Journal of Algebraic Systems
Vol. XX, No XX, (201X), pp XX-XX

# On Relative Central Extensions and Covering Pairs

A. Pourmirzaei∗, M. Hassanzadeh and B. Mashayekhy

###### Abstract.

Let $(G,N)$ be a pair of groups. In this article, first we construct a
relative central extension for the pair $(G,N)$ such that special types of
covering pair of $(G,N)$ are homomorphic image of it. Second, we show that
every perfect pair admits at least one covering pair. Finally, among extending
some properties of perfect groups to perfect pairs, we characterize covering
pairs of a perfect pair $(G,N)$ under some extra assumptions.

MSC(2010): Primary: 20E34; Secondary: 20E22, 20F05
Keywords: Pair of groups, Covering pair, Relative central extension,
Isoclinism of pairs of groups.
Received: 2 March 2015, Accepted: 26 February 2016.
$*$Corresponding author

In this paper the well-known notion of relative central extension of a pair of
groups is used. By a pair of groups we mean a group $G$ and a normal subgroup
$N$ and this is denoted by $(G,N)$. Let $M$ be another group on which an
action of $G$ is given. The $G$-commutator subgroup of $M$ is defined by the
subgroup $[M,G]$ of $M$ generated by all the $G$-commutators

$[m,g]=m^{g}m^{-1},$

in which $g\in G$, $m\in M$ and $m^{g}$ is the action of $g$ on $m$. Ellis [1]
defined the $G$-center of $M$ to be the subgroup

$Z(M,G)=\\{m\in M|m^{g}=m,\forall g\in G\\}$

Now we recall the definition of relative central extension of a pair of
groups.

######  Definition 0.1.

([2]) Let $(G,N)$ be a pair of groups. A relative central extension of the
pair $(G,N)$ consists of a group homomorphism $\sigma:M\rightarrow G$,
together with an action of $G$ on $M$ such that
$(i)\sigma(M)=N$;
$(ii)\sigma(m^{g})=g^{-1}\sigma(m)g$, for all $g\in G$, $m\in M$;
$(iii)m^{\prime\sigma(m)}=m^{-1}m^{\prime}m$, for all $m,m^{\prime}\in M$;
$(iv)Ker(\sigma)\subseteq Z(M,G)$,

Note that for every relative central extension $\sigma:M\rightarrow G$ of a
pair of groups $(G,N)$, we have $Inn(M)\subseteq Im\sigma$, by the property
$(iii)$ in Definition 0.1. Moreover, every relative central extension
$\sigma:M\rightarrow G$ yields an exact sequence

$1\rightarrow ker\sigma\rightarrow
M\stackrel{{\scriptstyle\sigma}}{{\rightarrow}}N\rightarrow 1.$

Conversely, for every exact sequence

$1\rightarrow ker\sigma\rightarrow
M^{\prime}\stackrel{{\scriptstyle\sigma}}{{\rightarrow}}N\rightarrow 1$

with properties $(i)$-$(iv)$, $\rho:M^{\prime}\rightarrow G$ is a relative
central extension. For this reason we do not differ between a relative central
extension and its associated exact sequence.

Let $\sigma:M\rightarrow G$, $\sigma^{\prime}:M^{\prime}\rightarrow G$ be two
relative central extensions of $(G,N)$. In 1998 Eliss [2] introduced a
morphism between these relative central extensions which is a group
homomorphism $\varphi:M\rightarrow M^{\prime}$ satisfying
$\sigma^{\prime}\varphi(m)=\sigma(m)$ and $\varphi(m^{g})=(\varphi(m))^{g}$
for all $g\in G$, $m\in M$. In particular, if $\varphi$ is a surjective
homomorphism, then $\sigma^{\prime}$ is called a homomorphic image of
$\sigma$. The resulting category is the category of relative central
extensions which Eliss denoted it by $\mathcal{R}\mathcal{C}\mathcal{E}$
$(G,N)$.

Now we recall the definition of a covering pair.

######  Definition 0.2.

([2]) A relative central extension $\sigma:M^{*}\rightarrow G$ of the pair
$(G,N)$ is called a covering pair for $(G,N)$ if there exists a subgroup $A$
of $M^{*}$ such that
$(i)A\subseteq Z(M^{*},G)\cap[M^{*},G]$;
$(ii)A\cong M(G,N)$;
$(iii)N\cong M^{*}/A$,
where $M(G,N)$ is the Schur multiplier of the pair $(G,N)$.

A covering pair $\sigma:G^{*}\rightarrow G$ of the pair $(G,G)$ coincides with
the usual notion of a covering group $G^{*}$ of the group $G$. In 1998 Ellis
[2] showed that every pair of finite groups admits at least one covering pair.

In this paper we consider $P$ as a perfect group that satisfies in maximal
condition which means $\phi(P)\neq P$ such that $\phi(P)$ is the Frattini
subgroup of $P$. We mention that if $\sigma:P\rightarrow G$ is a covering pair
of $(G,N)$, then we have $P^{\prime}=[P,G]$ and therefore

$N=\sigma(P)=\sigma([P,G])=[\sigma(P),G]=[N,G].$

Let $(G,N)$ be a pair of groups with free presentation $G\cong F/R$ and
$N\cong S/R$ for a normal subgroup $S$ of $F$, where $F$ is the free group on
the set $G$. In Section 1, we introduce a relative central extension
$\delta:S/[R,F]\rightarrow G$ which has an important role throughout the
paper. As a consequence, we show that for every covering pair
$\sigma:P\rightarrow G$, $P$ is a homomorphic image of $S/[R,F]$. We remark
that in 2007 Salemkar, Moghaddam and Chiti [7, Lemma 2.5] claimed that there
exists an epimorphism from $\delta$ to any relative central extension without
any condition. We present a counterexample to show that their claim is not
correct and hence one of their main results [7, Theorem 2.6] is not valid.

In 1978 Loday [5] extended the notion of perfect group to perfect pair in the
sense that a pair of groups $(G,N)$ is perfect if $[G,N]=N$. Moreover he
proved that $(G,N)$ is a perfect pair of groups if and only if
$\mathcal{R}\mathcal{C}\mathcal{E}$ $(G,N)$ has a universal object. To prove
this he used some cohomological methods. In Section 2, by restriction of
$\delta$ to $[S,F]/[R,F]$ we obtain a universal object in
$\mathcal{R}\mathcal{C}\mathcal{E}$ $(G,N)$ when $(G,N)$ is perfect. This is
also a covering pair of $(G,N)$. It is worth to mention that we use only
presentation methods instead of cohomological methods which seems easier.
In sequel we extend a result of Schur on covering groups to covering pairs.

##  1\. Some Results for Relative Central Extensions

Let $(G,N)$ be a pair of groups with a free presentation $1\rightarrow
R\rightarrow F\stackrel{{\scriptstyle\pi}}{{\rightarrow}}G\rightarrow 1$ such
that $N\cong S/R$ for a normal subgroup $S$ of $F$. Define the group
homomorphism

$\displaystyle\delta:\frac{S}{[R,F]}$ $\displaystyle\rightarrow$
$\displaystyle G,$ (1.1) $\displaystyle s[R,F]$ $\displaystyle\mapsto$
$\displaystyle\pi(s).\ \ \ \ $ (1.2)

It is straightforward to check that $\delta$ is a relative central extension
by the following action

$\displaystyle\bar{\delta}:\frac{S}{[R,F]}\times\frac{F}{R}$
$\displaystyle\rightarrow$ $\displaystyle\frac{S}{[R,F]},$ (1.3)
$\displaystyle(s[R,F],fR)$ $\displaystyle\mapsto$ $\displaystyle s^{f}[R,F].\
\ \ \ \ \ \ \ $ (1.4)

In the following theorem we present a relation between the above relative
central extension to anyone of the pair $(G,N)$, especially to any covering
pair $\sigma:M\rightarrow G$ of $(G,N)$.

######  Theorem 1.1.

Let $G\cong F/R$, where $F$ is the free group on the set $G$ and $N$ be a
normal subgroup of $G$ with $N\cong S/R$ for a normal subgroup $S$ of $F$. If
$\sigma:M\rightarrow G$ is a relative central extension of the pair $(G,N)$,
then there exists a homomorphism $\beta:S/[R,F]\rightarrow M$ such that the
following diagram is commutative:

$\begin{array}[]{ccccccccc}1&\rightarrow&\frac{R}{[R,F]}&\rightarrow&\frac{S}{[R,F]}&\stackrel{{\scriptstyle\delta}}{{\rightarrow}}&N&\rightarrow&1\\\
&&\beta|\downarrow&&\beta\downarrow&&\parallel&&\\\
1&\rightarrow&A&\rightarrow&M&\stackrel{{\scriptstyle\sigma}}{{\rightarrow}}&N&\rightarrow&1,\\\
\end{array}$

where $\delta$ is the relative central extension $(4)$. In particular, if M is
a perfect group with $\phi(M)\neq M$, then the above homomorphism $\beta$ is
an epimorphism.

###### Proof.

Consider the map $f:G\rightarrow N$ by the rule $f(g)=g$ if $g\in N$,
otherwise $f(g)=1$. Now $f$ induces a homomorphism $\theta:F\rightarrow N$.
Using the projective property of free groups, there is a homomorphism
$\alpha:F\rightarrow M$ such that $\sigma\alpha=\theta$. By the restriction of
$\theta$ and $\alpha$ to $S$ we have the following commutative diagram:

$\begin{array}[]{ccccc}&&S&&\\\ &\alpha|\swarrow&\downarrow\theta|&&\\\
M&\stackrel{{\scriptstyle\sigma}}{{\longrightarrow}}&N&\longrightarrow&1.\\\
\end{array}$

For $x\in F$ and $r\in R$, $\alpha([r,x])=[\alpha(r),\alpha(x)]=1$ since
$\alpha(r)\in ker\sigma\leq Z(M,G)\leq Z(M)$. Therefore $[R,F]\leq ker\alpha|$
which implies the existence of

$\beta:\frac{S}{[R,F]}\rightarrow M.$

Clearly $\beta(R/[R,F])\leq ker\sigma,$ which gives the restriction of $\beta$
to $R/[R,F]$, $\beta|$. Now let $A=ker\sigma\leq Z(M,G)$ and $\phi(M)\neq M$.
For every $m\in M$, there exists $\bar{s}\in S/[R,F]$ such that
$\sigma(m)=\delta(\bar{s})=\sigma\beta(\bar{s})$. Hence $m=\beta(\bar{s})a$,
for some $a\in ker\sigma$. Thus

$M=\langle\beta(\bar{s}),a\ \ |\ \ \bar{s}\in\frac{S}{[R,F]},a\in A\rangle.$

Since $M=M^{\prime}$ so

$A\leq Z(M,G)\cap M^{\prime}\leq Z(M)\cap M^{\prime}\leq\phi(M).$

Hence

$M=\langle\beta(\bar{s})\ \ |\ \ \bar{s}\in\frac{S}{[R,F]}\rangle$

which implies that $\beta$ is an epimorphism. ∎

######  Corollary 1.2.

With the assumptions and notations of the previous theorem, for every covering
pair $\sigma:P\rightarrow G$ of $(G,N)$, $P$ is a homomorphic image of
$S/[R,F]$.

Note that in Theorem 1.1 if

$\beta(\bar{s}^{g})=\beta(\bar{s})^{g}$

for every $\bar{s}\in S/[R,F]$ and $g\in G$ (in other words $\beta$ preserves
the action), then we can say that every covering pair $\sigma:P\rightarrow G$
is a homomorphic image of $\delta:S/[R,F]\rightarrow G$.

######  Remark 1.3.

In 2007 Salemkar, Moghaddam and Chiti [7, Lemma 2.5] claimed that in the above
theorem $\beta$ is always an epimorphism for any free presentation $G\cong
F/R$ and $N\cong S/R$ without any condition. The following counterexample
shows that their claim is not true and hence one of the main results of their
paper [7, Theorem 2.6] is not valid.

######  Example 1.4.

Put $G=N=\textbf{Z}\oplus...\oplus\textbf{Z}$ a free abelian group of finite
rank with the free presentation $G\cong F/F^{\prime}\cong N$, where
$F^{\prime}$ is the commutator subgroup of the free group $F$. Consider the
following commutative diagram:

$\begin{array}[]{ccccccccc}1&\longrightarrow&\frac{F^{\prime}}{[F^{\prime},F]}&\stackrel{{\scriptstyle\subseteq}}{{\longrightarrow}}&\frac{F}{[F^{\prime},F]}&\stackrel{{\scriptstyle\delta}}{{\longrightarrow}}&G=N=\textbf{Z}\oplus...\oplus\textbf{Z}&\longrightarrow&1\\\
&&\beta|\downarrow&&\beta\downarrow&&\parallel&&\\\
1&\longrightarrow&A=\textbf{Z}&\stackrel{{\scriptstyle\alpha}}{{\longrightarrow}}&M=G\oplus\textbf{Z}&\stackrel{{\scriptstyle\sigma}}{{\longrightarrow}}&G=N=\textbf{Z}\oplus...\oplus\textbf{Z}&\longrightarrow&1,\\\
\end{array}$

where $\alpha(x)=(1,x)$ and $\sigma(g,x)=g$, for all $x\in A$, $g\in G$. Then
we have

$\beta(\frac{F^{\prime}}{[F^{\prime},F]})=[\beta(\frac{F}{[F^{\prime},F]}),\beta(\frac{F}{[F^{\prime},F]})]\leq
M^{\prime}=1.$

Hence $\beta$ is not an epimorphism.

At the end of this section by a result of Ellis [2, Corollary 1.2] on the
Schur multiplier of a pair $(G,N)$ of finite nilpotent groups we present a
covering pair of the pair $(G,N)$.

######  Theorem 1.5.

Let $(G,N)$ be a pair of finite nilpotent groups. Assume that
$G=\prod^{k}_{i=1}S_{i}$, where $S_{i}$ is the Sylow $p_{i}$-subgroup of $G$
and $\sigma_{i}:M_{i}\rightarrow S_{i}$ for each $i=1,...,k$, is an arbitrary
covering pair of the pair $(S_{i},S_{i}\cap N)$. Then
$\sigma:\prod^{k}_{i=1}M_{i}\rightarrow G$ defined by
$\sigma(\\{m_{i}\\}_{i=1}^{k})=\\{\sigma_{i}(m_{i})\\}_{i=1}^{k}$ is a
covering pair of $(G,N)$.

###### Proof.

It is readily seen that $\sigma$ is a relative central extension of $(G,N)$.
By [2, Corollary 1.2] $M(G,N)\cong\Pi_{i=1}^{k}M(S_{i},S_{i}\cap N)$. hence
the proof is straightforward. ∎

##  2\. Some Properties of Covering Pairs

The aim of this section is to introduce a covering pair for a perfect pair of
groups $(G,N)$.

######  Theorem 2.1.

Let $(G,N)$ be a perfect pair of groups. Then for every covering pair
$\sigma:M\rightarrow G$, we have $[M,G]=M$.

###### Proof.

Since $\sigma:M\rightarrow G$ is a covering pair, we have $M/ker\sigma\cong N$
and $ker\sigma\leq[M,G]$. Hence

$\frac{M/ker\sigma}{[M,G]/ker\sigma}\cong\frac{N}{[N,G]}.$

The result is a consequence of $(G,N)$ being perfect. ∎

It is known that if $G\cong F/R$ is a group with a covering group $G^{*}$,
then there exists a normal subgroup $S$ of $F$ such that
$R/[R,F]=(F^{\prime}\cap R)/[R,F]\times S/[R,F]$, and we have an isomorphism
$F/S\cong G^{*}$ [6, Theorem 2.4.6(iv)(b)]. In the following theorem we intend
to extend this result to pairs of groups.

######  Theorem 2.2.

Let $(G,N)$ be a pair of groups with $G\cong F/R$ and $N\cong S/R$, where $F$
is the free group on the set $G$ and $S\unlhd F$. Then for every relative
central extension $\sigma:P\rightarrow G$ with $A\cong ker\sigma$ there is a
normal subgroup $T$ of $F$ such that

$\frac{R}{[R,F]}=\frac{R\cap[S,F]}{[R,F]}\times\frac{T}{[R,F]}.$

Moreover, there exists an isomorphism $S/T\cong P$ which carries $R/T$ onto
$A$.

###### Proof.

Let $\pi:F\rightarrow G$ be a surjective homomorphism with $R=ker\pi$. As the
proof of Theorem 1.1, there exists a homomorphism $\psi:S\rightarrow P$ which
induces the epimorphism $\beta:S/[R,F]\rightarrow P$. Thus $\psi$ is in fact
an epimorphism. By setting $T=ker\psi$, we have $S/T\cong P$. Clearly
$\psi(R)=A$ and by Theorem 1.1, $\psi([R,F])=1$ so $[R,F]\leq T$ which gives
the induced epimorphism $\bar{\psi}:R/[R,F]\rightarrow A$. Considering

$\bar{\psi}|:\frac{R\cap[S,F]}{[R,F]}\rightarrow A,$

we claim that $\bar{\psi}|$ is an isomorphism. To prove this, let $a\in A$. By
$\psi(R)=A$ there exists $x\in R$ such that $a=\psi(x)$. On the other hand for
every $p\in P$ and $g\in G$,

$[p,g]=[\beta(\bar{s}),g]=\beta(\bar{s}^{-1})\beta(\bar{s}^{g}).$

Easily it can be seen that

$\beta({\bar{s}^{g}})=\left\\{{\begin{array}[]{*{20}{c}}{\beta{{(\bar{s})}^{g}}}&{ifg\in
N}\\\ {\beta(\bar{s})\beta(\bar{r})}&{ifg\notin N}\\\ \end{array}}\right.$

for some $\bar{r}\in R/[R,F]$.
Thus by the proof of Theorem 1.1,

$P=\langle\beta(\bar{s})\ \ |\ \
\bar{s}\in\frac{S}{[R,F]}\setminus\frac{R}{[R,F]}\rangle.$

Since $P$ is perfect and so $(G,N)$ is perfect, we have $[S,F]R=S$. On the
other hand $A=\psi(R)\leq\phi(P)$. Therefore $[P,G]\leq\psi([S,F])$. This fact
and $A\leq[M,G]$ imply that $a=\psi(y)$ for some $y\in[S,F]$. Thus
$\psi(x)=a=\psi(y)$, hence $x^{-1}y\in ker\psi\leq R$, so that $y\in
R\cap[S,F]$. Since $A\cong M(G,N)\cong(R\cap[S,F])/[R,F]$ and $A$ is finite,
$\bar{\psi}|$ is an isomorphism. By surjectivity of $\bar{\psi}$, for every
$\bar{r}\in R/[R,F]$ there exists $\bar{x}\in(R\cap[S,F])/[R,F]$ such that
$\bar{\psi}(\bar{x})$=$\bar{\psi}(\bar{r})$, thus $\bar{x}^{-1}\bar{r}\in
ker\psi$. Consequently

$\frac{R}{[R,F]}=ker\bar{\psi}\ \frac{R\cap[S,F]}{[R,F]}.$

Let $\bar{x}\in ker\bar{\psi}\cap(R\cap[S,F])/[R,F]$. Then $\bar{x}\in
ker\bar{\psi}|=1$, so we have the following isomorphism

$\frac{R}{[R,F]}\cong
ker\bar{\psi}\times\frac{R\cap[S,F]}{[R,F]}=\frac{T}{[R,F]}\times\frac{R\cap[S,F]}{[R,F]}.$

∎

######  Theorem 2.3.

Let $\sigma_{i}:M_{i}\rightarrow G$, $i=1,2$ be two covering pairs of a finite
pair $(G,N)$. If $M_{i}=M^{\prime}_{i}$ and $\phi(M_{i})\neq M_{i}$ for
$i=1,2$, then
$(i)M_{1}\cong M_{2}$;
$(ii)M_{1}/Z(M_{1},G)\cong M_{2}/Z(M_{2},G)$;
$(iii)Z(M_{1},G)/ker\sigma_{1}\cong Z(M_{2},G)/ker\sigma_{2}$.

###### Proof.

$(i)$ Let $\sigma:M^{*}\rightarrow G$ be a fixed arbitrary covering pair with
the following relative central extension

$1\rightarrow A\rightarrow M^{*}\rightarrow N\rightarrow 1$

such that $A\subseteq Z(M^{*},G)\cap[M^{*},G]$ and $A\cong M(G,N)$. Since
$[N,G]=N$ and $[M^{*},G]=M^{*}$ so $\sigma:[M^{*},G]\rightarrow[N,G]$ is an
epimorphism. Therefore

$\begin{array}[]{*{20}{c}}{|[M^{*},G]|}&=&{|\ker\sigma|}&{|[N,G]|}\\\
{}\hfil&=&{|A|}&{|[N,G]|}\\\ {}\hfil&=&{|M(G,N)|}&{|[N,G]|}.\\\
{}\hfil&{}\hfil&{}\hfil&{}\hfil\\\ \end{array}$

Suppose that $G\cong F/R$ such that $F$ is the free group on the set $G$ with
$N\cong S/R$. By Theorem 1.1, we have the following diagram

$\begin{array}[]{ccccccccc}1&\rightarrow&\frac{R}{[R,F]}&\rightarrow&\frac{S}{[R,F]}&\stackrel{{\scriptstyle\delta}}{{\rightarrow}}&N&\rightarrow&1\\\
&&\beta|\downarrow&&\beta\downarrow&&\parallel&&\\\
1&\rightarrow&A&\rightarrow&M^{*}&\stackrel{{\scriptstyle\sigma}}{{\rightarrow}}&N&\rightarrow&1.\\\
\end{array}$

Since $[S/[R,F],G]=[S,F]/[R,F]$ and $\delta_{|}:[S,F]/[R,F]\rightarrow[N,G]$
is an epimorphism, thus

$|\frac{[S,F]}{[R,F]}|=|M(G,N)|\\\ |[N,G]|.$

Therefore

$|M^{*}|=|[M^{*},G]|=|\frac{[S,F]}{[R,F]}|.$

If $\bar{s}\in S/[R,F]$ and $g\in G$,

$\beta([\bar{s},g])=\beta{(\bar{s})^{-1}}\beta({\bar{s}^{g}})=\left\\{{\begin{array}[]{*{20}{c}}{\beta{{(\bar{s})}^{-1}}\beta{{(\bar{s})}^{g}}=[\beta(\bar{s}),g]}&{ifg\in
N}\\\ {\beta{{(\bar{s})}^{-1}}\beta(\bar{s})=1}&{ifg\notin N}\\\
\end{array}}\right.$

which yields that $\beta([S,F]/[R,F])\leq[M^{*},G].$ By Theorem 2.1,
$M^{*}=[M^{*},G]\leq\beta([S,F]/[R,F])$ which implies that
$M^{*}\cong[S,F]/[R,F]$.
$(ii)$ By Theorem 2.2 there exists a normal subgroup $T$ of $F$ with
$M^{*}\cong S/T.$ This yields naturally an action of $G$ on $S/T$ which
implies $Z(S/T,G)=Z(M^{*},G).$ Put $Z(S/T,G)=L/T.$ Let $x\in Z(S/[R,F],G)$.
Then $[x,g]\in[R,F]\leq T$ and so $Z(S/[R,F],G)\leq L/[R,F]$. To prove the
reverse containment, assume that $x[R,F]\in L/[R,F]$, then $[x,g]\in T\leq
R\cap[S,F]\cap T=[R,F]$, hence

$\frac{L}{[R,F]}=Z(\frac{S}{[R,F]},G).$

Consequently, it may be inferred that

$\frac{M^{*}}{Z(M^{*},G)}\cong\frac{S/T}{L/T}\cong\frac{S}{L}.$

The desired assertion is now a consequence of the fact that $S/L$ is
determined by the presentation of $G$ and $N$.
$(iii)$ Owing to Theorem 1.1 we have

$\frac{Z(M^{*},G)}{A}\cong\frac{L/T}{R/T}\cong\frac{L}{R}.$

The result is now established. ∎

Now, we are in a position to state and prove one of the main results of this
section.

######  Theorem 2.4.

Let $(G,N)$ be a perfect pair of groups with a free presentation $G\cong F/R$
and $N\cong S/R$. Then
$(i)\ \ \delta:[S,F]/[R,F]\rightarrow G$ by $\delta(x[R,F])=xR$, is a covering
pair of $(G,N)$
$(ii)$ The relative central extension $\delta:[S,F]/[R,F]\rightarrow G$ is the
universal object in the category $\mathcal{R}\mathcal{C}\mathcal{E}$ $(G,N)$.

###### Proof.

$(i)$ Consider the following action of $G$ on $[S,F]/[R,F]$

$\frac{[S,F]}{[R,F]}\times G\rightarrow\frac{[S,F]}{[R,F]}$
$([s,f][R,F],g)\mapsto([s,f][R,F])^{g}=[s^{t},f^{t}][R,F],$

where $\pi:F\rightarrow G$ is the epimorphism with $ker\pi=R$ and $\pi(t)=g$.
By the assumption $G=NQ$, let $Q\cong T/R$. Since $(G,N)$ is a perfect pair,
$[N,G]=N$ and therefore $([S,F]R)/R=S/R$. It can be shown that
$\delta:[S,F]/[R,F]\rightarrow G$ is a relative central extension of the pair
$(G,N)$. On the other hand, since

$ker\delta=(R\cap[S,F])/[R,F]\cong M(G,N),$

we have

$\frac{[S,F]/[R,F]}{ker\delta}\cong\frac{[S,F]}{R\cap[S,F]}\cong\frac{[S,F]R}{R}=\frac{S}{R}\cong
N.$

It is enough to show that

$ker\delta\leq[\frac{[S,F]}{[R,F]},G].$

For this we show that $[[S,F]/[R,F],G]=[S,F]/[R,F]$. Clearly

$[[S,F]/[R,F],G]\leq[S,F]/[R,F].$

So we only need to show the reverse containment. We know that

$[\frac{[S,F]}{[R,F]},G]=\langle[x,g]|x\in\frac{[S,F]}{[R,F]},g\in G\rangle$

such that

$[x,g]=x^{-1}x^{g}=[s,f][R,F]^{-1}([s,f][R,F])^{g}=[s,f]^{-1}[s,f]^{t}[R,F],$

for some $s\in S$ and $f\in F$, where $\pi(t)=g$. Since $S=[S,F]R$, for every
$s\in S$ there exist $s^{\prime}\in S$, $l\in F$ and $r\in R$ such that

$[s,f]=[[s^{\prime},l]r,f]=[[s^{\prime},l],f][[s^{\prime},l],f,r][r,f].$

Hence $[S,F]/[R,F]\leq[[S,F]/[R,F],G]$.
$(ii)$ Let $\sigma:M\rightarrow G$ be a relative central extension of the pair
$(G,N)$, and $F$ be the free group on the set $G$ and $S\unlhd F$ such that
$G\cong F/R$ and $N\cong S/R$. Then consider $\delta:[S,F]/[R,F]\rightarrow G$
as mentioned in the previous part. By Theorem1.1, there exists a homomorphism

$\varphi:[S,F]/[R,F]\rightarrow M$

such that $\varphi=\beta|_{[S,F]/[R,F]}$. Now for every $g\in G$ and
$\bar{x}\in[S,F]/[R,F]$,

$\varphi({{\bar{x}}^{g}})=\beta({{\bar{x}}^{g}})=\left\\{{\begin{array}[]{*{20}{c}}{\beta{{\left({\bar{x}}\right)}^{g}}=\varphi{{\left({\bar{x}}\right)}^{g}}}&{ifg\in
N}\\\ {\beta(\bar{x})=\varphi\left({\bar{x}}\right)}&{ifg\notin N}\\\
\end{array}}\right.$

We have to prove $\varphi$ preserves the action. To do this it is enough to
show that $\varphi(\bar{x})^{g}=\varphi(\bar{x})$ for every $g\in G$ and
$\bar{x}\in[S,F]/[R,F]$. Define

$\displaystyle\ f:\frac{S}{[R,F]}$ $\displaystyle\rightarrow$ $\displaystyle\
M$ $\displaystyle\bar{s}$ $\displaystyle\mapsto$
$\displaystyle\beta(\bar{s})^{g}\beta(\bar{s})^{-1}.$

For every $g\in G$ and $\bar{x}\in[S,F]/[R,F]$, $f(\bar{s})\in Z(G,M)$ so

$f({{\bar{s}}^{g}})=\beta{({{\bar{s}}^{g}})^{g}}\beta{({{\bar{s}}^{g}})^{-1}}=\left\\{{\begin{array}[]{*{20}{c}}{{{(\beta{{(\bar{s})}^{g}})}^{g}}{{(\beta{{(\bar{s})}^{-1}})}^{g}}=f{{(\bar{s})}^{g}}}&{ifg\in
N}\\\ {\beta{{(\bar{s})}^{g}}\beta{{(\bar{s})}^{-1}}}&{ifg\notin N}\\\
\end{array}}\right.$

Thus,

$f([\bar{s},g])=f({\bar{s}^{-1}}{\bar{s}^{g}})=\left\\{{\begin{array}[]{*{20}{c}}{f{{(\bar{s})}^{-1}}f{{(\bar{s})}^{g}}=[f(\bar{s}),g]=1}&{ifg\in
N}\\\ {f{{(\bar{s})}^{-1}}f(\bar{s})=1}&{ifg\notin N}\\\ \end{array}}\right.$

By part $(i)$

$f(\frac{[S,F]}{[R,F]})=f([\frac{[S,F]}{[R,F]},G])=1.$

Therefore

$\varphi(\bar{x}^{g})=\beta(\bar{x}^{g})=\beta(\bar{x})^{g}=\varphi(\bar{x})^{g}.$

We claim that $\varphi$ is unique. By contrary, let
$\varphi_{i}:[S,F]/[R,F]\rightarrow M$, $i=1,2$, be homomorphisms such that

$\sigma\varphi_{1}(\bar{s})=\delta(\bar{s})=\sigma\varphi_{2}(\bar{s}).$

Define

$\displaystyle\psi:\frac{[S,F]}{[R,F]}$ $\displaystyle\rightarrow$
$\displaystyle M$ $\displaystyle\bar{x}$ $\displaystyle\mapsto$
$\displaystyle\varphi_{1}(\bar{x}){\varphi_{2}}^{-1}(\bar{x}).$

If $g\in G$ and $\bar{x}\in[S,F]/[R,F]$,

$\psi({\bar{x}}^{g})=\varphi_{1}({\bar{x}}^{g}){\varphi_{2}}^{-1}({\bar{x}}^{g})={\varphi_{1}(\bar{x})}^{g}({\varphi_{2}}^{-1}(\bar{x}))^{g}=(\varphi_{1}(\bar{x}){\varphi_{2}}^{-1}(\bar{x}))^{g}=\psi(\bar{x})^{g}.$

Consequently $\psi([\bar{x},g])=[\psi(\bar{x}),g]=1.$ Now by $(i)$ we have
$[[S,F]/[R,F],G]=[S,F]/[R,F]$ which implies that $\psi=1$. Hence the results
holds. ∎

Note that if $(G,N)$ is a pair of groups with a free presentation $G\cong F/R$
and $N\cong S/R$, then $[S,F]/[R,F]$ is independent of the choice of the free
presentation of $(G,N)$.

The following theorem states a necessary and sufficient condition for a pair
of groups to be perfect. It should be noted that Loday [5] proved the
following result using homological method but our proof is different and seems
elementary.

######  Theorem 2.5.

A pair of groups $(G,N)$ is perfect pair if and only if the category
$\mathcal{R}\mathcal{C}\mathcal{E}$ $(G,N)$ has a universal relative central
extension.

###### Proof.

Necessity is Theorem 2.4. For sufficiency, let

$1\rightarrow A\rightarrow
M\stackrel{{\scriptstyle\rho}}{{\rightarrow}}G\rightarrow 1$

be a universal relative central extension of the pair $(G,N)$. Consider the
following relative central extension

$1\rightarrow A\times\frac{N}{[N,G]}\rightarrow
M\times\frac{N}{[N,G]}\stackrel{{\scriptstyle\psi}}{{\rightarrow}}N\rightarrow
1,$

where $\psi(m,\bar{n})=\rho(m)$, $m\in M$, $\bar{n}\in N/[N,G]$. Define
homomorphisms

$\theta_{i}:M\rightarrow M\times\frac{N}{[N,G]},$

$i=1,2$, by $\theta_{1}(m)=(m,1)$ and $\theta_{2}(m)=(m,\overline{\rho(m)})$.
Then we have $\psi\theta_{i}=\rho$, $i=1,2$, where $\theta_{1}=\theta_{2}$.
Consequently, $N=[N,G]$ and hence $(G,N)$ is a perfect pair. ∎

Finally, we intend to consider a covering pair as a pair of groups. If
$\sigma:M\rightarrow G$ is a covering pair of $(G,N)$, then there exists a
subgroup $A$ of $M$ such that $A\subseteq Z(M,G)\cap[M,G]$, $A\cong M(G,N)$
and $M/A\cong N$. Now, we can consider $(M,A)$ as a covering pair of $(G,N)$.
It is known that any two covering groups of a finite group $G$ are isoclinic.
Moreover, if $G$ is a finite perfect pair of groups, then a covering group of
$G$ is also perfect. We intend to investigate these facts for a covering pair
as our new point of view of a covering pair of groups. In this way, the notion
of isoclinism for the pair of groups is needed which we review it of [8].

######  Definition 2.6.

Let $(G,N)$ and $(H,K)$ be two pairs of groups. The pairs $(G,N)$ and $(H,K)$
are said to be isoclinic if there exists isomorphisms
$\epsilon:G/Z(G,N)\rightarrow H/Z(H,K)$ and $\eta:[G,N]\rightarrow[H,K]$ such
that $\epsilon(N/Z(G,N))=K/Z(H,K)$ and $\eta([g,n])=[h,k]$ whenever
$\epsilon(gZ(G,N))=hZ(H,K)$ and $\eta(nZ(G,N))=kZ(H,K)$. This concept is
denoted as $(G,N)\sim(H,K)$.

It is a well-known fact that all covering groups of a given finite group are
mutually isoclinic (see [4]). The following lemma helps us to prove this fact
for some special types of covering pairs for arbitrary pair of groups.

######  Lemma 2.7.

Let (G,N) and (H,K) be two pairs of groups. If $\theta:G\rightarrow H$ is an
epimorphism such that $\theta(N)=K$ and $ker\theta\cap N=1$, then
$(G,N)\sim(H,K).$

###### Proof.

Define $\epsilon:G/Z(G,N)\rightarrow H/Z(H,K)$ such that
$\epsilon(gZ(G,N))=\theta(h)Z(H,K)$ and $\varphi:[N,G]\rightarrow[K,H]$ by
$\varphi([n,g])=[\theta(n),\theta(g)]$. We can easily see that $\epsilon$ and
$\varphi$ are isomorphisms which yields that $(G,N)\sim(H,K)$. ∎

######  Theorem 2.8.

Let $(G,N)$ be a pair of groups. Then every two covering pairs
$(P_{i},A_{i})$, $i=1,2$, where $P_{i}=P^{\prime}_{i}$ and $\phi(P_{i})\neq
P_{i}$ of $(G,N)$ are isoclinic.

###### Proof.

By the proof of Theorem 2.2 if $(P,A)$ is a covering pair of $(G,N)$ then
there exists an epimorphism $\beta:S/[R,F]\rightarrow P$ such that
$\psi(([S,F]\cap R)/[R,F])=A$. Clearly

$ker\psi\cap[\frac{S}{[R,F]},\frac{[S,F]\cap R}{[R,F]}]=1.$

Using Lemma 2.7 we obtain

$(P,A)\sim(\frac{S}{[R,F]},\frac{[S,F]\cap R}{[R,F]}).$

∎

The following results can be easily obtained if we use the notion of
isoclinism for a perfect pair of groups. Note that these results are
generalizations of similar results for perfect groups to perfect pairs ( see
[4, Lemma 2.4, Corollary 2.5 and Proposition 2.6]).

######  Lemma 2.9.

Let $(G,N)$ be a finite perfect pair of groups, and $(H,K)$ be any finite pair
of groups which is isoclinic to $(G,N)$. Then $K$ is isomorphic to

$\frac{N\times Z(H,K)}{Z(G,N)\times(Z(H,K)\cap[H,K])}.$

######  Corollary 2.10.

Let $(G,N)$ be a finite perfect pair of groups with trivial center. Then for
any pair of groups isoclinic to $(G,N)$, say $(H,K)$, we have $K$ is
isomorphic to direct product of $N$ and an abelian group.

######  Proposition 2.11.

Let $(G,N)$ be a finite perfect pair of groups, and $(H,K)$ be any pair which
is isoclinic to $(G,N)$ such that $|N|=|K|$. Then $N\cong K$.

###### Proof.

The result follows from Lemma 3.9 and the assumption that $|K|=|N|.$ ∎

## References

  * [1] G. Ellis, Capability, homology and central series of a pair of groups, J. Algebra (1996), 179:31–46.
  * [2] G. Ellis, The Schur multiplier of a pair of groups, Appl. Categ. Structures 6(1998), 355–371.
  * [3] N. S. Hekster, On the structure of n-isoclinism classes of groups, J. Pure and Applied Algebra 40(1986), 63–85.
  * [4] M. R. Jones, J. Wiegold, Isoclinisms and covering groups, Bull. Austral. Math. Soc. (1974), 11:71 -76.
  * [5] J. L. Loday, Cohomologie et group de Steinberg relatif, J. Algebra 54 (1978), 178–202.
  * [6] G. Karpilovsky, The Schur Multiplier, Oxford Clarendon Press, 1987.
  * [7] A. R. Salemkar, M. R. R. Moghaddam, K. Chiti, Some properties on the Schur multiplier of a pair of groups, J. Algebra 312 (2007), 1–8.
  * [8] A. R. Salemkar, F. Saeedi and T. Karimi, The structure of isoclinism of pairs of groups, Southeast Asian Bull. Math. 31 (2007), 1173-1182.

Azam Pourmirzaei
Department of Mathematics, Hakim Sabzevari University, P. O. Box 96179-76487,
Sabzevar, Iran
<EMAIL_ADDRESS>

Mitra Hassanzadeh
Department of Mathematics, Department of Mathematics, Ferdowsi University of
Mashhad, P.O.Box 1159-91775, Mashhad, Iran.
<EMAIL_ADDRESS>

Behrooz Mashayekhy
Department of Mathematics, Center of Excellence in Analysis on Algebraic
Structures, Ferdowsi University of Mashhad, P.O.Box 1159-91775, Mashhad, Iran.
<EMAIL_ADDRESS>