DikSha}.

The proof of \cite[Lemma 6.4]{BanDimGut} implies the following

\begin{lemma} \label{FA} (TT) Let $G$ be a free abelian group generated by the set $\cc$.
There exists a Hausdorff group topology on $G$
such that for each countable infinite subset $M$ of the group $G$ there exists an element
$\alpha \in \overline M\cap\cc$ such that $M\subset\langle \alpha \rangle$.\qed
\end{lemma}

\begin{example} \label{CCNotTG} (TT) There exists a functionally Hausdorff countably compact
free abelian paratopological group $(G,\sigma)$ such that $(G,\sigma)$ is not a topological group.
\end{example}
\begin{proof}
Let $G$ be a free abelian group generated by the set $\cc$. Lemma \ref{FA} implies that
there exists a Hausdorff group topology $\tau$ on the group $G$
such that for each countable infinite subset $M$ of the group
$G$ there exists an element $\alpha \in \overline M\cap\cc$ such that
$M\subset \langle \alpha \rangle$. Each non-zero element $x\in G$ has a unique representation
$x=\sum_{\alpha\in A_x} n_\alpha \alpha$ such that the integer number $n_\alpha$ is
non-zero for each $\alpha\in A_x$.
Put $S_0=\{x\in G\setminus\{0\}:n_{\sup A_x}<0$ and $S=S_0\cup\{0\}$. Clearly,
$S$ is a subsemigroup of the group $G$. It is easy to check that the family $\{U\cap S:U\in\tau, U\ni 0\}$
is a base at the zero of a semigroup topology on $G$.
Denote this semigroup topology by $\sigma$.

Let $M$ be an arbitrary countable infinite subset of the group
$G$. There exists an element $\alpha \in \overline{M}^\tau\cap\cc$ such that
$M\subset \langle \alpha \rangle$.
Since $M\subseteq\alpha+S_0$, we see that $\alpha \in \overline{M}^\sigma$.
Hence $(G,\sigma)$ is a countably compact
Hausdorff paratopological group.

Suppose $(G,\sigma)$ is a topological group; then there exists a neighborhood $U\in\tau$ of the zero such that
$U\cap S\subset S\cap(-S)=\{0\}$. Fix any element $\alpha\in\cc$. Then $-\alpha\in S$. Since
the topological group $(G,\tau)$ is countably compact, there exists a number $n<0$ such that
$n\alpha\in U$. Then $0\not=n\alpha\in U\cap S$. This contradiction proves
that $(G,\sigma)$ is not a topological group.
\end{proof}

Example~\ref{CCNotTG} negatively answers Problem 1 from \cite{Gur} under TT.
Since each countably compact left $\omega$-precompact paratopological group is a topological group \cite[p.
82]{Rav3}, we see that Example \ref{CCNotTG} implies the negative answers to Question A from \cite{AlaSan} and
to Problem 2 from \cite{Gur} under TT.

\begin{problem}\label{prob1} Is there a ZFC-example of a paratopological group $G$ such that $G$ is not a topological
group but $G$ satisfies one of the following conditions:
\begin{itemize}
\item $G$ is Hausdorff countably compact,
\item $G$ is Hausdorff countably pracompact Baire,
\item $G$ is $T_1$ countably compact.
\end{itemize}
\end{problem}

If $G$ is such a group then $G$ is not topologically periodic, not saturated, not quasiregular,
and not left $\omega$-precompact. Moreover, $G\times G$ is not countably compact.
By Theorem 2.2 from \cite{AlaSan}, a group $(G,\sigma)$ from Example~\ref{CCNotTG}
is Baire. Therefore under TT there is a group $G$ such that $G$ satisfies all conditions listed in Problem~\ref{prob1}.

The following proposition answers Questions B from \cite{AlaSan}.
Also this and the next propositions positively answer
a special case for 2-pseudocompact groups of Problem 8.1 from \cite{ArhChoKen}.

\begin{proposition}\label{CountPseudoCh} Each 2-pseudocompact paratopological group of countable pseudocharacter
is a topological group.
\end{proposition}
\begin{proof} Let $G$ be such a group. By induction we can build a sequence
$\{V_i:i\in\omega\}$ of open neighborhoods of the unit $e$ of $G$ such that
$V_{i+1}^2\subset V_i$ for each $i\in\omega$ and $\bigcap\{V_i:i\in\omega\}=\{e\}$.
Suppose $G$ is not a topological group; then there exists a neighborhood $U\in\mathcal B^G$ such that
$V_i\not\subset\ol{U^{-1}}\subset (U^{-1})^2$ for each $i\in\omega$. Since the group $G$ is
2-pseudocompact, there is a point $x\in G$ such that
$xW^{-1}\cap (V_i\bs\ol{U^{-1}})\not=\0$ for each $W\in\mathcal B^G$ and each $i>0$.
Then $xW^{-1}\cap (V_i\bs U^{-1})\not=\0$ and $Wx^{-1}\cap (V_i^{-1}\bs U)\not=\0$.
Therefore $x^{-1}\in\ol{V^{-1}_i}\subset V_{i-1}^{-1}$. Hence
$x=e$. But $e\in U^{-1}$ and $U^{-1}\cap (V_1\bs\ol{U^{-1}})=\0$. This contradiction proves
that $G$ is a topological group.
\end{proof}

%check
%\begin{proposition}\label{CountPseudoCh} Each 2-pseudocompact paratopological group in which a set $(\ol{\{e\}})^{-1}$
%is a $G_\delta$-set is a topological group.
%\end{proposition}
%\begin{proof} Let $G$ be such the group. Put $H=(\ol{\{e\}})^{-1}$. Then $H=\bigcap\{U:U\in\mathcal
%B\}$.
%By induction we can build a sequence $\{V_i:i\in\omega\}$ of open neighborhoods of
%$e$ such that $V_{i+1}^2\subset V_i$ for each $i\in\omega$ and $\bigcap V_i=H$.
%Now we are going to show that
%$G$ is a topological group. Let $U\in\mathcal B$ be an arbitrary neighborhood.  Then
%$\bigcap V_i\subset\ol{U^{-1}}$. If there exists a number $i$ such that $V_i\subset\ol{U^{-1}}$
%then $V_i\subset (U^{-1})^2$ and we are done. In the opposite case the 2-pseudocompactness
%implies that there is a point $x\in G$ such that for each $W\in\mathcal B$ and each $i$ holds
%$xW^{-1}\cap (V_i\bs\ol{U^{-1}})\not=\0$. Then $xW^{-1}\cap (V_i\bs U^{-1})\not=\0$ and so
%$Wx^{-1}\cap (V_i^{-1}\bs U)\not=\0$. Thus $x^{-1}\in\ol{V^{-1}_i}\subset V_{i-1}^{-1}$. Hence
%$x=e$, which is impossible.
%\end{proof}
%checked\begin{corollary} Each 2-pseudocompact paratopological group of countable pseudocharacter
%is a topological group.
%\end{corollary}
%checked\begin{corollary} Each 2-pseudocompact first countable paratopological group
%is a topological group.
%\end{corollary}
%Moreover, we can generalize this proposition as follows.
We shall use the following lemmas to obtain a counterpart of Proposition~\ref{CountPseudoCh} for Hausdorff 2-pseudocompact
paratopological groups.

\begin{lemma}~\label{CancComp}\cite[Th. 0.5]{Rez} A compact Hausdorff semigroup with separately continouous
multiplication and two-sides cancellations is a topological group.
\end{lemma}

\begin{lemma}~\label{MySwellingLemma} Let $K$ be a compact non-empty subset of a Hausdorff semitopological group
$G$. Then the set $G_K=\{x\in G: xK\subset K\}$ is a compact topological group.
\end{lemma}
\begin{proof} Clearly, $G_K$ is a subsemigroup of the group $G$. If $x\in G\setminus G_K$ then
there exists a point $y\in K$ such that $xy\not\in K$. Since the group $G$ is Hausdorff,
the set $G\setminus K$ is open. Since the multiplication on the group $G$ is separately continuous,
there exists an open neighborhood $O_x$ of the point $x$ such that $O_xy\subset G\setminus K$.
Thus $Ox\subset G\setminus G_K$. Therefore $G_K$ is a closed subset of the space $G$.
Let $y\in K$ be an arbitrary element. If $x\in G_K$ then $x\in xKy^{-1}\subset Ky^{-1}$.
Therefore $G_K\subset Ky^{-1}$. Since the set $Ky^{-1}$ is compact, the set $G_K$ is compact too.
So, $G_K$ is a Hausdorff compact cancellative semitopological semigroup. By Lemma~\ref{CancComp},
$G_K$ is a topological group.
\end{proof}
\begin{remark} Lemma~\ref{MySwellingLemma} does not necessarily hold for $T_1$ paratopological groups,
as the following example shows. Let $G=(\mathbb Z,+)$ be the group of integers endowed with a
topology with the base $\{\{x\}\cup \{z\in\mathbb Z:z\ge y\}:x,y\in\mathbb Z\}$. It is easy to
check that $G$ is a $T_1$ paratopological group. Put $K=\{x\in\mathbb Z:x\ge 0\}$. Then $K$ is a compact
subset of the group $G$ and $G_K=K$ is not a group.
\end{remark}

\begin{lemma}\label{SquareCompNeighb}\cite[Pr. 1.8]{Rav} Let $G$ be a paratopological group,
$K\subset G$ be a compact subspace, $F\subset G$ be a~closed set, and $K\cap F=\0$.
Then there exists a neighborhood $V$ of the unit such that $VK\cap F=\0$.
\end{lemma}

\begin{proposition}\label{CountPseudoChComp} Each 2-pseudocompact Hausdorff paratopological group
containing a compact non-empty $G_\delta$-set is a topological group.
\end{proposition}
\begin{proof} Let $G$ be such a group and $K$ be a compact non-empty $G_\delta$-set of the space
$G$. Suppose $G$ is not a topological group;
then there exists a neighborhood $U\in\mathcal B^G$ such that
$V\not\subset\ol{U^{-1}}\subset (U^{-1})^2$ for each neighborhood $V\in\mathcal B^G$.

By induction using Lemma~\ref{SquareCompNeighb} we can build a sequence
$\{V_i:i\in\omega\}$ of open neighborhoods of the unit $e$ of $G$ such that $V_0=U$,
$V_{i+1}^2\subset V_i$ for each $i\in\omega$ and $\bigcap\{V_iK:i\in\omega\}=K$.
Put $H=\bigcap\{V_i:i\in\omega\}$. The construction of $H$ implies that $H$ is a semigroup.
 Moreover, since $\overline{V_{i+1}^{-1}}\subset V_{i+1}^{-1}V_{i+1}^{-1}\subset V_{i}^{-1}$ for each
$i\in\omega$, we see that $H^{-1}$ is a closed subset of $G$. Moreover,
$HK\subset \bigcap\{V_iK:i\in\omega\}=K$, so $H\subset
G_K$. Since $G_K$ is a group then $H^{-1}\subset G_K$ too, so $H^{-1}$ is a
compact cancellative topological semigroup. By Lemma~\ref{CancComp},
$H^{-1}$ is a group.

We have
$V_i\not\subset\ol{U^{-1}}$ for each $i\in\omega$. Since the group $G$ is
2-pseudocompact, there is a point $x\in G$ such that
$xW^{-1}\cap (V_i\bs\ol{U^{-1}})\not=\0$ for each $W\in\mathcal B^G$ and each $i>0$.
Then $xW^{-1}\cap (V_i\bs U^{-1})\not=\0$ and $Wx^{-1}\cap (V_i^{-1}\bs U)\not=\0$.
Therefore $x^{-1}\in\ol{V^{-1}_i}\subset V_{i-1}^{-1}$. Hence
$x\in H^{-1}$. But $H^{-1}=H\subset U\cap U^{-1}$ and $U^{-1}\cap (V_1\bs\ol{U^{-1}})=\0$. This contradiction proves
that $G$ is a topological group.
\end{proof}

The author does not know do counterparts of Proposition~\ref{CountPseudoChComp} hold for $T_1$
2-pse\-u\-do\-compact or countably compact paratopological groups.
But a $T_0$ sequentially compact paratopological group $G_S$ from Example 2 from~\cite{Rav6}
contains an open compact subsemigroup $S$. Nevertheless, $G_S$ is not a topological group.

Example~\ref{NonStandardArrowedCircle} is an extension of Example 5.17 from~\cite{Rav3}.
Example~\ref{NonStandardArrowedCircle} shows that counterparts of Proposition \ref{CountPseudoCh}
and Implication ($4\Rightarrow 1$) of Proposition \ref{PseTG}
do not hold for countably pracompact paratopological groups.
Also Example~\ref{NonStandardArrowedCircle} negatively answers Problem 6.2 from \cite{Tka2}
and Problem 8.1 from \cite{ArhChoKen}.

\begin{example}~\label{NonStandardArrowedCircle}
There exists a functionally Hausdorff second countable
pa\-ra\-to\-po\-lo\-gi\-cal group $G$ such that each power of $G$
is countably pracompact but $G$ is not a topological group.
\end{example}
\begin{proof} Let $\T=\{z\in\C:|z|=1\}$ be the unit circle. Denote by $\tau$ the standard topology on the group $\T$.
For each $\eps>0$ put $B_\eps(0_\T)=\{z\in\T:|\arg z|<\eps\}$. Let $s:\T\to\Q$ be a homomorphism onto.
For each $a\in\R$ put $\T_{a}=\{x\in\T:s(x)>a\}$ and
$\T_{a}^0=\{x\in\T:s(x)\ge a\}$. Put $\mathcal B'_1=\{\T_a:a<0\}$, $\mathcal B'_2=\{\T^0_0\}$,
$\mathcal B'_3=\{\{0_\T\}\cup \T_0\}$ and $\mathcal B'_4=\{\{0_\T\}\cup \T_a:a>0\}$.
It is easy to check that $\mathcal B'_i$ is a base at the unit of a first countable semigroup
topology on the group $\T$ for every $i$.  Therefore for every $i$ the family
$\mathcal B_i=\{S\cap B_\eps(0_\T):S\in \mathcal B'_i,\eps>0\}$ is
a base at the unit of a first countable functionally Hausdorff semigroup
topology on the group $\T$. Denote this topology by $\tau_i$ and put $G_i=(\T,\tau_i)$.
Clearly, $\tau\subset\tau_1\subset\tau_2\subset\tau_3\subset\tau_4$.
%In \cite[Ex. 5.17]{Rav3}, the author constructed the following group $G$.
%Let $\T=\{z\in\C:|z|=1\}$ be the unit circle and let $\pi:\R\to\T$ be the homomorphism
%such that $\pi(x)=e^{2\pi xi}$ for
%each $x\in\R$. Since $\R$ is a vector space over the field $\Q$,
%there is a basis $B\ni\{1\}$ of the vector space $\R$ over $\Q$.
%Choose an arbitrary element $x'_0\in B\bs\{1\}$. Put $x_0=\pi(x'_0)$.
%Let $s':\R\to\Q$ be the homomorphism of vector spaces over $\Q$ such that $s'(x'_0)=1$ and
%$s'(B\bs\{x'_0\})=\{0\}$. It is easy to check that there exists a unique map $s:\T\to\Q$
%such that $s'=s\pi$ and that $s$ is a homomorphism.
%Put $S=\{0_{\T}\}\cup\{x\in\T:s(x)>0\}$. There exists a semigroup topology $\sigma$ on $\T$ such that
%the family $\{S\cap \pi((-1/n;1/n)):n>0\}$ is a base at the zero of $\T$ of the topology $\sigma$
%(see \cite[Ex. 5.17]{Rav3}). Put $G=(\T,\sigma)$. Denote by $\tau$ the standard topology on the group $\T$.
Choose an element $x_0\in \T_0$.
It is easy to check that the set $A=\{nx_0:n\in\Z, n\ge 0\}$ is dense in $G_4$. Let $\kappa$ be an arbitrary cardinal.
Then $A^\kappa$ is dense in $G_4^\kappa$.
Let $\{y^n:n\in\omega\}$ be an arbitrary sequence of points from $A^\kappa$. Since the space
$(\T,\tau)^\kappa$ is compact, we see that there is a point $y\in\T^\kappa$ such that the set
$\{n\in\omega:y^n\in U\}$ is infinite for each open in $(\T,\tau)^\kappa$ set $U\ni y$.

We claim that each open in $G_4^\kappa$ set $U\ni y$ has a similar property. Fix such a
neighborhood $U$. For each $\alpha\in\kappa$ let $\pi_\alpha:\T^\kappa\to\T$ be the projection onto the $\alpha$-th coordinate.
There exist a finite subset $F$ of $\kappa$, a family $\{U_\alpha:\alpha\in F\}$ of open in
$(\T,\tau)$ sets, and a number $a>0$ such that $V'\subset U$, where
$V'=\bigcap_{\alpha\in F}\pi_\alpha^{-1}((y_\alpha+\T_a)\cap U_\alpha)$.
Decreasing neighborhoods $U_\alpha$, if necessary, we can assume that for each
$\alpha\in F$ and for every integer $0<n\le (s(y_\alpha)+a)/s(x_0)$ if $nx_0\not=y_\alpha$,
then $nx_0\not\in U_\alpha$.

Put $V=\bigcap_{\alpha\in F}\pi^{-1}(U_\alpha)$.
Now let $z$ be an arbitrary point of $A^\kappa\cap V$.  Then $z_\alpha=y_\alpha$
or $s(z_\alpha)>s(y_\alpha)$ for each $\alpha\in F$. Therefore $z\in V'$. Hence the
set $\{n\in\omega:y^n\in U\}\supset \{n\in\omega:y^n\in V'\} = \{n\in\omega:y^n\in V\}$ is infinite.
Therefore the space $G_4^\kappa$ is countably compact at its dense subset $A^\kappa$.
Thus the space $G_4^\kappa$ is countably pracompact.

Let $i=1$ or $i=2$. A set $s^{-1}(0)$ is a subgroup of $\T$ of countable index and
$\tau_i|s^{-1}(0)=\tau|s^{-1}(0)$.
Since the group $(\T,\tau)$ has the countable weight, the group $G_i$ has a countable
network. Since $\chi(G_i)=\omega$, by~\cite[Pr. 2.3]{Rav}, the group $G_i$ is second countable.

We claim that the group $G_2$ has no countable network of closed subsets of $G_2$. Assume the converse.
Let $\mathcal N$ be a countable network such that for each set $X\in\mathcal N$ there exists an
element $x\in G$ with $x+\T^0_0\supset X$. We claim that the set $X$ is nowhere dense in $(G,\tau)$.
Assume the converse. Since the set $\{nx_0:n\in\Z, n<0\}$ is dense in $(G,\tau)$,
there exists an element $y=nx_0\in\operatorname{\int}_{\tau}\overline X^\tau$ such that $s(y)<s(x)$.
Let $\eps>0$ be an arbitrary number such that $y+B_\eps(0_\T)\subset \overline X^\tau$.
There exists a point $z\in (y+B_\eps(0_\T))\cap X$. Since $z\in X\subset x+\T^0_0$, we see that
$s(z)\ge s(x)>s(y)$. Thus $z\in y+\T^0_0$. Hence $z\in y+(B_\eps(0_\T)\cap\T^0_0)$.
Therefore $y\in \overline X^{\tau_2}=X\subset x+\T^0_0$. Thus $s(y)\ge s(x)$.
This contradiction proves that the set $X$ is nowhere dense in $(G,\tau)$.
Then $(G,\tau)=\bigcup\mathcal N$ is a union of a countable family of its closed nowhere
dense sets. This contradiction proves that the group $G_2$ has no countable network of closed subsets
of $G_2$.

The set $\T\bs \T_0$ is a closed discrete subset of $G_4$. Thus $e(G_4)=\mathfrak c$.
The set $\{(x,-x):x\in G\}$ is a closed discrete subset of $G_3\times G_3$.
Thus $e(G_3\times G_3)=\mathfrak c$.

Extent and Lindel\"of number of the group $G_3$ depend on properties of the homomorphism $s$.
The following two examples belong to T.~Banakh. We delay their constructions to
Section~\ref{Cons}.
\begin{example}~\label{OstapEx1N} There is a homomorphism $s:(\T,\tau)\to\Q$ onto such that the set $s^{-1}(0)$
contains a compact subset $K$ of cardinality $\mathfrak c$.
\end{example}
\begin{example}~\label{OstapEx3N} There is a homomorphism $s:(\T,\tau)\to\Q$ onto such that
for each $a\in\Q$ each closed subset of the set $\T_a$ is countable.
\end{example}

Suppose that the homomorphism $s$ satisfies the conditions of Example~\ref{OstapEx1N}.
Since $\tau\subset\tau_3$, $K$ is a closed subset of $G_3$.
Since $(x+\T_0)\cap K=\{x\}$ for each $x\in K$, $K$ is discrete.
Thus $\mathfrak c\le e(G_3)\le l(G_3)=\mathfrak c$.

Suppose that the homomorphism $s$ satisfies the conditions of Example~\ref{OstapEx3N}.
Let $\mathcal U$ be an open cover of the group $G_3$. For each point $x\in G_3$
choose a neighborhood $O_x\in\tau$ of $x$ and an element $U_x\in\mathcal U$
such that such that $(\{x\}\cup(x+\T_0))\cap O_x\subset U_x$. For each
$a\in\Q$ put $A_a=\bigcup\{Ox:x\in\T\bs\T_a\}$ and $B_a=\T\bs A_a$.
Then $B_a$ is closed in $(\T,\tau)$ subset of $\T_a$ and hence $B_a$ is countable.
Since $(\T,\tau)$ is hereditarily Lindel\"of there is a countable subset $C_a$ of
$\T\bs\T_a$ such that $A_a=\bigcup\{Ox:x\in C_a\}$. Put $\mathcal U_a=\{U_x:x\in B_a\cup C_a\}$.
Let $y\in \T_a$. If $y\in B_a$ then $y\in U_y\in\mathcal U_a$. If $y\not\in B_a$ then
there is a point $x\in C_a$ such that $y\in O_x$. Since $y\in\T_a$ and $x\in\T\bs\T_a$
we see that $y\in (x+\T_0)\cap O_x\subset U_x\in\mathcal U_a$.
Hence $\T_a\subset\bigcup\mathcal U_a$ and therefore $\T=\bigcup\{\T_a:a\in\Q\}\subset
\bigcup\{\bigcup \mathcal U_a:a\in\Q\}$. Therefore $l(G_3)\le\omega$.
\end{proof}

Moreover, Manuel Sanchis and Mikhail Tkachenko
constructed a Hausdorff 2-pseudo\-compact Fr\'echet-Urysohn
paratopological group which is not a topological
group \cite[Th.1]{SanTka}, answering an author's
question from a previous version of the manuscript.

Oleg Gutik asked the following question. Let $G$ be a countably compact paratopological group and $H$
be a subgroup of $G$. Is $\overline{H}$ a group? The next proposition exclaims the situation.

\begin{proposition} Let $G$ be a countably compact paratopological group such that the closure
of each cyclic subgroup of $G$ is a group. Then $G$ is a topological group.
\end{proposition}
\begin{proof} Let $x$ be an arbitrary point of $G$. Put $H=\overline{\langle x\rangle}$. Then $H$ is a separable
countably compact paratopological group. Since each countably compact left $\omega$-precompact paratopological group is a
topological group \cite[p. 82]{Rav3}, we see that $H$ is a topological group.
Therefore $H$ is a topologically periodic group. Hence
$G$ is a topologically periodic group. By \cite{BokGur}, $H$ is a topological group.
\end{proof}

The following proposition answers a question of Igor Guran.

\begin{proposition}~\label{T2SigCPsTG} Each Hausdorff $\sigma$-compact pseudocompact paratopological
group is a compact topological group.
\end{proposition}
\begin{proof} Let $(G,\tau)$ be such a group and let $(G,\tau_r)$ be the
regularization of the group $(G,\tau)$. Then $(G,\tau_r)$ is a regular
pseudocompact topological group. Let $G=\bigcup \{K_n:n\in\omega\}$ be
a union such that the set $K_n$ is compact in $(G,\tau)$ for each $n\in\omega$.
Then for each $n\in\omega$ the set $K_n$ is compact in $(G,\tau_r)$.
Since the group $(G,\tau_r)$ is Baire there exists a number $n\in\omega$
such that the set $K_n$ has nonempty interior in $(G,\tau_r)$. Since $\tau_r\subset\tau$
then the set $K_n$ has nonempty interior in $(G,\tau)$. Therefore the group
$(G,\tau)$ is locally compact. By ~\cite[Pr.5.5]{Rav3},
$(G,\tau)$ is a topological group. Since $(G,\tau)$ is precompact, we see that
$(G,\tau)$ is compact.
\end{proof}

\section{The Constructions of Examples~\ref{OstapEx1N} and \ref{OstapEx3N}}\label{Cons}

{\it Example~\ref{OstapEx1N}}. By~\cite[19.2]{Kech} there is a Cantor set $C\subset\R$ whose
members are linearly independent over $\Q$. Pick arbitrary distinct numbers $a,b\in C$.
Let $B'\supset C$ be a
maximal linearly independent over $\Q$ subset of $\R$ (which exists by Zorn lemma).
Then $B'$ is a basis of the vector space $\R$ over $\Q$.
Let $s':\R\to\Q$ be a homomorphism map of vector spaces over $\Q$ such that $s'(a)=1$ and
$s'(B'\bs\{a\})=\{0\}$.
Let $\pi:\R\to\T$ be the homomorphism such that $\pi(x)=e^{2\pi xi/b}$ for each $x\in\R$.
It is easy to check that there exists a unique map $s:\T\to\Q$ such that
$s'=s\pi$. Moreover, $s$ is a homomorphism and $\ker s\supset \pi(B'\bs\{a\})$.
There exists an uncountable compact set $B\subset C\bs\{a\}$.
Since $s$ is continuous and $\pi|C$ is injective, $\pi(B)$ is an uncountable compact
subset of $\ker s$. By \cite[3.3.2 and 6.5]{Kech}, $|\pi(B)|=\frak c$.
$\qed$

{\it Example~\ref{OstapEx3N}}.
We shall proceed similarly to the construction of a Bernstein set.
It is easy to check that $\T$ contains exactly $\frak c$ uncountable closed subsets.
Let $\{K_\alpha:\alpha<\frak c\}$ be an enumeration of all uncountable closed subsets of $\T$.
By \cite[3.3.2 and 6.5]{Kech}, each set $K_\alpha$ has cardinality continuum.

For a set $B\subset\T$ by $[B]$ we shall denote a set $\{x\in\T: (\exists k\in\Z\setminus\{0\}):kx\in B\}$.
It is easy to check that $|[B]|\le \aleph_0|B|$.

Now we shall construct the homomorphism $s$ by transfinite recursion. For this purpose
we construct a non-decreasing transfinite sequence $\{B_\alpha:\alpha<\frak c\}$
of subgroups of $\T$. Let $\alpha<\frak c$ be an ordinal.
Suppose that we already have constructed the subgroups $B_\beta$
(such that $B_\beta=[B_\beta]$ and $|B_\beta|\le\aleph_0\beta$) and the restrictions $s|B_\beta$ for all $\beta<\alpha$.
(To assure the recursion base we formally put $B_{-1}=[\{0\}]$ and $s(B_{-1})=\{0\}$).
Put $C_\alpha=\bigcup\{B_\beta:\beta<\alpha\}$.
Then $|C_\alpha|\le\aleph_0\alpha
<\frak c=|K_\alpha|$. Therefore we can inductively construct a sequence $\{c_n:n\in\omega\}$ of points
of $K_\alpha$ such that $c_n\not\in C_{\alpha,n}$ for each $n$, where $C_{\alpha,0}=C_{\alpha}$
and $C_{\alpha,n+1}=[C_{\alpha,n}\cup\{c_n\}]$ for each $n$.
Put $B_\alpha=\bigcup\{C_{\alpha,n}:n\in\omega\}$. Then $|B_\alpha|\le\aleph_0\alpha$.
We define the restriction $s|B_\alpha$ inductively. The restriction $s|C_{\alpha,0}$ is already
defined. Suppose that we already defined the restriction $s|C_{\alpha,n}$.
Let $x$ be an element of $C_{\alpha,n+1}$. Then there exist integers
$k\ne 0$ and $l$  such that $kx=c+lc_n$,
where $c\in C_{\alpha,n}$. % Since $x\not\in C_{\alpha,n}$, we see that $l\ne 0$.
Put $s(x)=(s(c)-ln)/k$. Now we have to show that this definition is
correct, that is it does not depend on the integers $k$ and $l$.
Suppose that $k'x=c'+l'c_n$, for element $c'\in C_{\alpha,n}$ and integers $k'\ne 0$ and $l'$.
Then $(lk'-l'k)c_n=kc'-k'c\in C_{\alpha,n}$. Since
$c_n\not\in C_{\alpha,n}=[C_{\alpha,n}]$, we see that
$lk'=l'k$ and thus $kc'=k'c$. Then $(s(c)-ln)/k=(s(c')-l'n)/k'$.
In particular, we have that $(s|C_{\alpha,n+1})|C_{\alpha,n}=s|C_{\alpha,n}$.
It is easy to check that $s|C_{\alpha,n+1}$ is a homomorphism.
Then $s|B_{\alpha}$ is a homomorphism too and $K_\alpha\not\subset\T_a$ for
each $a\in\Q$ because $\{c_n\}\subset K_\alpha$ and $s(c_n)=-n$ for each $n$.
The construction implies that $s|B$ is a homomorphism and
$\inf s|B(K)=-\infty$ for each uncountable closed subset $K$ of $\T$.
Since $B$ is a divisible subgroup of an abelian group $\T$, there
exists an extension $s$ of a homomorphism $s|B$ onto $\T$ (see, for instance,
\cite[$\S$ 23]{Kur}).$\qed$

\section{Acknowledgements}
The author thanks to Artur Tomita and to Manuel Sanchis who provided him their
articles, to Taras Banakh who turned his attention to the class of countably pracompact spaces,
intermediate between the classes of countably compact and pseudocompact spaces
and for Examples ~\ref{BanakhEx}, ~\ref{OstapEx1N}, and ~\ref{OstapEx3N}
and to Mikhail Tkachenko who have inspired him to prove Proposition~\ref{2PseP}.
Also the author thanks to Minnan Normal University (Fujian, China) for the hospitality
and possibility to give lectures about the present paper.


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\end{document}
%\begin{question} Is every $(T_1/Hausdorff/T_3)(abelian)$ pseudocompact topologically periodic
%paratopological group a topological group?
%\end{question}
%%\begin{question} Is every $T_3$ pseudocompact left $\omega$-bounded paratopological group
%%a topological group?
%%\end{question}
%\begin{lemma}\label{Le46.0} \cite[p. 71]{Rav3} (or see \cite[Lemma 1.2]{ArhRez} for a slightly weaker claim)
%A paratopological group $G$ is a
%topological group if and only if $\int(\ol U\cap\ol{U^{-1}})\not=\0$
%for every neighborhood $U$ of the unit of the group $G$.
%\end{lemma}
%\begin{proposition} Every 2-pseudocompact paratopological group is a topological group. CHECK PROOF
%\end{proposition}
%\begin{proof} The proof is very similar to that from \cite[Th. 1.7]{ArhRez}.
%Assume the contrary. Then by Lemma\ref{Le46.0} there exists an open neighborhood $U$
%of the neutral element $e$ in $G$ such that $U\cap U^{-1}$ is nowhere dense. Let $W$ be an open
%neighborhood of $e$ in G such that $W^3\subset U$. Put $O=W\bs\ol{U\cap U^{-1}}$.
%Then, clearly, $O\subset W\subset\ol{O}$ and $O^{-1}\cap U =\0$.
%We are going to define a sequence $\{V_n : n\in\omega\}$ of open neighborhoods of
%the unit and a sequence $\{x_n:n\in\omega\}$ of elements of $G$. Let $x_0$ be any point of $O$.
%and put $V_0=x_0^{-1}O$.
%Assume now that, for some $n\in \omega$ an open subset $V_n$ and $x_n$
%are already defined.Since $e\in W\subset\ol{O}$, we have $x_n\in x_n\ol{O}=\ol{x_nO}$.
%Since $x_nV_n$ is an open neighborhood of $x_n$, it follows that $x_nV_n\cap x_nO\not=\0$.
%We take $x_{n+1}$ to be any point of $x_n(V_n\cap O)$. Choose an open neighborhood $V_{n+1}$ of
%the unit such that $x_{n+1}V_{n+1}^2\subset x_n(V_n\cap O)$.
%Using regularity of X
%\end{proof}
%\begin{proposition} Every Hausdorff pseudocompact quasiregular paratopological group is a
%topological group. CHECK PROOF
%\end{proposition}
%\begin{proof} The proof is very similar to that from \cite[Th. 1.7]{ArhRez}.
%Assume the contrary. Then by Lemma\ref{Le46.0} there exists an open neighborhood $U$
%of the neutral element $e$ in $G$ such that $U\cap U^{-1}$ is nowhere dense. Let $W$ be an open
%neighborhood of $e$ in G such that $\ol{WW}\subset U$. Put $O=W\bs\ol{U\cap U^{-1}}$.
%Then, clearly, $O\subset W\subset\ol{O}$ and $O^{-1}\cap U =\0$.
%We are going to define a sequence {
%\end{proof}
%%\begin{question} Is every $T_3$ pseudocompact paratopological group a topological group?
%%\end{question}
%\begin{question} Is every $T_0/T_1$ Baire pseudocompact left $\omega$-bounded saturated
%paratopological group a topological group?
%\end{question}
%\begin{question} Is every pseudocompact saturated paratopological group a topological group?
%\end{question}
%\begin{proposition}\label{SatPse} Let $G$ be a saturated paratopological group. Then $G$ is 2-pseudocompact
%if and only if $G$ is pseudocompact.
%\end{proposition}
%\begin{proof} {\it Sufficiency.} Suppose that $G$ is a 2-pseudocompact paratopological group. Let
%$\{U_n\}$ be an infinite sequence of nonempty open subsets of $G$. For each $n$ choose a nonempty
%open subset $V_n$ of $G$ such that $V_n^{-1}\subset U_n$. Since $G$ is 2-pseudocompact then there is
%a point $x\in\bigcap_{i\in\omega}\ol{\bigcup_{n\ge i} V_n^{-1}}$. So each neighborhood of the
%point $x$ intersects infinitely many sets of the sequence $\{U_n\}$.
%{\it Necessity.} Suppose that $G$ is a pseudocompact paratopological group.
%Let $\{U_n\}$ be a non-increasing infinite sequence of nonempty open subsets of $G$. For each $n$ choose a nonempty
%open subset $V_n$ of $G$ such that $V_n^{-1}\subset U_n$. Since $G$ is pseudocompact then there is
%a point $x$ such that each neighborhood of the
%point $x$ intersects infinitely many sets of the sequence $\{V_n\}$.
%So $x\in\bigcap \ol{U_n^{-1}}$ thus $G$ is a  2-pseudocompact paratopological group
%\end{proof}
%����� ��� � ���� ������
see p. 5 of RP3
��� �� ���� ������
and there are even still some open porblems.
{\bf subtle} Q from TkaSurvey
Since, by
K\"onig's Lemma, $\frak c$ has
uncountable cofinality,
Since $\R$ is a vector space over the field $\Q$,
there is a basis $B\ni\{1\}$ of the vector space $\R$ over $\Q$.
Choose an arbitrary element $x'_0\in B\bs\{1\}$. Put $x_0=\pi(x'_0)$.
Let $s':\R\to\Q$ be the homomorphism of vector spaces over $\Q$ such that $s'(x'_0)=1$ and
$s'(B\bs\{x'_0\})=\{0\}$. It is easy to check that there exists a unique map $s:\T\to\Q$
Taking, if needed a proper uncountable compact subset $B$ of this set,
we can assure that $1$ does not belong to a linear subspace spanned over $B$ over $\Q$.
�������� ������������ ��������� ��������...

1406.2001.extracted_bib
[AlaSan]{AlaSan Ofelia T. Alas, Manuel Sanchis, {Countably Compact Paratopological Groups, Semigroup Forum, {74 (2007), 423-438.
1406.2001.extracted_bib
[Arh]{Arh Alexander V. Arhangel'ski\u{\i, {Topological function spaces, Kluwer Publ., Dordrecht, 1992.
1406.2001.extracted_bib
[ArhChoKen]{ArhChoKen Alexander V. Arhangel'skii, Mitrofan M. Choban, and Petar S. Kenderov, {Topological games and topologies on groups, Math. Maced., {8 (2010), 1-19.
1406.2001.extracted_bib
[ArhRez]{ArhRez Alexander V. Arhangel'ski\u{\i, E. A. Reznichenko, {Paratopological and semitopological groups versus topological groups, Topology Appl. {151 (2005), 107--119.
1406.2001.extracted_bib
[ArhTka]{ArhTka Alexander V. Arhangel'skii, Mikhail Tkachenko, {Topological groups and related structures, Atlantis Press, Paris; World Sci. Publ., NJ, 2008.
1406.2001.extracted_bib
[Ban]{Ban B. Banashewski, {Minimal topological algebras, Math. Ann, {211 (1974), 107--114.
1406.2001.extracted_bib
[BanDimGut]{BanDimGut Taras O. Banakh, Svetlana Dimitrova, Oleg V. Gutik {Embedding the bicyclic semigroup into countably compact topological semigroups, Topology Appl. {157:18 (2010) 2803--2814. http://arxiv.org/abs/0811.4276
1406.2001.extracted_bib
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