\documentclass[11pt]{article} \usepackage{amsfonts} \usepackage{amsmath,amscd} \usepackage{amssymb} \usepackage{amsthm} \usepackage{newlfont} \textheight=22cm \textwidth=14cm \hoffset=-1cm \voffset=-1.5cm \newcommand{\f}{\frac} \newtheorem{thm}{Theorem}[section] \newtheorem{cor}[thm]{Corollary} \newtheorem{lem}[thm]{Lemma} \newtheorem{prop}[thm]{Proposition} \theoremstyle{definition} \newtheorem{defn}[thm]{Definition} \theoremstyle{remark} \newtheorem{rem}[thm]{Remark} \newcommand{\Z}{{\it\sf Z}\hspace{-.31cm}{\it\sf Z}} \newcommand{\ds}{\displaystyle} \newcommand{\st}{\stackrel} \renewcommand{\baselinestretch}{1.8} \begin{document} \title{On The Relative Non-Abelian Tensor Product of a Prime Power Group} \author{ S. Hadi Jafari \\ {\small Department of Mathematics, Islamic Azad University, Mashhad-Branch, } \vspace{-.4cm}\\ {\small Mashhad, Iran} %\vspace{.1cm}\\ \vspace{.1cm}\\ }\maketitle \begin{abstract} Let $(G,N)$ be a pair of prime power groups. We give a new upper bound for $|N\otimes G|$, where $N\otimes G$ is the non-abelian tensor product of $N$ and $G$. Among other results, the relative Schur multiplier of free product of groups is determined under some conditions. \footnotetext{2000 Mathematics Subject Classification(Amer. Math. Soc.): 19C09 \\Keywords: Non-abelian tensor product, Schur multiplier \\ } \end{abstract} \section{Introduction}\ \ \ \ When the concept of non-abelian tensor product of groups introduced by R. Brown and J.-L. Loday \cite{BL} in 1987, many people interested to study and apply it to different scopes of group theory. One of these attempts is to find upper bound for the order of this group. For instance, by the conjugation action it is always meaningful to consider the non-abelian tensor product of a group and it's normal subgroup. G. Ellis \cite{E} has shown that when $N$ is a normal subgroup of a $d$-generator finite $p$-group $G$ and $|N|=p^n$, then $|N\otimes G|\leq p^{nd}$. In this article we obtain a new upper bound $p^{(n-s)d+m}$, where the group $\displaystyle\frac{N}{[N,G]}$ has exponent $p^s$ and $|G^{ab}|=p^m$. In \cite{E1} Ellis introduced the Schur multiplier of the pair of groups to yield sharper results of the usual multiplier ${\mathcal M}(G)$, more study on the pairs of groups and provide non-trivial information on the third integral homology of a group. To see the relation of this subject with the non-abelian tensor product of groups suppose $(G,N)$ is a pair of groups and denote by $J_2(N,G)$ the kernel of epimorphism $N\otimes G\longrightarrow G$ which maps $n\otimes g$ to $[n,g]$ for all $n\in N$ and $g\in G$. In \cite{E1} it is established that the quotient group $\displaystyle\frac{J_2(N,G)}{\nabla(N,G)}$ is isomorphic to the Schur multiplier of the pair $(G,N)$ where $\nabla(N,G)=\langle n\otimes n | n\in N\rangle$ is a subgroup of $N\otimes G$. Our aim is to compute $\nabla(N,G)$ and give the order of non-abelian tensor product of a pair of groups with respect to the order of it's Schur multiplier under some conditions. One of the suggested problems about the non-abelian tensor product of groups in \cite{BJ} was the verifying the treatment of tensor product on the free product of groups. For this purpose N. D. Gilbert \cite{G} computed $J_2(G,G)$ when $G$ is the free product of some groups. To generalize Gilbert's result we will determine $J_2(N_1\ast N_2,G_1\ast G_2)$ where $N_i$ is a normal subgroup of $G_i$, $i=1,2$. %\includegraphics[scale=1]{image1.png} %\begin{figure}[h] %\centering %\includegraphics[width=13cm,height=7cm]{C:\Documents and Settings\s.h.jafari\Desktop\personal/k.jpg} %\caption{Result for the day} %\label{fig:151106b_measure} %\end{figure} \section{Upper bound}\ \ \ \ Let start this section with following lemma. \begin{lem}(\cite{BJ})\label{m0} Let $N$ be a normal subgroup of a group $G$. Let $Z$ be a central subgroup of $G$ contained in $N$. Then the following sequence is exact: \[(N\otimes Z)\times (Z\otimes G)\longrightarrow N\otimes G\longrightarrow N/Z\otimes G/Z\longrightarrow 1,\] if in addition $Z\subseteq [N,G]$, then the sequence \[Z\otimes G\longrightarrow N\otimes G\longrightarrow N/Z\otimes G/Z\longrightarrow 1~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(*)\] is exact. \end{lem} \begin{thm}\label{m1} Let $G$ be a d-generator finite p-group with $G^{ab}$ of order $p^m$. If $N$ is a normal subgroup of $G$ of order $p^n$ and $\displaystyle\frac{N}{[N,G]}$ has exponent $p^s$, then \[|N\otimes G|\leq p^{(n-s)d+m}.\] \end{thm} \begin{proof} If $G$ is of order $p$, then the result holds. Suppose that $G$ is a finite $p$-group and the inequality holds for all $p$-groups of order less than $|G|$. If $G\cong C_{p^{m_1}}\times C_{p^{m_2}}\times...\times C_{p^{m_d}}$, where $0\leq m_1\leq m_2\leq...\leq m_d$ and $m_1+m_2+...+m_d=m$ and also $N\cong C_{p^{n_1}}\times C_{p^{n_2}}\times...\times C_{p^{n_d}}$ where $0\leq n_1\leq n_2\leq...\leq n_d$ and $n_1+n_2+...+n_d=n$, then $|N\otimes G|=p^t$ in which \[\begin{array}{lcl} t&=&\ds\sum_{i=1}^{d}n_i+\ds\sum_{i=1}^{d-1}in_{d-i}+\ds\sum_{j=2}^{d}(\sum_{i=1}^{j-1}min\{n_j,m_i\})\vspace{.3cm}\\ &\leq&d(n_1+n_2+...+n_{d-1})+n_d+m_1+m_2+...+m_{d-1}\vspace{.3cm}\\ &\leq&d(n-n_d)+m.\vspace{.3cm}\\ \end{array}\] If $G$ is not abelian choose a subgroup $Z$ in $[N,G]\cap Z(G)$ of order $p$. By the exact sequence ($*$) and the isomorphism $Z\otimes G\cong Z\otimes G^{ab}\cong (C_p)^d$ together with induction hypotheses it follows that \[\begin{array}{lcl}|N\otimes G|&\leq&|N/Z\otimes G/Z||Z\otimes G|\vspace{.3cm}\\ &=&p^{(n-1-s)d+m+d}\vspace{.3cm}\\ &\leq&p^{(n-s)d+m}.\vspace{.3cm}\\ \end{array}\] \end{proof} Note that when $e(\displaystyle\frac{N}{[N,G]})\geq e(G^{ab})$ this bound is better than that given in \cite{E}. For example if $G$ is a finite $d$-generator extra special $p$-group, then $|N\otimes G|\leq p^{2d}$ for all cyclic normal subgroups $N$ of $G$. Suppose $N\otimes^{c}G=(((N\otimes G)\otimes G)...\otimes G)$ is the power tensor of $N$ with $c-1$ copies of $G$ and \[\gamma_1(N,G)\supseteq \gamma_2(N,G)\supseteq ...\supseteq \gamma_c(N,G) \supseteq... \] is the central series defined in \cite{E}, where $\gamma_1(N,G)=N$ and $\gamma_c(N,G)=[N,_{c-1}G]$. Then there is an epimorphism $N\otimes^{c}G\longrightarrow \gamma_c(N,G)$ with kernel $J_c(N,G)$. \begin{cor} Let $N$ be a normal subgroup of a $d$-generator finite $p$-group $G$ with $|G^{ab}|=p^m$. Suppose that $|\gamma_i(N,G)|=p^{n_i}$ and $e(\displaystyle\frac{\gamma_i(N,G)}{\gamma_{i+1}(N,G)})=p^{s_i}$. Then for any $c\geq 1$ \[|N\otimes^{c+1}G|\leq p^{t},\] in which $t=\ds \sum_{i=1}^c(n_i-s_i)d^{c-i+1}+m(1+(c-1)d)$. \end{cor} \begin{proof} The case $c=1$ obtains from Theorem \ref{m1}. By exact sequence \[J_c(N,G)\otimes G\longrightarrow (N\otimes^cG)\otimes G\longrightarrow \gamma_c(N,G)\otimes G\longrightarrow 1\] and inequality $|J_c(N,G)\otimes G|\leq |J_c(N,G)|^d\leq |N\otimes^cG|^d$ we have \[|N\otimes^{c+1} G|\leq p^{(n_c-s_c)d+m}|N\otimes^cG|^d =p^{t}. \] \end{proof} \section{The Schur multiplier of pair of groups}\ \ \ \ Let $(G,N)$ be a pair of groups. The non-abelian exterior product $N\wedge G$ is obtained from the non-abelian tensor product $N\otimes G$ by imposing the additional relations $n\otimes n=1$ for all $n\in N$. Ellis \cite{E1} showed that the Schur multiplier of the pair $(N,G)$, ${\mathcal M}(N,G)$ is isomorphic to $Ker(N\wedge G\longrightarrow G)$. In particular if $N$ is central, then \[{\mathcal M}(N,G)\cong \frac{N\otimes G^{ab}}{\nabla(N,G)}.\] Results in \cite{BL,E1} give a commutative diagram with exact rows and central extensions as columns: \begin{equation}\begin{CD} @. @. 0 @. 0\\ @. @. @VVV @VVV\\ @.\Gamma(\displaystyle\frac{N}{[N,G]}) @>\psi>>J_2(N,G)@>>>{\mathcal M}(N,G)@>>>0\\ @. @| @VVV @VVV\\ @.\Gamma(\displaystyle\frac{N}{[N,G]}) @>\psi>>N\otimes G @>>>N\wedge G@>>>1 @.\\ @. @. @V VV @V VV\\ @. @. [N,G] @= [N,G]\\ @. @. @VVV @VVV\\ @. @. 1 @. 1\\ \end{CD}\end{equation} where $\Gamma$ is the Whithead's quadratic functor (\cite{W}) and the homomorphism $\Gamma(\displaystyle\frac{N}{[N,G]})\stackrel{\psi}\longrightarrow N\otimes G$ assigns $\gamma(n[N,G])$ to $n\otimes n$ for all $n\in N$. \begin{thm}\label{m3} Let $N$ be a normal subgroup of a finite group $G$ and $|\displaystyle\frac{N}{[N,G]}|$ be odd. $i)$ If $N$ has a complement and $[N,G]=[G,G]$, then $\nabla(N,G)$ is isomorphic to $\Gamma(\displaystyle\frac{N}{[N,G]})$ and \[|N\otimes G|=|N||{\mathcal M}(N,G)||{\mathcal M}(\displaystyle\frac{N}{[N,G]})|.\] If $\displaystyle\frac{N}{[N,G]}$ has $r$ cyclic component of even order, then the right hand of formula may be multiplied by $2^i$, $0\leq i\leq r$. $ii)$ If $G^{ab}$ is elementary abelian $p$-group and $\displaystyle\frac{N}{[N,G]}\cong \prod_{i=1}^{d'}\langle \widehat{n_i}\rangle\cong(C_p)^{d'}$, in which $\widehat{n_i}$ denotes the corresponding element in $\displaystyle\frac{N}{[N,G]}$ and $n_i\in [G,G]$ for $0\leq i\leq k\leq d'$, then \[\nabla(N,G)\cong \prod_{i=k}^{d'}(n_i\otimes n_i)\times \prod_{i=1}^{d'-1}(\prod_{j=k+1}^{d'}(n_i\otimes n_j)(n_j\otimes n_i))\cong (C_p)^{\frac{1}{2}(d'-k)(d'+k+1)}\] In particular $\nabla(N,G)\cong \Gamma(\displaystyle\frac{N}{[N,G]})$ if and only if $k=0$. \end{thm} \begin{proof} $i)$ Suppose $N$ has a complement in $G$. It follows that the exact sequence \[0\rightarrow \displaystyle\frac{N}{[N,G]}\rightarrow \displaystyle\frac{N}{[N,G]}\rightarrow G/N\rightarrow1\] splits. So if $[N,G]=[G,G]$ then $\displaystyle\frac{N}{[N,G]}\otimes \displaystyle\frac{N}{[N,G]}\leq \displaystyle\frac{N}{[N,G]}\otimes \displaystyle\frac{G}{[N,G]}$. On the other hand there is a surjection homomorphism \[\nabla(N,G)\longrightarrow \nabla(\displaystyle\frac{N}{[N,G]},\displaystyle\frac{G}{[N,G]}).~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)\] Therefore the result holds by \cite{J1}. $ii)$ The image of $n_i$, $k< i\leq d'$ in $G^{ab}$, say $\bar{n_i}$, is of order $p$ and \[p=o(\widehat{n_i}\otimes \bar{n_i})\leq o(n_i\otimes n_i)\leq o(\gamma(\widehat{n_i}))=p\] The first inequality is because of epimorphism (2) %\[N\otimes G\stackrel{\pi} \longrightarrow N/[N,G]\otimes G/[N,G]\cong N/[N,G]\otimes G/[G,G]\] and the last inequality is because of $\psi$. Also \[p=o((\widehat{n_i}\otimes \bar{n_j})(\hat{n_j}\otimes \overline{n_i}))\leq o((n_i\otimes n_j)(n_j\otimes n_i))\leq o(\widehat{n_i}\otimes \hat{n_j})=p.\] The last inequality is because of the homomorphism $\displaystyle\frac{N}{[N,G]}\otimes \displaystyle\frac{N}{[N,G]}\longrightarrow N\otimes G$ given by $ \widehat{n_i} \otimes \hat{n_j}\mapsto (n_i\otimes n_j)(n_j\otimes n_i)$. Note that all elements $n_i\otimes n_i$ and $(n_i\otimes n_j)(n_j\otimes n_i)$ are distinct and if both of $i$ and $j$ are less than or equal $k$, then $(n_i\otimes n_j)(n_j\otimes n_i)=0$. \end{proof} Invoking Theorem \ref{m3} for example if $G$ is a finite $d$-generator non-abelian $p$-group of nilpotency class 2, $p$ is odd, $G^{ab}$ an elementary abelian of order $p^{n-c}$ and $Z=Z(G)$ an elementary abelian of order $p^{r}$, then $\nabla(Z,G)\cong (C_p)^{\frac{1}{2}(r-c)(r+c+1)}$ and consequently \[{\mathcal M}(Z,G)\cong (C_p)^{rd-\frac{1}{2}[r^2+r-(c^2+c)]}.\] Gilbert \cite{G} computed the non-abelian tensor square of free product of groups. We here generalize his result and also determine the schur multiplier of a pair of free product of groups. \begin{thm}\label{free} Let $N_i$ be a normal subgroup of group $G_i$, $i=1,2$ and $(G_1\ast G_2,N_1\ast N_2)$ be a pair of groups. Then \begin{itemize} \item[(i)] \[J_2(N_1\ast N_2,G_1\ast G_2)\cong J_2(N_1,G_1)\oplus J_2(N_2,G_2)\oplus (\displaystyle\frac{N_1}{[N_1,G_1]}\otimes \displaystyle\frac{N_2}{[N_2,G_2]}),\] \item[(ii)] \[\nabla (N_1\ast N_2,G_1\ast G_2)\cong \nabla (N_1\times N_2,G_1\times G_2),\] \item[(iii)] If $N_1$ has a complement in $G_1$ and $[N_1,G_1]=[G_1,G_1]$, then \[{\mathcal M}(N_1\ast N_2,G_1\ast G_2)\cong {\mathcal M}(N_1,G_1)\oplus {\mathcal M}(N_2,G_2).\] \end{itemize} \end{thm} \begin{proof} (i) There are homomorphisms \[i:N_1\otimes G_1\rightarrow (N_1\ast N_2)\otimes (G_1\ast G_2),~~~j:N_2\otimes G_2\rightarrow (N_1\ast N_2)\otimes (G_1\ast G_2)\] and the function \[(N_1\otimes G_1)\times (N_2\otimes G_2)\longrightarrow (N_1\ast N_2)\otimes (G_1\ast G_2)\] given by $(x,y)\mapsto i(x)j(y)$ which restricts to an injective homomorphism \[J_2(N_1,G_1)\oplus J_2(N_2,G_2)\longrightarrow J_2(N_1\ast N_2,G_1\ast G_2).\] On the other hand there is a homomorphism \[\displaystyle\frac{N_1}{[N_1,G_1]}\otimes \displaystyle\frac{N_2}{[N_2,G_2]}\longrightarrow J_2(N_1\ast N_2,G_1\ast G_2)~~~~~~~~~~~~~~~~~~~~~~~~~~(3)\] which maps $\hat{n_1}\otimes \hat{n_2}$ to $(n_1\otimes n_2)(n_2\otimes n_1)$ for all $n_i\in N_i$, $i=1,2$ and hence the homomorphism \[\xi:J_2(N_1,G_1)\oplus J_2(N_2,G_2)\oplus (\displaystyle\frac{N_1}{[N_1,G_1]}\otimes \displaystyle\frac{N_2}{[N_2,G_2]})\longrightarrow J_2(N_1\ast N_2,G_1\ast G_2).\] arises so that $\xi$ is injective. Because by the surjection \[\alpha: (N_1\ast N_2)\otimes (G_1\ast G_2)\longrightarrow ({N_1}\otimes {G_1})\times({N_1}\otimes {G_2})\times ({N_2}\otimes {G_1})\times ({N_2}\otimes {G_2})\] if $\xi(x,y,z)=1$ then $\alpha\xi(x,y,1)=\alpha\xi(1,1,z^{-1})$ whence $\alpha\xi(1,1,z^{-1})=1$. So $z=1$ and $\xi(x,y,1)=1$ implies that $x=y=1$. Now let $t\in (N_1\ast N_2)\otimes (G_1\ast G_2)$. Write $t=uvw$ where $u\in \langle n_1\otimes g_1| n_1\in N_1,g_1\in G_1\rangle$, $w\in \langle n_2\otimes g_2| n_2\in N_2,g_2\in G_2\rangle$ and $v\in V=\langle n_1\otimes g_2, n_2\otimes g_1| n_i\in N_i,g_i\in G_i\rangle$. (See \cite{G}) Then $\kappa(t)=acb$ where $a\in [N_1,G_1]$, $b\in [N_2,G_2]$ and $c\in [G_1,G_2]$, the Cartesian subgroup of $G_1\ast G_2$. Note that $\kappa$ is the commutator map. If $\kappa(t)=1$ then $a=b=1$. Thus $\kappa(t)=\kappa(v)=c=1$. So $u\in J_2(N_1,G_1)$ and $w\in J_2(N_2,G_2)$. But if $v$ contains no subword $y=(n_1\otimes n_2)(n_2\otimes n_1)$ or $y=(n_2\otimes n_1)(n_1\otimes n_2)$, then it's image $\kappa(v)$ is a freely reduced word in $[G_1,G_2]$ and so $\kappa(v)\neq 1$. Thus we can write $v=x_0yx_1$ with $x_i\in V$. But $\kappa(v)=\kappa(x_0x_1)=1$. So by induction on the number of $n_1\otimes g_2$, $n_2\otimes g_1$ needed to express $v$ and by epimorphism (3) we should have $\xi$ is surjective. (ii) The proof is similar to the special case $N_i=G_i$, $i=1,2$ in \cite{J}. (iii) The isomorphism \[(N_1\times N_2)\otimes (G_1\times G_2)\cong ({N_1}\otimes {G_1})\times({N_1}\otimes {G_2})\times ({N_2}\otimes {G_1})\times ({N_2}\otimes {G_2})\] implies that \[\nabla (N_1\times N_2,G_1\times G_2)\cong \nabla(N_1,G_1)\oplus \nabla(N_2,G_2)\oplus U\] where $U=\langle(n_1\otimes n_2)(n_2\otimes n_1)~|~n_1\in N_1, n_2\in N_2\rangle\leq (N_1\otimes G_2)(N_2\otimes G_1)$. If $N_1$ has a complement, the homomorphism (3) extends to an isomorphism on to $U$. Because the restriction of the composition map \[(N_1\otimes G_2)\times (N_2\otimes G_1)\stackrel{\pi_2}\longrightarrow N_2\otimes G_1 \longrightarrow \displaystyle\frac{N_2}{[N_2,G_2]}\otimes \displaystyle\frac{G_1}{[N_1,G_1]}\longrightarrow \displaystyle\frac{N_2}{[N_2,G_2]}\otimes \displaystyle\frac{N_1}{[N_1,G_1]} \] to the $U$ is a left inverse of it. So the result holds by $(ii)$ and diagram (1). \end{proof} \begin{thebibliography}{99} \bibitem{BL} R. Brown and J.-L. Loday, Van Kampen theorems for diagrams of spaces, {\small \it Topology} {\small \bf 26} (1987), 311-335 . \bibitem{BJ} R. Brown, D. L. Johnson and E. F. Robertson, Some computations of non-abelian tensor products of groups, {\small \it J. of Algebra} {\small \bf 111} (1987), 177-202. \bibitem{E} G. Ellis and A. McDermott, Tensor products of prime-power groups, {\small \it J. Pure Appl. Algebra} {\small \bf 132(2)} (1998), 119--128. \bibitem{E1} G. Ellis, The Schur multiplier of a pair of groups, {\small \it Appl. 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