\documentstyle[12pt]{amsart} % restore plain tex's definition of \cases \makeatletter \def\cases#1{\left\{\,\vcenter{\normalbaselines\m@th \ialign{$##\hfil$&\quad##\hfil\crcr#1\crcr}}\right.} \makeatother \author{Jie Wu} \address{MSRI\\ 1000 Centennial Drive\\ Berkeley, CA 94720-5070} \title{On Combinatorial Descriptions of Homotopy Groups of $\Sigma K(\pi,1)$} \thanks{Research at MSRI is supported in part by NSF grant DMS-9022140} \newtheorem{theorem}{Theorem}[section] \newtheorem{definition}[theorem]{Definition} \newtheorem{definitions}[theorem]{Definitions} \newtheorem{notation}[theorem]{Notation} \newtheorem{notations}[theorem]{Notations} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{example}[theorem]{Example} \newtheorem{examples}[theorem]{Examples} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \newtheorem{remarks}[theorem]{Remarks} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{sublemma}[theorem]{Sublemma} \newtheorem{otherlemma}[theorem]{Lemma} \begin{document} \maketitle \begin{abstract} We give a combinatorial description of homotopy groups of $\Sigma K(\pi,1)$. In particular, all of the homotopy groups of the $3$-sphere are combinatorially given. \end{abstract} \section{Introduction} In this article, we study homotopy groups and some related problems by using simplicial homotopy theory. The point of view here is that combinatorial aspects of group theory provide further information about homotopy groups.\\ \par The homotopy groups of the $3$-sphere, the suspension of $K(\pi,1)$ and wedges of $2$-spheres are shown to be the centers of certain groups with specific generators and specific relations. We list two group theoretical descriptions of $\pi_*(S^3)$ as follows. \begin{definition} {\bf A bracket arrangement of weight $n$} in a group is a set of elements defined recursively as follows: \par Let $G$ be a group and let $a_1,a_2,\cdots,a_n$ be a finite sequence of elements of $G$. Let $$ \beta^1(a_1)=a_1 $$ and if $n>1$, then let $$ \beta^n(a_1,\cdots,a_n)=[\beta^k(a_1,\cdots,a_k),\beta^l(a_{k+1},\cdots,a_n)] $$ for some $k$ and $l>0$. \end{definition} \begin{theorem} For $n\geq 1$, $\pi_{n+2}(S^{3})$ is isomorphic to the center of the group with generators $y_{0},\dots,y_{n}$ and relations $$[y_{i_{1}}^{\epsilon_{1}},y_{i_{2}}^{\epsilon_{2}},\dots,y_{i_{t}}^{\epsilon_{t}}]$$ with $\{i_{1},\dots,i_{t}\}=\{-1,0,\dots,n\}$ as sets in which the indices $i_{j}$ can be repeated, where $\epsilon_{j}=\pm 1, y_{-1}=(y_{0}\dots y_{n})^{-1}$ and the commutator bracket $[ \dots ]$ runs over all bracket arrangements of weight $t$ for each $t$. \end{theorem} \begin{notations} Let $G$ be a group and let $S$ be a subset of $G$. Let $<\!S\!>$ denote the normal subgroup generated by $S$. Let $H_{j}$ be a sequence of subgroups of $G$ for $1\leq j\leq k$. Let $[[H_{1},\dots,H_{k}]]$ denote the subgroup of $G$ generated by all of the commutators $[h_{i_1}^{(1)},\dots,h_{i_t}^{(t)}]$ with $\{i_1,\dots,i_t\}=\{1,\dots,k\}$ as sets in which the indices $i_j$can be repeated, where $h_j^{(s)}\in H_j$ and the commutator bracket $[ \dots ]$ runs over all of the bracket arrangements of weight $t$ for each $t$. \end{notations} \begin{theorem} Let $n\geq 1$. In the free group $F(y_0,\dots,y_n)$ freely generated by $y_0,\allowbreak \dots,\allowbreak y_n$ we have $$ ([[<\!y_0\!>,<\!y_1\!>,\dots,<\!y_n\!>]]\cap <\!y_{-1}\!>)/ [[<\!y_{-1}\!>,<\!y_0\!>,\dots,<\!y_n\!>]]\cong \pi_{n+2}(S^{3}) $$ where $y_{-1}=(y_0\dots y_n)^{-1}$. \end{theorem} The method of the proofs of these theorems is to study the Moore chain complex of Milnor's construction $F(S^1)$ for the $1$-sphere $S^1$. A group theoretical description of the homotopy groups $\pi_*(\Sigma K(\pi,1))$ is as follows. \begin{theorem} Let $\pi$ be any group and let $\{x^{(\alpha)}|\alpha\in J\}$ be a set of generators for $\pi$. Then, for $n\not=1$, $\pi_{n+2}(\Sigma K(\pi,1))$ is isomorphic to the center of the quotient group of the free product $${\coprod^{groups}}_{0\leq j\leq n}(\pi)_j$$ modulo the relations $$ [y^{(\alpha_1)\epsilon_1}_{i_1},y^{(\alpha_2)\epsilon_2}_{i_2},\cdots,y^{(\alpha_t)\epsilon_t}_{i_t}] $$ with $\{i_1,i_2,\cdots,i_t\}=\{-1,0,1,\cdots,n\}$ as sets, where $(\pi)_j$ is a copy of $\pi$ with generators $\{x^{(\alpha)}_j|\alpha\in J\}$, $\epsilon_j=\pm1$, $y^{(\alpha)}_{-1}={x^{(\alpha)}_0}^{-1}$, $y_j=x^{(\alpha)}_j{x^{(\alpha)}_{j+1}}^{-1}$ for $1\leq j\leq n-1$, $y^{(\alpha)}_n=x^{(\alpha)}_m$ and the commutator bracket $[\cdots]$ runs over all of the commutator bracket arrangment of weight $t$ for each $t$. \end{theorem} In particular, for $\pi={\bf Z}/m$, we have \begin{corollary} $\pi_{n+2}(\Sigma K({\bf Z}/m,1))$ is isomorphic to the center of the group with generators $x_{0},\dots,x_{n}$ and relations: (1). $x_j^m$ for $0\leq j\leq n$ and (2). $$[y_{i_{1}}^{\epsilon_{1}},y_{i_{2}}^{\epsilon_{2}},\dots,y_{i_{t}}^{\epsilon_{t}}]$$ with $\{i_{1},\dots,i_{t}\}=\{-1,0,\dots,n\}$ as sets in which the indices $i_{j}$ can be repeated, where $\epsilon_{j}=\pm 1$, $y_{-1}=x_0^{-1}$, $y_j=x_{j-1}x_j^{-1}$ for $-1\leq j\leq n-1$, $y_n=x_n$ and the commutator bracket $[ \dots ]$ runs over all bracket arrangements of weight $t$ for each $t$. \end{corollary} One of the features of Theorems 1.2 and 1.5 is that homotopy groups embed in certain 'enveloping groups'. These 'enveloping groups' have systematic and uniform structure. The centers of these groups are of course more complicated to analyze. Theorem 1.4 is very similar to the combinatorial decription of J. H. C. Whitehead's conjecture [see, Bo2, pp.317]. These decriptions give a combinatorial question how to give a computable way to understand the quotient groups $R\cap S/[R,S]$ for certain subgroups $R$ and $S$ of a finite generated free group. Few informations about this question are known [see, Bo1]. The article is organized as follows.\\ \par In Section 2, we study some general properties of simplicial groups. Central extensions in the Moore-Postnikov systems will be considered. In Section 3, we study the intersection of certain subgroups in free groups. The proofs of Theorems 1.2 and 1.4 is given in Section 4, where Theorem 4.5 is Theorem 1.2 and Theorem 4.10 is Theorem 1.4. The proof of Theorem 1.5 is given in Section 5, where Theorem 5.9 is Theorem 1.5. In Section 6, we give some applications of our descriptions. One example is to compute the homotopy groups of the Cohen's construction on the $1$-sphere. A direct corollary is to give a short proof of the Milnor's counter example for the minimal simplicial groups.\\ \par The author would like to thank F. Cohen, B. Gray, J. Harper, D. Kan, M. Mahowald, J. Moore, P. May, S. Priddy and many other mathematicians for their kindly encouragement and helpful suggestions. The author is indebted to helpful discussions with J. C. Moore and F. R. Cohen. \section{Central Extensions in simplicial group theory} In this section, we study some general properties of simplicial groups. A simplicial set $K$ is called a Kan complex if it satisfies the ${\it extension\, condition}$,i.e, for any simplicial map $f:\Lambda^k[n]\to K$ has an extension $g:\Delta[n]\to K$, where $\Delta[n]$ is the standard $n$ simplex, $\Lambda^k[n]$ is the subcomplex of $\Delta[n]$ generated by all $d_i(\sigma_n)$ for $i\not=k$ and $\sigma_n$ is the nondegenerate $n$ simplex in $\Delta[n]$,[K1, C2]. Recall that any simplicial group is a Kan complex [Mo1]. Given a simplicial group $G$, the Moore chain complex $(NG,d_0)$ is defined by $NG_n=\bigcap_{j\not=0}Kerd_j$ with $d_0:NG_n\to NG_{n-1}$. The classical Moore Theorem is that $\pi_n(G)\cong H_n(NG)$ [Mo1, Theorem 4;C2, Theorem 3.7 or K2, Proposition 5.4]. Now let ${\cal Z}_n={\cal Z}_n(G)=\bigcap_jKerd_j$ denote the cycles and let ${\cal B}_n={\cal B}_n(G)=d_0(NG_{n+1})$ denote the boundaries. It is easy to check that ${\cal B}_n$ is a normal subgroup of $G_n$ for any simplicial group $G$. \begin{lemma} Let $G$ be a simplicial group. Then the homotopy group $\pi_n(G)$ is contained in the center of $G_n/{\cal B}G_n$ for $n\geq1$. \end{lemma} \noindent{\em Proof.} Notice that $\pi_n(G)\cong{\cal Z}G_n/{\cal B}G_n$. It suffices to show that the commutator $[x,y]\in{\cal B}G_n$ for any $x\in{\cal Z}G_n$ and $y\in G_n$. \par Notice that $x$ is a cycle. There is a simplicial map $f_x: S^n\rightarrow G$ such that $f_x(\sigma_n)=x$, where $S^n$ is the standard $n$-sphere with a nondegenerate $n$-simplex $\sigma_n$. Now let the simplicial map $f_y:\Delta[n]\rightarrow G$ be the representative of $y$, i.e, $f_y(\tau_n)=y$ for the nondegenerate $n$-simplex $\tau_n$. Let $\phi$ be the composite $$ S^n\stackrel{j}{\rightarrow} S^n\wedge\Delta[n]\stackrel{[f_x,f_y]}{\rightarrow}G, $$ where $j(\sigma_n)=\sigma_n\wedge\tau_n$ and $[f_x,f_y](a\wedge b)=[f_x(a),f_y(b)]$, the commutator of $f_x(a)$ and $f_y(b)$. Notice that $\phi(\sigma_n)=[x,y]$ and $S^n\wedge\Delta[n]$ is contactible. Thus $[x,y]\in{\cal B}_n$. The assertion follows. \begin{definitions} Let $G$ be a simplicial group. The subsimplicial group {\bf $R_nG$} is defined by setting $$ (R_nG)_q=\{x\in G_q|f_x(\Delta[q])^{[n]}=1\}, $$ where $f_x$ is the representative of $x$ and $X^{[n]}$ is the $n$-skeleton of the simplicial set $X$. The subsimplicial group {\bf $\overline{R}_nG$} is defined by setting $$ (\overline{R}_nG)_q=\{x\in G_q|f_x((\Delta[q])_n)\subseteq{\cal B}G_n\}. $$ Let {\bf $P_nG$} denote $G/R_nG$ and let {\bf $\overline{P}_nG$} denote $G/\overline{R}_nG$. \end{definitions} It is easy to check that both $R_nG$ and $\overline{R}_nG$ are normal subsimplicial group of $G$. Thus both $P_nG$ and $\overline{P}_nG$ are quotient simplicial group of $G$. Notice that $$ R_nG\subseteq\overline{R}_nG\subseteq R_{n-1}G. $$ The quotient simplicial group $\overline{P}_nG$ is between $P_nG$ and $P_{n-1}G$. By checking the definition of Moore-Post\-nikov systems of a simplicial set [Mo1;C2], we have \begin{lemma} The quotient simplicial group $P_nG$ is the standard $n$-th Moore-Postnikov system of the simplicial group $G$. \end{lemma} The quotient simplicial group $\overline{P}_nG$ has the same homotopy type of $P_nG$. \begin{proposition} The quotient simplicial homomorphism $q_n:P_nG\rightarrow\overline{P}_nG$ is a homotopy equivalence for each $n$. \end{proposition} \noindent{\em Proof:} The Moore chain complex of $P_nG$ is as follows: $$ N(P_nG)_q=\cases{1&for\,$q>n+1$,\cr NG_{n+1}/{\cal Z}G_{n+1}&for\,$q=n+1$,\cr NG_q&for\,$qn$,\cr NG_n/{\cal B}G_n &for\, $q=n$,\cr NG_q &for $q0$. \begin{proposition} The simplicial group $\overline{F}_n$ is the minimal simplicial group $K(\pi_nG,n)$ for each $n>0$. \end{proposition} \noindent{\em Proof:} It is directly to check that $$ N(\overline{F}_nG)_q=\cases{1&for\,$q\not=n$,\cr \pi_nG&for\, $q=n$.\cr} $$ The assertion follows. \begin{proposition} The short exact sequence of simplicial groups $$ 0\rightarrow\overline{F}_nG\rightarrow\overline{P}_nG\rightarrow P_{n-1}G\rightarrow0 $$ is a central extension for each $n>0$. \end{proposition} \noindent{\em Proof:} Consider the relative commutator sussimplicial group $[\overline{F}_nG,\overline{P}_nG]$. By Lemma 2.1, $[\overline{F}_nG,\overline{P}_nG]_n=1$. Notice that $[\overline{F}_nG,\overline{P}_nG]$ is a subsimplicial group of minimal simplicial group $\overline{F}_nG\cong K(\pi_nG,n)$. Thus $[\overline{F}_nG,\overline{P}_nG]=1$ and the assertion follows. \begin{definition} A simplicial group is said to be {\it $r$-centerless} if the center $Z(G_n)=\{1\}$ for $n\geq r$. \end{definition} \begin{proposition} Let $G$ be a reduced $r$-centerless simplicial group. Then $\pi_n(G)\cong Z(G_n/{\cal B}_n)$ for $n\geq r+1$ \end{proposition} \noindent{\em Proof.} By Lemma 2.1, ${\cal Z}_n/{\cal B}_n\subseteq Z(G_n/{\cal B}_n)$. It suffices to show that $Z(G_n/{\cal B}_n)\subseteq{\cal Z}_n/{\cal B}_n$ for $n\geq r+1$. Now let ${\tilde{x}}\in Z(G_n/{\cal B}_n)$. Choose $x\in G_n$ with $p(x)={\tilde x}$, where $p:G_n\to G_n/{\cal B}_n$ is the quotient homomorphism. To check ${\tilde x}\in{\cal Z}_n/{\cal B}_n$, it suffices to show that $x\in {\cal Z}_n$ or $d_jx=1$ for all $j$. Since $Z(G_{n-1})=\{1\}$, $d_jx=1$ if and only if $[d_jx,y]=1$ for all $y\in G_{n-1}$. Now $[d_jx,y]=d_j[x,s_{j-1}y]$ for $j>0$ and $[d_0x,y]=d_0[x,s_0y]$. Since ${\tilde x}\in Z(G_n/{\cal B}_n)$, $[x,z]\in{\cal B}_n\subseteq{\cal Z}_n$ for all $z\in G_n$ and therefore $[d_jx,y]=1$ for all $y\in G_{n-1}$. The assertion follows.\\ \par By inspecting the proof, we also have \begin{proposition} Let $G$ be a reduced $r$-centerless simplicial group. Then $$Z(G_n/{\cal Z}_n)=\{1\}$$ for $n\geq r+1$. \end{proposition} \begin{lemma} Let $G$ be a reduced simplicial group so that $G_n$ is cyclic or centerless for each n. Then there exists a unique integer $\gamma_G>0$ so that $G_n=\{1\}$ for $n<\gamma_G$ and $Z(G_n)=\{1\}$ for $n>\gamma_G$. \end{lemma} \noindent{\em Proof.} Let $\gamma_G=max\{\gamma|G_n=\{1\}\,\,for\,\,n<\gamma\}$. Then $\gamma_G>0$. If $\gamma_G<\infty$. Then $G_{\gamma_G}\neq\{1\}$. We show that $G_{\gamma_G+q}$ is centerless for each $q>0$. Notice that $d_0^q\circ s_0^q:G_n\rightarrow G_{n+q}\rightarrow G_n$ and $d_1^q\circ s_1^q:G_n\rightarrow G_{n+q}\rightarrow G_n$ are identities for $n>0$ Thus $s_0^q(G_{\gamma_G})$ and $s_1^q(G_{\gamma_G})$ are nontrivial summands of $G_{\gamma_G+q}$. Now let $x=s_0^qy=s_1^qz\in s_0^q(G_{\gamma_G})\cap s_1^q(G_{\gamma_G})$. Then $d_{q+1}x=d_{q+1}s_0^qy=s_0^qd_1y=1=d_{q+1}s_1^qz=s_1^{q-1}d_2s_1z=s_1^{q-1}z$. Thus $x=s_1^qz=1$. And therefore $s_0^q(G_{\gamma_G})\cap s_1^q(G_{\gamma_G})=\{1\}$. The assertion follows. \begin{corollary} Let $G$ be a reduced simplicial group such that $G_n$ is cyclic or centerless for each $n$. Then $\pi_n(G)\cong Z(G_n/{\cal B}_n)$ for $n\neq\gamma_G+1$, where $\gamma_G$ is defined as above. \end{corollary} Notice that, for any free group $F$, $rank(F)\geq2\Leftrightarrow Z(F)=\{1\}$ and $F\neq\{1\}$. We have \begin{lemma} Let $G$ be a reduced simplicial group such that $G_n$ is a free group for each $n$. Then there exits a unique integer $\gamma_G>0$ so that $G_n=\{1\}$ for $n<\gamma_G$ and $rank(G_n)\geq2$ for $n>\gamma_G$. \end{lemma} \begin{example} The $1$-stem is determined in this example. \end{example} Let $G=F(S^n)$, Milnor's $F$-construction on the standard n-sphere for $n\geq1$. Then $G_n\cong F(\sigma)\cong {\bf Z}(\sigma)$, the free abelian group generated by $\sigma$, $$G_{n+1}\cong F(s_0\sigma,s_1\sigma,\dots,s_n\sigma)$$ and $G_{n+2}\cong F(s_is_j\sigma |0\leq j1$. Consider the following equations $$ d_k([s_{j-1}s_i\sigma,s_{j+1}s_j\sigma])=\cases{1&for\,$k\neq j,$\cr[s_i\sigma,s_j\sigma]&for\,k=j,\cr} $$ for $i+11$, where $y^{\epsilon_1}_1\cdots y_t^{\epsilon_t}$ is a reduced word. Let $y\in T$ and let $\epsilon=\pm1$. If $y_1^{\epsilon_1}\cdots y_t^{\epsilon_t}y^{\epsilon}$ is a reduced word, then $[w,y^{\epsilon}]\in F({\cal A}_T(k))$ by definition. Suppose that $y_1^{\epsilon_1}\cdots y_t^{\epsilon_t}y^{\epsilon}$ is not a reduced word. Then $t>0$, $y=y_t$ and $\epsilon=-\epsilon_t$. Let $w'$ denote $[[x,y_1^{\epsilon_1}],\cdots,],y_{t-1}^{\epsilon_{t-1}}]\in{\cal A}_T(k-2)$. Then $w=[w',y_t^{\epsilon_t}]$. By the Witt-Hall identities, there is an equation $$ 1=[w',y_t^{-\epsilon_t}]\cdot w\cdot [w,y_t^{-\epsilon_t}]. $$ By induction, $[w',y_t^{-\epsilon_t}]\in F({\cal A}_T(k-1))\subseteq F({\cal A}_T(k))$ and so $$[w,y_t^{-\epsilon_t}]=w^{-1}[w',y_t^{-\epsilon_t}]^{-1}\in F({\cal A}_T(k-1))\subseteq F({\cal A}_T(k)).$$ The second induction is finished and so the first induction is finished. The assertion follows.\\ \par \noindent{\em Step 3.} ${\tilde i_k}: F({\cal A}_T(k))\rightarrow F({\cal B}_T(k))$ is an isomorphism.\\ \par By Step 2, ${\bf Z}({\cal A}_T(k))\rightarrow{\bf Z}({\cal B}_T(k))$ is an epimorphism. Notice that $$rank(F({\cal B}_T(k)))=|{\cal B}_T(k)|=|{\cal A}_T(k)|=rank(F({\cal A}_T(k))).$$ Thus ${\tilde i_k}:{\bf Z}({\cal A}_T(k))\rightarrow{\bf Z}({\cal B}_T(k))$ is an isomorphism and so ${\tilde i_k}: F({\cal A}_T(k))\rightarrow F({\cal B}_T(k))$ is a monomorphism. Thus ${\tilde i_k}$ is an isomorphism.\\ \par \noindent{\em Step 4.} Since $F({\cal A}_T(k))\rightarrow F({\cal B}_T(k))$ is an isomorphism for each $k$, $F({\cal A}_T)=colim_kF({\cal A}_T(k))\rightarrow F({\cal B}_T)=colim_kF({\cal B}_T(k))$ is an isomorphism.\\ \par Now consider the general case. By Lemma 3.6, it suffices to show that ${\tilde i}: F({\cal A}_T)\rightarrow F({\cal B}_T)$ is an isomorphism. To check that $F({\cal A}_T)\rightarrow F({\cal B}_T)$ is an epimorphism. Let $w\in{\cal B}_T$, there exist finite subsets $S'$ and $T'$ of $S$ and $T$, respectively, so that $w\in{\cal B}_{T'}$. By the special case as above, ${\tilde i}|_{F({\cal A}_{T'})}: F({\cal A}_{T'})\rightarrow F({\cal B}_{T'})$ is an isomorphism and $w\in Im{\tilde i}|_{F({\cal A}_{T'})}$. Thus ${\tilde i}$ is an epimorphism. To check that $F({\cal A}_T)\rightarrow F({\cal B}_T)$ is a monomorphism. Let $w\in Ker{\tilde i}$, there exist finite subsets $S'$ and $T'$ of $S$ and $T$, respectively, so that $w\in F({\cal A}_{T'})$. Notice that ${\tilde i}|_{F({\cal A}_{T'})}$ is an isomorphism. Thus $w=1$ and the assertion follows.\\ \par Now let's consider the intersection of kernels of projection homomorphisms. Let $S$ be a set and let $T_j$ be a subset of $S$ for $1\leq j\leq k$. Let $\pi_j:F(S)\rightarrow F(T_j)$ be the projection homomorphism for $1\leq j\leq k$. We construct a subset ${\cal A}(T_1,\cdots,T_k)$ of $F(S)$ by induction on $k$ as follows. $$ {\cal A}(T_1)={\cal A}_{T_1}, $$ where ${\cal A}_T$ is defined in Definition3.1. Let $$ T_2^{(2)}=\{w\in{\cal A}(T_1)|w=[[x,y_1^{\epsilon_1},\cdots,y_t^{\epsilon_t}]\, with\,x,y_j\in T_2\, for\,all\,j\} $$ and define $$ {\cal A}(T_1,T_2)={\cal A}(T_1)_{T_2^{(2)}}. $$ Suppose that ${\cal A}(T_1,T_2,\cdots,T_{k-1})$ is well defined so that all of the elements in ${\cal A}(T_1,T_2,\cdots,T_{k-1})$ are written down as certain commutators in $F(S)$ in terms of elements in $S$. Let $T_k^{(k)}$ be the subset of ${\cal A}(T_1,T_2,\cdots,T_{k-1})$ defined by $$ T^{(k)}_k=\{w\in{\cal A}(T_1,T_2,\cdots,T_{k-1})|w=[x_1^{\epsilon_1},\cdots,x_l^{\epsilon_l}]\,with\,x_j\in T_k\,for\,all\,j\}, $$ where $[x_1^{\epsilon_1},\cdots,x_l^{\epsilon_l}]$ are the elements in ${\cal A}(T_1,T_2,\cdots,T_{k-1})$ which are written down as commutators. Then define $$ {\cal A}(T_1,T_2,\cdots,T_k)={\cal A}(T_1,T_2,\cdots,T_{k-1})_{T_k^{(k)}} $$ \begin{theorem} Let $S$ be a set and let $T_j$ be a subset of $S$ for $1\leq j\leq k$. Let $\pi_j:F(S)\rightarrow F(T_j)$ be the projection homomorphism for $1\leq j\leq k$. Then the intersection $\bigcap_{j=1}^kKer\pi_j$ is a free group freely generated by ${\cal A}(T_1,T_2,\cdots,T_k)$. \end{theorem} \noindent{\em Proof.} The proof is given by induction on $k$. If $k=1$, the assertion follows from the above lemma. Suppose that $\bigcap_{j=1}^{k-1}Ker\pi_j=F({\cal A}(T_1,T_2,\cdots,T_{k-1}))$ and consider $\pi_k:F(S)\rightarrow F(T_k)$. Then $$\bigcap_{j=1}^kKer\pi_j=Ker({\bar\pi_k}:F({\cal A}(T_1,T_2,\cdots,T_{k-1}))\rightarrow F(T_k))$$, where ${\bar\pi_k}$ is $\pi_k$ restricted to the subgroup $F({\cal A}(T_1,T_2,\cdots,T_{k-1}))$. Let $$w=[x_1^{\epsilon_1},\cdots,x_l^{\epsilon_l}]\in{\cal A}(T_1,T_2,\cdots,T_{k-1}).$$ If $w\notin T_k^{(k)}$, then $x_j\notin T_k$ for some $j$ and ${\bar\pi_k}(w)=1$. Thus ${\bar\pi_k}$ factors through $F(T_k^{(k)})$, i.e, there is a homomorphism $j:F(T_k^{(k)})\rightarrow F(T_k)$ so that ${\bar\pi_k}=j\circ\pi$, where $\pi:F({\cal A}(T_1,T_2,\cdots,T_{k-1}))\rightarrow F(T_k^{(k)})$ is the projection homomorphism. We claim that $j:F(T_k^{(k)})\rightarrow F(T_k)$ is a monomorphism. Consider the commutative diagram $$ \hspace{0.0in} \begin{array}{cccccccccc} &F({\cal A}(T_1,T_2,\cdots,T_{k-1}))&\stackrel{{\bar\pi_k}}{\rightarrow}& F(T_k)& \hookrightarrow& F(S) \\ & \uparrow & & \uparrow j & & \uparrow \\ & F(T_k^{(k)}) &\stackrel{=}{\rightarrow} & F(T_k^{(k)}) & \hookrightarrow & F({\cal A}(T_1,T_2,\cdots,T_{k-1})), \end{array} $$ where $F(T_k^{(k)})\rightarrow F({\cal A}(T_1,T_2,\cdots,T_{k-1}))$ and $F({\cal A}(T_1,T_2,\cdots,T_{k-1}))\rightarrow F(S)$ are inclusions of subgroups. Thus $j:F(T_k^{(k)})\rightarrow F(T_k)$ is a monomorphism and $$ Ker{\bar\pi_k}=Ker(F({\cal A}(T_1,T_2,\cdots,T_{k-1}))\rightarrow F(T_k^{(k)})=F({\cal A}(T_1,T_2,\cdots,T_k)). $$ The assertion follows. \begin{corollary} Let $\pi_j$ be the projection homomorphisms as in Theorem 3.5. Then the intersection subgroup $\bigcap_{j=1}^kKer\pi_j$ equals the commutator subgroup $[[,\cdots,]]$ which is defined in Notations 1.3. \end{corollary} \section{On the Homotopy Groups of the $3$-sphere} In this section, we study the Moore chain complex of the Milnor's construction $F(S^1)$. The proofs Theorems 1.2 and 1.4 are given in this section, where Theorem 4.5 is Theorem 1.2 and Theorem 4.10 is Theorem 1.4. Recall that the simplicial $1$-sphere $S^1$ is a free simplicial set generated by a $1$-simplex $\sigma$. Thus $S^1_0=\{*\}$, $S^1_1=\{\sigma,*\}$ and $S^1_{n+1}=\{*,s_n\cdots s_{i+1}s_{i-1}\cdots s_0\sigma|0\leq j\leq n\}$. Let $x_i$ denote $s_n\cdots s_{i+1}s_{i-1}\cdots s_0\sigma$. Then $F(S^1)_{n+1}=F(x_0,x_1,\cdots, x_n)$ the free group freely generated by $x_0,\cdots,x_n$. Let $y_i$ denote $x_{i-1}x_i^{-1}$ for $-1\leq i\leq n$, where we put $x_{-1}=x_{n+1}=1$ in $F(S^1)_{n+1}=F(x_0,\cdots, x_n)$. By direct calculation, we have \begin{lemma} $F(S^1)_{n+1}=F(y_0,\cdots,y_n)$ with $$ d_jy_k=\cases{y_{k-1}&$j\leq k$,\cr 1&$j=k+1$,\cr y_k&$j>k+1$,\cr} $$ and $$ s_jy_k=\cases{y_{k+1}&$j\leq k$,\cr y_ky_{k+1}&$j=k+1$,and\cr y_k&$j>k+1$\cr} $$ for $0\leq j\leq n+1$, where $y_{-1}=(y_0\cdots y_{n-1})^{-1}$ in $F(S^1)_n$. \end{lemma} Now let $C_{n+1}$ denote the subgroup of $F(y_0,\cdots,y_n)$ generated by all of the commutators $[y_{i_1}^{\epsilon_1},\cdots,y_{i_t}^{\epsilon_t}]$ with $\{i_1,\cdots,i_t\}=\{0,1,\cdots,n\}$ as sets, i.e. each $j$ $(0\leq j\leq n)$ appears in the index set $\{i_1,\cdots, i_t\}$ at least one time, where $\epsilon_j=\pm1$ and the commutator $[y_{i_1}^{\epsilon_1},\cdots,y_{i_t}^{\epsilon_t}]$ runs over all of the commutator bracket arrangements of weight $t$ for $y_{i_1}^{\epsilon_1},\cdots,y_{i_t}^{\epsilon_t}$. \begin{lemma} The group $C_{n+1}$ is a subgroup of $NF(S^1)_{n+1}$,i.e $C_{n+1}\subseteq \cap_{j\neq 0}Ker(d_j)$. \end{lemma} \noindent{\em Proof.} Notice that $d_jy_{j-1}=1$ for $1\leq j\leq n+1$. Since $\{y_{i_1},\cdots, y_{i_t}\}=\{y_0,y_1,\cdots,y_n\}$, $d_j[y_{i_1}^{\epsilon_1},\cdots,y_{i_t}^{\epsilon_t}]=1$ for each $j>0$. The assertion follows.\\ \par \begin{theorem} $NF(S^1)_{n+1}=C_{n+1}$. \end{theorem} \noindent{\em Proof.} For $1\leq j\leq n+1$, let $S=\{y_0,y_1,\cdots,y_n\}$ and let $T_j=\{y_0,\cdots,{\hat y_j}\cdots,y_n\}$. By Lemma 4.1, there is a commutative diagram $$ \hspace{0.0in} \begin{array}{cccccc} & F(S) & \stackrel{\pi_j}{\rightarrow} & F(T_{j-1}) \\ & \downarrow d_j & & {\bar d_j}\downarrow \cong \\ & F(y_0,\cdots,y_{n-1}) &\stackrel{=}{\rightarrow} & F(y_0,\cdots,y_{n-1}),\\ \end{array} $$ where $$ {\bar d_j}(y_k)=\cases{y_k&$0\leq k\leq j-2,$\cr y_{k-1}&$j\leq k\leq n.$\cr} $$ Thus $Kerd_j=Ker\pi_j$ and $NF(S^1)_{n+1}=\bigcap_{j=1}^{n+1}Kerd_j=\bigcap_{j=1}^{n+1}Ker\pi_j$. By Theorem 3.5, $NF(S^1)_{n+1}=F({\cal A}(T_0,T_1,\cdots,T_n))$, where the notation ${\cal A}(T_0,T_1,\cdots,T_n)$ is given in Section 3. To check that $$ F({\cal A}(T_0,T_1,\cdots,T_n))\subseteq C_{n+1}, $$ it suffices to show that ${\cal A}(T_0,T_1,\cdots,T_n)\subseteq C_{n+1}$. This will follow from the following statement.\\ \par \noindent{\em Statement.} For each $0\leq j\leq n$ and $w=[y_{i_1}^{\epsilon_1},y_{i_2}^{\epsilon_2},\cdots,y_{i_t}^{\epsilon_t}]\in{\cal A}(T_0,T_1,\cdots,T_j)$, $\{y_0,y_1,\cdots,y_j\}\subseteq\{y_{i_1},y_{i_2},\cdots,y_{i_t}\}$.\\ \par We show this statement by induction on $j$. Note that $F(T_0)=F(y_1,\cdots,y_n)$. For $j=0$, $$ {\cal A}(T_0)=\{[[y_0,y_{i_1}^{\epsilon_1}],\cdots,],y_{i_t}^{\epsilon_t}]|i_s>0, y_{i_1}^{\epsilon_1}\cdots y_{i_t}^{\epsilon_t}\,a\,reduced\,word\, in\,F(T_0)\}. $$ Thus the assertion holds for $j=0$. Suppose that the assertion holds for $j-1$ with $j\leq n$. Notice that $T_j=\{y_0,\cdots,{\hat y_j},\cdots, y_n\}$. Thus $$ T_j^{(j)}=\{w\in{\cal A}(T_0,T_1,\cdots,T_{j-1})|w=[y_{i_1}^{\epsilon_1},\cdots,y_{i_t}^{\epsilon_t}]\, with\,y_j\notin\{y_{i_1},\cdots,y_{i_t}\}\}. $$ and so $y_j\in\{y_{i_1},\cdots,y_{i_t}\}$ for $w=[y_{i_1}^{\epsilon_1},\cdots,y_{i_t}^{\epsilon_t}]\in{\cal A}(T_0,T_1,\cdots,T_{j-1})-T_j^{(j)}$. Hence, by induction, $$ \{y_0,y_1,\cdots,y_j\}\subseteq\{y_{i_1},y_{i_2},\cdots,y_{i_t}\} $$ for $w=[y_{i_1}^{\epsilon_1},\cdots,y_{i_t}^{\epsilon_t}]\in{\cal A}(T_0,T_1,\cdots,T_{j-1})-T_j^{(j)}$. Notice that $${\cal A}(T_0,T_1,\cdots,T_j)={\cal A}(T_0,T_1,\cdots,T_{j-1})_{T_j^{(j)}}.$$ Thus $$ \{y_0,y_1,\cdots,y_j\}\subseteq\{y_{i_1},y_{i_2},\cdots,y_{i_t}\} $$ for each $w=[y_{i_1}^{\epsilon_1},\cdots,y_{i_t}^{\epsilon_t}]\in{\cal A}(T_0,T_1,\cdots,T_{j})$. The induction is finished and the theorem follows. \begin{corollary} ${\cal A}(T_0,T_1,\cdots,T_n)$ is a set of free generators for $NF(S^1)_{n+1}$. \end{corollary} \begin{theorem}[Theorem 1.2] For $n\geq1$, $\pi_{n+2}(S^3)$ is isomorphic to the center of the group with generators $y_0,y_1,\cdots,y_n$ and relations $[y_{i_1}^{\epsilon_1},y_{i_2}^{\epsilon_2},\cdots,y_{i_t}^{\epsilon_t}]$ with $$ \{i_1,i_2,\cdots,i_t\}=\{-1,0,1,\cdots,n\} $$ as sets, where $\epsilon_j=\pm1$,$y_{-1}=(y_0y_1\cdots y_n)^{-1}$ and the commutator bracket $[\cdots]$ runs over all bracket arrangements of weight $t$ for each $t$. \end{theorem} \noindent{\em Proof.} Notice that $\pi_{n+2}(S^3)\cong\pi_{n+1}F(S^1)$ for $n\geq1$ and ${\cal B}_{n+1}=d_0(NF(S^1)_{n+2})$. By Lemma 4.1 and Theorem 4.3, ${\cal B}_{n+1}$ is generated by $[y_{i_1}^{\epsilon_1},y_{i_2}^{\epsilon_2},\cdots,y_{i_t}^{\epsilon_t}]$ so that $\{y_{i_1},y_{i_2},\cdots,y_{i_t}\}=\{y_0,y_1,\cdots,y_n\}$, where $\epsilon_j=\pm1$. By Proposition 2.8, it suffices to check that $\pi_2(F(S^1))\cong Z(F(S^1)_2/{\cal B}_2)=Z(F(y_0,y_1)/{\cal B}_2)$. By Example 2.13, ${\cal Z}_2=\Gamma^2F(S^1)_2=\Gamma^2F(y_0,y_1)$ and $\Gamma^3F(y_0,y_1)\subseteq{\cal B}_2$. Now , by the construction of ${\cal B}_2$, ${\cal B}_2\subseteq\Gamma^3F(y_0,y_1)$ and therefore $$\pi_2(F(S^1))\cong{\cal Z}_2/{\cal B}_2=\Gamma^2F(y_0,y_1)/\Gamma^3F(y_0,y_1)=Z(F(y_0,y_1)/{\cal B}_2).$$ The assertion follows. \begin{remark} The relations in above theorem are not minimal, i.e, many of them can be cancelled out. \end{remark} \begin{proposition} Let ${C'}_{n+1}$ be the subgroup of $F(S^1)_{n+1}$ generated by all commutators given by $[\cdots[y_{i_1}^{\epsilon_1},y_{i_2}^{\epsilon_2}],\cdots,],y_{i_t}^{\epsilon_t}]$ with $\{i_1,\cdots,i_t\}=\{0,1,\cdots,n\}$ and $\epsilon_j=\pm1$. Then $$ NF(S^1)_{n+1}/\Gamma^s\cap NF(S^1)_{n+1}\cong C'_{n+1}/\Gamma^s\cap C'_{n+1} $$ for each $s$, where $\Gamma^s=\Gamma^sF(S^1)_{n+1}$ is the $s$-term in the lower central series of $F(S^1)_{n+1}$. \end{proposition} \noindent{\em Proof.} Notice that $C'_{n+1}\subseteq NF(S^1)_{n+1}$. The induced homomorphism $f_s:C'_{n+1}/\Gamma^s\cap C'_{n+1}\rightarrow NF(S^1)_{n+1}/\Gamma^s\cap NF(S^1)_{n+1}$ is a monomorphism. We check that $f$ is an epimorphism. It suffices to show that, for each $w\in NF(S^1)$, there exists a sequence of elements $\{x_j\}$ so that $x_j\in\Gamma^j\cap C'_{n+1}$ and $wx_1x_2\cdots x_s\in\Gamma^{s+1}$ for each $s$. In fact, if this statement holds, $wx_1x_2\cdots x_{s-1}\equiv 1$ $mod$ $\Gamma^s\cap NF(S^1)_{n+1}$ for each $s$ and $w=(wx_1\cdots x_{s-1})\cdot(x_1x_2\cdots x_{j-1})^{-1}\in C'_{n+1}$ $mod\,\Gamma^s$. Now we construct $x_j$ by induction, which depends on $w$. Notice that, for $n\geq1$, $NF(S^1)_{n+1}\in\Gamma^{n+1}\subseteq\Gamma^2$. Choose $x_j=1$ for $j\leq n$. Suppose that there are $x_1,\cdots,x_{s-1}$ so that $x_j\in\Gamma^j\cap C'_{n+1}$ and $wx_1\cdots x_{s-1}\in\Gamma^s$. Since $C'_{n+1}\subseteq NF(S^1)_{n+1}$, $wx_1\cdots x_{s-1}\in\Gamma^s\cap NF(S^1)_{n+1}$ and $d_j(wx_1\cdots x_{s-1})=1$ for $j>1$. Let $\pi:\Gamma^s\rightarrow\Gamma^s/\Gamma^{s+1}$ be the quotient homomorphism. $\pi(wx_1\cdots x_{s-1})$ is a linear combination of basic Lie products. We claim that $\pi(wx_1\cdots x_{s-1})$ is a linear combination of basic Lie products in which each $y_j$ appears in the Lie product for $0\leq j\leq n$. If not, there exists $j$ so that $\pi(wx_1\cdots x_{s-1})=b+c$, where $b$ is a nontrivial linear combination of basic Lie products in which $y_j$ does not appear and $c$ is a linear combination of basic Lie products in which $y_j$ appears. Now the face homomorphism $d_{j+1}:F(y_0,\cdots,y_n)\rightarrow F(y_0,\cdots,y_{n-1})$ induces a homomorphism $$ {\bar d}_{j+1}:\Gamma^sF(y_0,\cdots,y_n)/\Gamma^{s+1}F(y_0,\cdots,y_n)\rightarrow\Gamma^sF(y_0,\cdots,y_{n-1})/\Gamma^{s+1}F(y_0,\cdots,y_{n-1}) $$ and $$1={\bar d}_{j+1}\pi(wx_1\cdots x_{s-1})={\bar d}_{j+1}(b)+{\bar d}_{j+1}(c)={\bar d}_{j+1}(b).$$ Notice that $$d_{j+1}|_{F(y_0,\cdots,y_{j-1},y_{j+1},\cdots,y_n)}:F(y_0,\cdots,y_{j-1},y_{j+1},\cdots,y_n)\rightarrow F(y_0,\cdots,y_{n-1})$$ is an isomorphism. Thus $b=1$. This contradicts to that $b$ is a nontrivial linear combination of a basis. Thus $\pi(wx_1\cdots x_{s-1})$ is a linear combination of basic Lie products in which all of the $y_j$ appear. By Theorem 5.12 in [MKS,pp.337], there exists $x_s$ in $C'_{n+1}$ so that $\pi(wx_1\cdots x_s)=1$, or $wx_1\cdots x_s\in\Gamma^{s+1}$. The induction is finished now. \\ \par By Corollary 3.6, we have \begin{theorem} In the free group $F(y_0,\cdots,y_n)$, $NF(S^1)_{n+1}=[[,\cdots,]]$ and ${\cal B}F(S^1)_{n+1}=[[,,\cdots,]]$. \end{theorem} Thus Theorem 4.5 can be rewritten as follows. \begin{theorem} In the free group $F(y_0,\cdots,y_n)$ for $n\geq1$, the center $$ Z(F(y_0,\cdots,y_n)/[[,,\cdots,]])\cong\pi_{n+2}(S^3) $$ \end{theorem} \par By Lemma 4.1 and Proposition 3.5, $Kerd_0==$ and therefore $$ {\cal Z}_{n+1}=[[,\cdots,]]\cap. $$ Thus we have \begin{theorem}[Theorem 1.4] In the free group $F(y_0,\cdots,y_n)$ with $n\geq1$, $$ [[,\cdots,]]\cap]]/[[,,\cdots,]]\cong\pi_{n+2}(S^3). $$ \end{theorem} \section{On the Homotopy Groups of $\Sigma K(\pi,1)$} In this section, we give group theoretical descriptions for $\pi_*(\Sigma K(\pi,1))$ for any group $\pi$. The proof of Theorem 1.5 is given in this section, where Theorem 5.9 is Theorem 1.5. We will use the notations defined in Section 3. First we extend our description for $\pi_*(S^2)$ to the case $\pi_*(\vee_{\alpha\in J}S^2)$. Recall that $(\vee_{\alpha\in J}S^1)_0=*$, $(\vee_{\alpha\in J}S^1)_1=\{\sigma_{\alpha},*|\alpha\in J\}$ and $(\vee_{\alpha\in J}S^1)_{n+1}=\{s_n\cdots{\hat s_i}\cdots s_0\sigma_{\alpha},*|\alpha\in J,0\leq i\leq n\}$. Let $x^{(\alpha)}_i$ denote $s_n\cdots{\hat s_i}\cdots s_0\sigma_{\alpha}$. Then $$ F(\vee_{\alpha\in J}S^1)_{n+1}=F(x^{(\alpha)}_0,x^{(\alpha)}_1,\cdots,x^{(\alpha)}_n|\alpha\in J). $$ Let $y^{(\alpha)}_j=x^{(\alpha)}_j\cdot x^{(\alpha)}_{j+1}$ for $0\leq j\leq n-1$ and $y^{(\alpha)}_n=x^{(\alpha)}_n$. By Lemma 4.1, we have \begin{lemma} $F(\vee_{\alpha\in J}S^1)_{n+1}=F(y^{(\alpha)}_j|0\leq j\leq n,\alpha\in J)$ with $$ d_j(y^{(\alpha)}_k)=\cases{y^{(\alpha)}_{k-1}& $j\leq k$, \cr 1& $j=k+1$, \cr y^{(\alpha)}_k & $j>k+1$, \cr} $$ and $$ s_j(y^{(\alpha)}_k)=\cases{y^{(\alpha)}_{k+1}& $j\leq k$, \cr y^{(\alpha)}_k\cdot y^{(\alpha)}_{k+1}& $j=k+1$, \cr y^{(\alpha)}_k & $j>k+1$, \cr} $$ for $0\leq j\leq n+1$, where $y^{(\alpha)}_{-1}=(y^{(\alpha)}_0\cdots y^{(\alpha)}_{n-1})^{-1}$ in $F(\vee_{\alpha\in J}S^1)$ \end{lemma} Let $C^J_{n+1}$ denote the subgroup of $F(\vee_{\alpha\in J}S^1)_{n+1}$ generated by all of the commutators $[y^{(\alpha_1)\epsilon_1}_{i_1},\cdots, y^{(\alpha_t)\epsilon_t}_{i_t}]$ with $\{i_1,\cdots,i_t\}=\{0,1,\cdots,n\}$ as sets, where $\epsilon_j=\pm1$, $\alpha_j\in J $ and the commutator bracket $[\cdots]$ runs over all of the commutator bracket arrangements of weight $t$ for each $t$. \begin{lemma} $C^J_{n+1}\subseteq NF(\vee_{\alpha\in J}S^1)_{n+1}$. \end{lemma} \noindent{\em Proof:} For each $1\leq j\leq n+1$, there exists some $i_s=j-1$. Thus $d_j(y^{(\alpha_s)\epsilon_s}_{i_s})=1$ for some $i_s$ and therefore $$ d_j([y^{(\alpha_1)\epsilon_1}_{i_1},\cdots, y^{(\alpha_t)\epsilon_t}_{i_t}])=1 $$ for each $j>0$. The assertion follows. \begin{lemma} For each $1\leq j\leq n+1$, $$ Kerd_j\cap F(\vee_{\alpha\in J}S^1)_{n+1}=, $$ the normal subgroup generated by $y_{j-1}^{(\alpha)}$ with $\alpha\in J$. \end{lemma} \noindent{\em Proof:} By the definition of $d_j$, there is a commutative diagram $$ \hspace{0.0in} \begin{array}{cccccc} & F(y^{(\alpha)}_j|0\leq j\leq n,\alpha\in J) & \stackrel{p}{\rightarrow} & F(y^{(\alpha)}_0\cdots{\hat y^{(\alpha)}_{j-1}}\cdots y_n^{(\alpha)}|\alpha\in J) \\ & d_j\downarrow & &\cong \downarrow {\bar d_j} \\ & F(y^{(\alpha)}_j|0\leq j\leq n-1,\alpha\in J)&\stackrel{=}{\rightarrow} & F(y^{(\alpha)}_j|0\leq j\leq n-1,\alpha\in J) \\ \end{array} $$ where $p$ is the projection and $$ {\bar d_j}y^{(\alpha)}_k=\cases{y^{(\alpha)}_{k-1}& $j\leq k$,\cr y^{(\alpha)}_k& $j>k+1$.\cr} $$ The assertion follows. \begin{theorem} Let $C^J_{n+1}$ be defined as above. Then $$ NF(\vee_{\alpha\in J}S^1)_{n+1}=C^J_{n+1} $$ \end{theorem} \noindent{\em Proof:} By lemma 5.1, each $d_j$ with $j>0$ is a projection homomorphism. Thus, by Theorem 3.5, $$ NF(\vee_{\alpha\in J}S^1)_{n+1}=\bigcap_{j=1}^{n+1}Kerd_j=F({\cal A}(T_0,T_1,\cdots,T_n)) $$ where $T_j=\{y^{(\alpha)}_0,\cdots,{\hat y^{(\alpha)}_j},\cdots,y^{(\alpha)}_n|\alpha\in J\}$. It suffices to show that $${\cal A}(T_0,T_1,\cdots,T_n)\subseteq C^J_{n+1}.$$ This follows from the next lemma. \begin{lemma} For $0\leq j\leq n$, let $W=[y^{(\alpha_1)\epsilon_1}_{i_1},\cdots,y^{(\alpha_t)\epsilon_t}_{i_t}]\in {\cal A}(T_0,T_1,\cdots,T_j)$. Then $$ \{0,1,\cdots,j\}\subseteq\{i_1,\cdots,i_t\}. $$ \end{lemma} \noindent{\em Proof:} The proof is given by induction on $j$ for $0\leq j\leq n$. Notice that, by construction, each element in ${\cal A}(T_0,T_1,\cdots,T_j)$ is written as a certain commutator. If $j=0$, then $$ {\cal A}(T_0)=\{y^{(\alpha)}_0,[[y^{(\alpha)}_0,y^{(\alpha_1)\epsilon_1}_{i_1}],\cdots,],y^{(\alpha_t)\epsilon_t}_{i_t}]\} $$ where $\alpha,\alpha_j\in J$, $\epsilon_j=\pm1$ and $y^{(\alpha_1)\epsilon_1}_{i_1}\cdots y^{(\alpha_t)\epsilon_t}_{i_t}$ runs over all of the reduced words $\not=1$ in $F(y^{(\alpha)}_j|\alpha\in J,1\leq j\leq n)$. Thus the assertion holds for $j=0$. Suppose that the the assertion holds for $j-1$ with $j\leq n$. Recall that $$ T^{(j)}_j=\{W\in{\cal A}(T_0,T_1,\cdots,T_{j-1}|W=[y^{(\alpha_1)\epsilon_1}_{i_1},y^{(\alpha_2)\epsilon_2}_{i_2},\cdots,y^{(\alpha_t)\epsilon_t}_{i_t}],\,with\,y^{(\alpha_s)}_{i_s}\in T_j\}. $$ Notice that $y^{(\alpha_s)}_{i_s}\in T_j\Longleftrightarrow i_s\not=j$. Thus, for $W=[y^{(\alpha_1)\epsilon_1}_{i_1},y^{(\alpha_2)\epsilon_2}_{i_2},\cdots,y^{(\alpha_t)\epsilon_t}_{i_t}]\in{\cal A}(T_0,\cdots,T_{j-1})-T^{(j)}_j$, $j\in\{i_1,\cdots,i_t\}$. By induction, $\{0,\cdots,j-1\}\subseteq\{i_1,\cdots,i_t\}$. Hence $$\{0,1,\cdots,j\}\subseteq\{i_1,i_2,\cdots,i_t\}$$ for any $W=[y^{(\alpha_1)\epsilon_1}_{i_1},y^{(\alpha_2)\epsilon_2}_{i_2},\cdots,y^{(\alpha_t)\epsilon_t}_{i_t}]\in{\cal A}(T_0,\cdots,T_{j-1}-T^{(j)}_j$. Recall that, by construction, $$ {\cal A}(T_0,\cdots, T_j)={\cal A}_{T^{(j)}_j}. $$ $$\{0,1,\cdots,j\}\subseteq\{i_1,i_2,\cdots,i_t\}$$ for any $W=[y^{(\alpha_1)\epsilon_1}_{i_1},y^{(\alpha_2)\epsilon_2}_{i_2},\cdots,y^{(\alpha_t)\epsilon_t}_{i_t}]\in{\cal A}(T_0,\cdots,T_{j})$. This completes the proof. \begin{theorem} $\pi_{n+2}(\vee_{\alpha\in J}S^2)$ is isomorphic to the center of the group with generators $$ y^{(\alpha)}_0,y^{(\alpha)}_1,\cdots,y^{(\alpha)}_n $$ for $\alpha\in J$. and relations $$ [y^{(\alpha_1)\epsilon_1}_{i_1}, y^{(\alpha_2)\epsilon_2}_{i_2},\cdots,y^{(\alpha_t)\epsilon_t}_{i_t}] $$ with $\{i_1,i_2,\cdots,i_t\}=\{-1,0,1,\cdots,n\}$ as sets, where the indices $i_j$ can be repeated, $\epsilon_j=\pm1$, $\alpha_j\in J$ and the commutator bracket runs over all of the commutator bracket arrangements of weight $t$ for each $t$. \end{theorem} \noindent{\em Proof:} Notice that $\pi_{n+2}(\vee_{\alpha\in J} S^2)\cong\pi_{n+1}(F(\vee_{\alpha\in J}S^1))$. By the above theorem, ${\cal B}_{n+1}$ is generated by $$ [y^{(\alpha_1)\epsilon_1}_{i_1}, y^{(\alpha_2)\epsilon_2}_{i_2},\cdots,y^{(\alpha_t)\epsilon_t}_{i_t}] $$ By Theorem 2.8, the assertion holds for $n\geq1$. For $n=0$, ${\cal B}_1=\Gamma^2(F(y^{(\alpha)}_0|\alpha\in J))$ and $$ F(y^{(\alpha)}_0|\alpha\in J)/{\cal B}_1\cong \oplus_{\alpha\in J}{\bf Z}\pi_2(\vee_{\alpha\in J}S^2). $$ The assertion holds for all of $n$.\\ \par For the general case, we need a simplicial group construction. \begin{definition} Let $G$ be a simplicial group and let $X$ be a pointed simplicial set with a point $*$. The simplicial group {\bf $F^G(X)$} is defined by setting $$ F^G(X)_n=\coprod_{x\in X_n}(G_n)_x, $$ the free product, modulo the relations $(G_n)_*$, where $(G_n)_x$ is a copy of $G_n$. The faces and degeneracies homomorphisms in $F^G(X)$ is given in the canonical way by the universal property of the coproduct in the category of groups and group homomorphisms. \end{definition} \begin{lemma}[Ca, Theorem 9, pp.88] Let $G$ be a simplicial group and let $X$ be a pointed simplicial set. Then the goemetric realization $|F^G(X)|$ is homotopy equivalent to $\Omega(B|G|\wedge|X|)$. \end{lemma} A generalization of this lemma by using fibrewise simplicial groups is given in [Wu1]. \begin{theorem}[Theorem 1.5] Let $\pi$ be any group and let $\{x^{(\alpha)}|\alpha\in J\}$ be a set of generators for $\pi$. Then, for $n\not=1$, $\pi_{n+2}(\Sigma K(\pi,1))$ is isomorphic to the center of the quotient group of the free product $${\coprod^{groups}}_{0\leq j\leq n}(\pi)_j$$ modulo the relations $$ [y^{(\alpha_1)\epsilon_1}_{i_1},y^{(\alpha_2)\epsilon_2}_{i_2},\cdots,y^{(\alpha_t)\epsilon_t}_{i_t}] $$ with $\{i_1,i_2,\cdots,i_t\}=\{-1,0,1,\cdots,n\}$ as sets, where $(\pi)_j$ is a copy of $\pi$ with generators $\{x^{(\alpha)}_j|\alpha\in J\}$, $\epsilon_j=\pm1$, $y^{(\alpha)}_{-1}={x^{(\alpha)}_0}^{-1}$, $y_j=x^{(\alpha)}_j{x^{(\alpha)}_{j+1}}^{-1}$ for $1\leq j\leq n-1$ and $y^{(\alpha)}_n=x^{(\alpha)}_n$, the commutator bracket $[\cdots]$ runs over all of the commutator bracket arrangment of weight $t$ for each $t$. \end{theorem} \noindent{\em Proof:} Since $\{x^{(\alpha)}|\alpha\in J\}$ is a set of generators for $\pi$, $F(x^{(\alpha)}|\alpha\in J)\rightarrow\pi $ is an epimorphism. Thus $$ F(\vee_{\alpha\in J}S^1)\cong F^{F(x^{(\alpha)}|\alpha\in J)}(S^1)\rightarrow F^{\pi}(S^1) $$ is an epimorphism. Hence $$ NF(\vee_{\alpha\in J}S^1)\rightarrow NF^{\pi}(S^1) $$ and $$ {\cal B}F(\vee_{\alpha\in J}S^1)\rightarrow {\cal B}F^{\pi}(S^1) $$ are epimorphisms. The assertion follows from Theorem 5.6, Lemma 5.7 and Proposition 2.8. \section{Applications} In this section, we consider Cohen's $K$-construction. Our descriptions for the homotopy groups of $3$-sphere gives a calculation for the $K$-construction of $1$-sphere. \begin{definition} Let $X$ be set. The group ${\bf K(X)}$ is defined to the the quotient group of the free group $F(X)$ modulo the normal subgroup generated by all of the commutators $[[x_1,x_2],\cdots,],x_t]$ with $x_i\in X$ and $x_i=x_j$ for some $1\leq i2$. \end{corollary} Now we consider the Samelson product in $\pi_*(K(S^1))$. Let $x_j$ denote $s_n\cdots{\hat s_j}\cdots s_0\sigma$ in $S^1_{n+1}$. The following lemma follows directly from Lemma 6.6. \begin{lemma} Let $I=(i_1,i_2,\cdots,i_m)$ be a sequence with $i_1=\prod_{(a,b)}[s_b\sigma, s_a\tau]^{sign(a,b)}, $$ where $G$ is a simplicial group, $(a,b)=(a_1,\cdots,a_n,b_1,\cdots,b_m)$ runs over all shuffles of $(0,1,\cdots,m+n-1)$,i.e. all permutations, so that $a_1 $$ $$ =\sum_{(I,J)}sign(I,J)[[x_{i_{\sigma(0)}},\cdots,x_{i_{\sigma(n)}}],[x_{j_{\tau(0)}},\cdots,x_{j_{\tau(m)}}]] $$ for the commutators $$ [x_{\sigma(0)},x_{\sigma(1)},\cdots,x_{\sigma(n)}]\in\pi_{n+1}(KS^1))\cong Lie(n+1) $$ and $$ [x_{\tau(0)},x_{\tau(1)},\cdots,x_{\tau(m)}]\in\pi_{m+1}(K(S^1))\cong Lie(m+1) $$ where $$(I,J)=(i_0,i_1,\cdots,i_n,j_0,j_1,\cdots,j_m)$$ runs over all shuffles of $(0,1,\cdots,m+n+1)$ so that $i_00$. Notice that $F(S^1)_{n+1}=F(x_0,\cdots,x_n)$ and $K(S^1)_{n+1}=K(x_0,\cdots,x_n)$. Thus $H_{n+1}$ is a normal subgroup of $F(x_0,\cdots,x_n)$ generated by the commutators $$ [[x_{i_1},x_{i_2}],\cdots,x_{i_t}],x_{i_t}] $$ such that $i_p\not=i_q$ for $pj$. By induction, $[[x_{i'_1}, x_{i'_2}],\cdots,x_{i'_t}],x_{i'_t}]\in\overline{B}$. Thus $W=[[x_{i_1},x_{i_2}],\cdots,x_{i_t}],x_{i_t}]\in\overline{B}$. The induction is finished and the assertion follows.\\ \par The simplicial group $K(S^1)$ is homotopy eqivalent to a product of the Eilenberg-Maclane spaces with a different product structure. \begin{proposition} $\Omega K(S^1)$ is an abelian simplicial group. Therefore $K(S^1)$ is homotopy equivalent to a product of the Eilenberg-MacLane spaces as a simplicial set. \end{proposition} \noindent{\em Proof:} Consider $$ d_0:K(x_0,x_1,\cdots,x_n)\rightarrow K(x_0,x_1,\cdots,x_{n-1}) $$ $ d_0(x_0)=1 $ and $ d_0(x_j)=x_{j-1}. $ Thus $Kerd_0\cap K_{n+1}(S^1)\cong $ is the normal subgroup generated by $x_0$ which is abelian by Lemma 6.4. The assertion follows.\\ \par In the end of this section, we give some applications of $K(S^1)$ to minimal simplicial groups. \begin{proposition} Let $G$ be a minimal simplicial group such that the abelianlizer $G^{ab}$ is a minimal simplicial group $K(\pi,1)$ for a cyclic group $\pi$. Then $G$ is homotopy equivalent to a product of Eilenberg-Maclane spaces. \end{proposition} \noindent{\em Proof:} Notice that $G_1=\pi$. Let $x$ be a generator for the cyclic group $\pi$ and let $f_x:S^1\rightarrow G$ be a representive map of $x$, i.e, $f_x(\sigma)=x$. Let $g:F(S^1)\rightarrow G$ be the simplicial homomorphism induced by $f_x$. We need a lemma. \begin{lemma} The simplicial homomorphism $g:F(S^1)\rightarrow G$ is simplicial surjection. \end{lemma} \noindent{\em Proof:} It suffices to show that the subsimplicial group, denote by $H$, of $G$ generated by $G_1$ is $G$ itself. This is given by induction on the dimensions starting with $H_1=G_1$. Suppose that $H_{n-1}=G_{n-1}$ with $n>1$. By a result of Condule [see, e.g, Po, Proposition 1, pp.6], $G_n$ is generated by the degenerate images of lower order Moore chain complex terms and $NG_n$. Thus $G_n$ is genereted by $NG_n$ and $H_n$ by induction. Notice that $G$ is a minimal simplicial group. Thus $NG_n={\cal Z}G_n$, the cycles. By Proposition 2.1, ${\cal Z}G_n$ is contained in the center of $G_n$. Thus $H_n$ is a normal subgroup of $G_n$ and the composite $\phi:\pi_nG\cong{\cal Z}G_n\rightarrow G_n\rightarrow G_n/H_n$ is an epimorphism. Thus $G_n/H_n$ is an abelian group and so the quotient homomorphism $G_n\rightarrow G_n/H_n$ factors through $G_n^{ab}$. Notice that $G^{ab}=K(\pi,1)$. Thus $G^{ab}_n\cong G^{ab}_n/{\cal Z}G_n$ and so $\phi:\pi_nG\rightarrow G_n/H_n$ is trivial. Thus $G_n/H_n$ is trivial and the assertion follows.\\ \par \noindent{\em Continuation of Proof of Proposition 6.14:} Notice that $G$ is minimal. The simplicial epimorphism $g:F(S^1)\rightarrow G$ factors through $K(S^1)$ by Proposition 6.12. By Proposition 6.13, $\Omega K(S^1)$ is an abelian simplicial group. Thus $\Omega G$ is also an abelian simplicial group. Thus $G$ is homotopy equivalent to a product of Eilenberg-Maclane spaces, which is the assertion.\\ \par The following counter-example for minimal simplicial groups is due to J. W. Milnor (unpublished). \begin{proposition} $\Omega (S^{n+1}[n+1,n+2,n+3])$ does not have a homotopy type of a minimal simplicial group for $n>0$, where $S^{n+1}[n+1,n+2,n+3]$ is the 3-stage Postnikov system by taking the first three nontrivial homotopy groups of $S^{n+1}$. \end{proposition} \noindent{\em Proof:} Suppose that $G$ is a minimal simplicial group such that $G\simeq\Omega(S^{n+1}[n+1,n+2,n+3])$. Let $f:F(S^1)\rightarrow \Omega^{n-1}G$ be a simplicial homomorphism such that $f(\sigma)$ is a generator of $(\Omega^{n-1}G)_1\cong G_n\cong {\bf Z}$. Then $f_*:\pi_j(F(S^1))\rightarrow\pi_j(\Omega^{n-1}G)$ is an isomorphism for $j\leq3$. Notice that $\Omega^{n-1}G$ is also a minimal simplicial group. The simplicial homomorphism $f:F(S^1)\rightarrow\Omega^{n-1}G$ factors through $K(S^1)$. Notice that $\pi_3(F(S^1))\cong\pi_3\Omega^{n-1}G)\cong{\bf Z}/2$ and $\pi_3(K(S^1))\cong Lie(3)\cong{\bf Z}\oplus{\bf Z}$. There is a contradiction and the assertion follows.\\ \par More examples and counter-examples for minimal simplicial groups will be given in [Wu2]. It was know that there are still many counter-examples of two-stage Postnikov systems for minimal simplicial groups [Wu2]. \bigskip\begin{thebibliography}{Lellmannaa} \bibitem[A]{A} J. F. Adams, {\em On products in minimal complexes}, Trans. Amer. Math. Soc. 82 (1956), 180-189. \bibitem[AM]{AM} M. Artin and B. Mazur, {\em On the van Kampen theorem}, Topology 5(1966), 179-189. \bibitem[BK]{BK} A. K. Bousfield, D. 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