Let $S$ denote the sphere spectrum localized at $p$ and $p$ denote an odd prime. From [14], the homotopy group of $n$-dimensional sphere $\pi_{n+r}S^{n}$ $(r>0)$ is a finite group. So the determination of $\pi_{n+r}S^{n}$ has become one of the central problems in algebraic topology.

Ever since the introduction of the Adams spectral sequence (ASS) in the late 1950's (see [1]), the study of the homotopy groups of spheres $\pi_{\ast}S$ was split into algebraic and geometric problems, including the computation of ${\rm{Ext}}_A^{*, *}({\mathbb{Z}_p}, {\mathbb{Z}_p})$ and the detection which element of ${\rm{Ext}}_A^{*, *}({\mathbb{Z}_p}, {\mathbb{Z}_p})$ can survive to $E_{\infty}^{\ast, \ast}$, here $A$ is the mod $p$ Steenrod algebra, ${\rm{Ext}}_A^{*, *}({\mathbb{Z}_p}, {\mathbb{Z}_p})$ is the $E_{2}$-term of the ASS. By [2],

$E_2^{s, t}\cong{\rm{Ext}}_A^{s, t}({\mathbb{Z}_p}, {\mathbb{Z}_p})\Rightarrow\pi_{t-s}S, $

and the Adams differential is $d_{r}:E_{r}^{s, t}\mathop \to E_{r}^{s+r, t+r-1}$.

In addition, we also have the Adams-Novikov spectral sequence (ANSS) [12, 13] based on the Brown-Peterson spectrum BP in the determination of $\pi_{\ast}S$.

Many wonderful results were obtained, however, it is still far from the total determination of $\pi_{\ast}S$. After the detection of $\eta_{j}\in\pi_{p^{j}q+pq-2}S$ for $p=2$, $j\neq2$, by Mahowald in [11], which was represented by $h_{1}h_{j}\in{\rm{Ext}}_A^{2, p^{j}q+pq}({\mathbb{Z}_p}, {\mathbb{Z}_p})$, many nontrivial elements in $\pi_{\ast}S$ were found. Please see references [5-9] for details. In recent years, the first author established several convergence of elements by an arithmatic method, see [16-18, 21].

In [5], Cohen made the nontrivial secondary Adams differential $d_{2}(h_{i})=a_{0}b_{i-1}$ $(p>2, i>0)$ as geometric input, then, a nontrivial element $\xi_{i}\in\pi_{(p^{i+1}+1)q}S$ $(i\geqslant 0)$ is detected. In this paper, we also detect a new family in $\pi_{\ast}S$ by geometric method, the only geometric input used in the proof is the secondary nontrivial differential given in [20].

The main result is obtained as follows.

Theorem 1.1  Let $3\leqslant s< p-1$, $n>3$, $p\geqslant7$, then

$0\neq \widetilde{\gamma_{s}}h_{0}g_{n}\in {\rm{Ext}}_A^{s+3, {p^{n+1}}q+2{p^{n}}q+sp^{2}q+(s-1)pq+(s-1)q+s-3}(\mathbb{Z}_p, \mathbb{Z}_p) $

is a permanent cycle in the Adams spectral sequence and converges to a nontrivial element of order $p$ in $\pi_{{p^{n+1}}q+2{p^{n}}q+sp^{2}q+(s-1)pq+(s-1)q-6} S$.

The paper is organized as follows. After giving some necessary preliminaries and useful knowledge about the MSS in Section 2. The proof of Theorem $1.1$ and some results on Ext groups will be given in Section 3.

For the convenience of the reader, let us briefly indicate the necessary preliminaries in the proof of the propositions and theorems.

Let $M$ be the Moore spectrum modulo an odd prime $p$ given by the cofibration

${ S \xrightarrow{p} S \xrightarrow{i} M \xrightarrow{j} \Sigma S}. $

Let $\alpha$: $\sum^{q}M\rightarrow M$ be the Adams map and $V(1)$ be its cofibre given by the cofibration

${ \Sigma^{q}M\xrightarrow{\alpha } M \xrightarrow{{{i}^{\prime }}} V(1)\xrightarrow{{{j}^{\prime }}} \Sigma^{q+1} M.} $

Let $\beta$: $\sum^{(p+1)q}V(1)\rightarrow V(1)$ be the $\nu_2$-mapping and $V(2)$ be the cofibre of $\beta$ sitting in the cofibration

${ \Sigma^{(p+1)q}V(1)\xrightarrow{\beta } V(1) \xrightarrow{{{i}^{\prime \prime }}} V(2)\xrightarrow{{{j}^{\prime \prime }}} \Sigma^{(p+1)q+1} V(1).} $

Furthermore, $\gamma$: $\sum^{(p^{2}+p+1)q}V(2)\rightarrow V(2)$ is the $\nu_3$-mapping and the $\gamma$-element $\gamma_{s}=jj^{\prime}j^{\prime\prime}\gamma^{s}i^{\prime\prime}i^{\prime}i$ is a nontrivial element in $\pi_{sp^{2}q+(s-1)pq+(s-2)q-3}S$, where $p\geqslant7$ (see [15]).

From [19], we know that the third periodicity family $\gamma_{s}$ is represented by the third Greek letter family element

$\widetilde{\gamma}_{s}\in {\rm {Ext}}_{A}^{s, sp^{2}q+(s-1)pq+(s-2)q+s-3}(\mathbb{Z}_p, \mathbb{Z}_p)$

in the ASS, which is represented by the element

$s(s-1)(s-2)a_{3}^{s-3}h_{3, 0}h_{2, 1}h_{1, 2}$

in the May spectral sequence (MSS).

Let $L$ be the cofibre of $\alpha_{1}=j\alpha i$: $\sum^{q-1}S\rightarrow S$ given by the cofibration

${ \Sigma^{q-1}S\xrightarrow{{{\alpha }_{1}}} S \xrightarrow{{\bar{i}}} L\xrightarrow{{\bar{j}}} \Sigma^{q} S.} $

From [10], we can see that ${\rm{Ext}}_A^{1, \ast}(\mathbb{Z}_p, \mathbb{Z}_p)$ has $\mathbb{Z}_{p}$-bases

${{a}_{0}}\in \text{Ext}_{A}^{1, 1}({{\mathbb{Z}}_{p}}, {{\mathbb{Z}}_{p}}), {{h}_{i}}\in \text{Ext}_{A}^{1, {{p}^{i}}q}({{\mathbb{Z}}_{p}}, {{\mathbb{Z}}_{p}})(i\ge 0).$

${\rm{Ext}}_A^{2, \ast}(\mathbb{Z}_p, $ $\mathbb{Z}_p)$ has $\mathbb{Z}_p$-bases

$\alpha_{2}$, $a_0^2$, $a_{0}h_{i}$ $(i>0)$, $g_{i}$ $(i\geqslant0)$, $k_{i}$ $(i\geqslant0)$, $b_{i}$ $(i\geqslant0)$, and $h_{i}h_{j}$ $(j\geqslant{i+2}, i\geqslant0)$,

whose internal degrees are

$2q+1$, $2$, $p^{i}q+1$, $2p^{i}q+p^{i+1}q$, $2p^{i+1}+p^{i}q$, $p^{i+1}q$ and $p^{i}q+p^{j}q$,

respectively. Aikawa computed ${\rm{Ext}}_A^{3, \ast}(\mathbb{Z}_p, \mathbb{Z}_p)$ by $\lambda$-algebra in [3].

In the following, recall the Adams resolution of some spectra related to $S$ from [4]. Let

(2.1)

be the minimal Adams resolution of the sphere spectrum S which satisfles

(A) ${ E_{s} \xrightarrow{{{{\bar{b}}}_{s}}} KG_{s}\xrightarrow{{{{\bar{c}}}_{s}}} E_{s+1} \xrightarrow{{{{\bar{a}}}_{s}}} \Sigma E_{s} }$ are cofibrations for all $s\geqslant0$, which induce short exact sequences in $\mathbb{Z}_p$-cohomology

${ 0\xrightarrow{{}} H^{\ast}E_{s+1}\xrightarrow{\bar{c}_{s}^{*}} H^{\ast}KG_{s} \xrightarrow{\bar{b}_{s}^{*}} H^{\ast}E_{s}\xrightarrow{{}} 0.} $

(B) $KG_{s}$ are the graded wedge sums of Eilenberg-Maclane spectrum $K\mathbb{Z}_p$ of type $\mathbb{Z}_p$.

(C) $\pi_{t}KG_{s}$ are the $E_{1}^{s, t}$-terms of the ASS,

$(\overline{b}_{s}\overline{c}_{s-1})_{\ast}: \pi_{t}KG_{s-1}\longrightarrow \pi_{t}KG_{s}$

are the $d_{1}^{s-1, t}$-differentials of the ASS, and $\pi_{t}KG_{s} \cong \rm Ext_{A}^{s, t}(\mathbb{Z}_p, \mathbb{Z}_p)$. Then, an Adams resolution of an arbitrary spectrum $V$ can be obtained by smashing $V$ to (2.1).

Remark 2.1  In the ANSS, $h_{0}$ is a permanent cycle and converges to the corresponding homotopy element $i^{\prime}i\alpha_{1}$($\alpha_1=j\alpha i\in\pi_{q-1}S$) in $\pi_{q-1}K$. Furthermore, if some suppositions on Ext groups are given, then there exists $\overline{w}\in\pi_{p^{n+1}q+2p^{n}q-2}K$ such that $i^{\prime}i\xi=\alpha^{\prime\prime}\cdot\overline{w}$ (mod $F^{4}\pi_{\ast}K$) and $\overline{w}$ is represented by $(i^{'}i)_{\ast}(g_{n})\in{\rm {Ext}}_{A}^{2, p^{n+1}q+2p^{n}q}(H^{\ast}K, \mathbb{Z}_p)$ in the ASS, where $\xi\in\pi_{p^{n+1}q+2p^{n}q-4}S$ is the homotopy element which is represented by $h_{0}l_{n}\in {\rm {Ext}}_{A}^{4, p^{n+1}q+2p^{n}q+q}$ $(\mathbb{Z}_p, \mathbb{Z}_p)$ in the ASS and $F^{4}\pi_{\ast}K$ denotes the group consisting of all elements in $\pi_{\ast}K$ with filtration no less than 4.

To detect $\pi_{\ast}S$ with the ASS, we must compute the $E_{2}$-term of the ASS, ${\rm Ext}_{A}^{\ast, \ast}(\mathbb{Z}_p, \mathbb{Z}_p)$. The most successful method for computing it is the MSS.

From [13], there is a MSS $\left\{ E_{r}^{s, t, *}, {{d}_{r}} \right\}$, which converges to ${\rm Ext}_{A}^{s, t}(\mathbb{Z}_p, \mathbb{Z}_p)$ with $E_{1}$-term

$E_1^{\ast, \ast, \ast}=E(h_{i, j}\mid i>0, j\geqslant0)\otimes P(b_{i, j}\mid i>0, j\geqslant0)\otimes P(a_{i}\mid i\geqslant0), $

(2.2)

where $E( )$ denotes the exterior algebra, $P( )$ denotes the polynomial algebra, and

$h_{i, j}\in E_1^{1, 2(p^{i}-1)p^{j}, 2i-1}, b_{i, j}\in E_1^{2, 2(p^{i}-1)p^{j+1}, p(2i-1)}, a_{i}\in E_1^{1, 2p^{i}-1, 2i+1}. $

One has $d_{r}$: $E_r^{s, t, M}\rightarrow E_r^{s+1, t, M-r}$ $(r\geqslant1)$. If $x\in E_r^{s, t, \ast}$ and $y\in E_r^{s', t', \ast}$, then

$d_{r}(x\cdot y)=d_{r}(x)y+(-1)^{s}xd_{r}(y).$

(2.3)

Furthermore, the May $E_{1}$-term is graded commutative in the sense that

$\begin{align} & {{a}_{m}}{{h}_{n, j}}={{h}_{n, j}}{{a}_{m}}, {{h}_{m, k}}{{h}_{n, j}}=-{{h}_{n, j}}{{h}_{m, k}}, \\ & {{a}_{m}}{{b}_{n, j}}={{b}_{n, j}}{{a}_{m}}, \quad {{h}_{m, k}}{{b}_{n, j}}={{b}_{n, j}}{{h}_{m, k}}, \\ & {{a}_{m}}{{a}_{n}}={{a}_{n}}{{a}_{m}}, \quad \quad {{b}_{m, n}}{{b}_{i, j}}={{b}_{i, j}}{{b}_{m, n}}. \\ \end{align}$

The first May differential $d_{1}$ is given by

$\left\{\begin{array}{l} d_{1}(h_{i, j})=-\sum \limits_{0<k<i}h_{i-k, k+j}h_{k, j}, \\ d_{1}(a_{i})=-\sum \limits_{0<k<i}h_{i-k, k}, a_{k}, \\ d_{1}(b_{i, j})=0. \end{array} \right.$

(2.4)

For each element $x\in E_{1}^{s, t, \ast}$, if we denote dim $x=s$, deg $x=t$, we have

$\left\{\begin{array}{l} \dim h_{i, j}=\dim a_{i}=1, \dim b_{i, j}=2, \\ \deg h_{i, j}=2(p^{i}-1)p^{j}=(p^{i+j-1}+\cdots+p^{j})q, \\ \deg b_{i, j}=2(p^{i}-1)p^{j+1}=(p^{i+j}+\cdots+p^{j+1})q, \\ \deg a_{i}=2p^{i}-1=(p^{i-1}+\cdots+1)q+1, \\ \deg a_{0}=1. \end{array} \right.$

(2.5)

Remark 2.2  Any positive integer $t$ can be expressed uniquely as $t=q(c_{n}p^{n}+c_{n-1}p^{n-1}+\cdots+ c_{1}p+c_{0})+e, $ where $0\leqslant c_{i}<p (0\leqslant i <n)$, $0<c_{n}<p$, $0\leqslant e< q$.

Then, it is easy to get the following result from [16].

Proposition 2.3   In the MSS, we have $E_{1}^{s, t, \ast}=0$ for some $j$ $(0\leqslant j\leqslant n)$, $s<c_{j}$, where $s$ is also a positive integer with $0<s<p$.

In this section, we mainly give some important results about Adams $E_{2}$-terms. At the end, the proof of Theorem $1.1$ will be given.

Proposition 3.1   Let $3\leqslant s<p-1, n>3, p\geqslant7$, then

$0\neq \widetilde{\gamma}_{s}h_{0}g_{n}\in {\rm{Ext}}_A^{s+3, {p^{n+1}}q+2{p^{n}}q+sp^{2}q+(s-1)pq+(s-1)q+s-3}(\mathbb{Z}_p, \mathbb{Z}_p). $

Proof  Consider the structure of $E_1^{s+2, t, \ast}$ in the MSS, where $t=p^{n+1}q+2p^{n}q+sp^{2}q+(s-1)pq+(s-1)q+s-3$. Due to $3\leqslant s<p-1$, then $5\leqslant s+2<p+1$.

Case 1  $5\leqslant s+2<p$. Let $h=x_{1}x_{2}\cdots x_{m}$ be the generator of $E_1^{s+2, t, \ast}$, where $x_{i}$ is one of $a_{k}$, $h_{i, j}$ or $b_{u, z}$, $0\leqslant k\leqslant n+2$, $0< i+j\leqslant n+2$, $0< u+z\leqslant n+1$, $i>0$, $j\geqslant 0$, $u>0$, $z\geqslant 0$.

Assume that $\deg x_{i}=q(c_{i, n+1}p^{n+1}+\cdots+c_{i, 1}p+c_{i, 0})+e_{i}, $ where $c_{i, j}=0$ or 1, $e_{i}=1$ if $x_{i}=a_k$ or $e_{i}=0$, then

$\deg h =\sum\limits_{i=1}^{m}\deg x_{i} =q((\sum\limits_{i=1}^{m}c_{i, n+1})p^{n+1}+(\sum\limits_{i=1}^{m}c_{i, n})p^{n}+\cdot \cdot\cdot+(\sum\limits_{i=1}^{m}c_{i, 0}))+(\sum\limits_{i=1}^{m}e_{i})\\ \quad \quad =q(p^{n+1}+2p^{n}+sp^{2}+(s-1)p+(s-1))+s-3, \\ \dim h =\sum\limits_{i=1}^{m}\dim x_{i}=s+2. $

Note that $\dim x_{i}=1$ or 2, we can see that $m\leqslant s+2<p$ from $\sum\limits_{i=1}^{m}\dim x_{i}=s+2$. By the fact that $c_{i, j}=0$ or 1, $e_i=0$ or 1, $m\leqslant s+2<p$, we have

$ \sum\limits_{i=1}^{m}e_{i}=s-3, \sum\limits_{i=1}^{m}c_{i, 0}=s-1, \sum\limits_{i=1}^{m}c_{i, 1}=s-1, \sum\limits_{i=1}^{m}c_{i, 2}=s, \\ \sum\limits_{i=1}^{m}c_{i, 3}=\cdots=\sum\limits_{i=1}^{m}c_{i, n-1}=0, \sum\limits_{i=1}^{m}c_{i, n}=2, \sum\limits_{i=1}^{m}c_{i, n+1}=1. $

From the above results, we can see that $b_{1, n}b_{1, n-1}h_{1, n}$, $h_{2, n}h_{1, n}$, $h_{2, n}b_{1, n-1}$, $b_{2, n-1}h_{1, n}$, $b_{1, n-1}b_{2, n-1}$, $b_{1, n}b_{1, n-1}^{2}$, $h_{1, n+1}b_{1, n-1}^{2}$ and $h_{1, n+1}b_{1, n-1}h_{1, n}$ are contained in the $x_{i}$. By the commutativity of $E_1^{\ast, \ast, \ast}$, we can denote

$\begin{array}{l} {h_1} = {x_1}{x_2} \cdots {x_{m - 3}}{b_{1, n}}{h_{1, n}}{b_{1, n - 1}}, \quad h_1^\prime = {x_1}{x_2} \cdots {x_{m - 3}} \in E_1^{s - 3, {t^\prime }, * };\\ {h_2} = {x_1}{x_2} \cdots {x_{m - 3}}{b_{1, n}}b_{1, n - 1}^2, \quad \quad h_2^\prime = {x_1}{x_2} \cdots {x_{m - 3}} \in E_1^{s - 4, {t^\prime }, * };\\ {h_3} = {x_1}{x_2} \cdots {x_{m - 3}}{h_{1, n + 1}}b_{1, n - 1}^2, \quad \;\;h_3^\prime = {x_1}{x_2} \cdots {x_{m - 3}} \in E_1^{s - 3, {t^\prime }, * };\\ {h_4} = {x_1}{x_2} \cdots {x_{m - 2}}{h_{2, n}}{h_{1, n}}, \quad \quad \;\;h_4^\prime = {x_1}{x_2} \cdots {x_{m - 2}} \in E_1^{s, {t^\prime }, * };\\ {h_5} = {x_1}{x_2} \cdots {x_{m - 2}}{h_{2, n}}{b_{1, n - 1}}, \quad \quad \;h_5^\prime = {x_1}{x_2} \cdots {x_{m - 2}} \in E_1^{s - 1, {t^\prime }, * };\\ {h_6} = {x_1}{x_2} \cdots {x_{m - 2}}{b_{2, n - 1}}{h_{1, n}}, \quad \quad \;h_6^\prime = {x_1}{x_2} \cdots {x_{m - 2}} \in E_1^{s - 1, {t^\prime }, * };\\ {h_7} = {x_1}{x_2} \cdots {x_{m - 2}}{b_{2, n - 1}}{b_{1, n - 1}}, \quad \;\;h_7^\prime = {x_1}{x_2} \cdots {x_{m - 2}} \in E_1^{s - 2, {t^\prime }, * };\\ {h_8} = {x_1}{x_2} \cdots {x_{m - 3}}{h_{1, n + 1}}{h_{1, n}}{b_{1, n - 1}}, \quad h_8^\prime = {x_1}{x_2} \cdots {x_{m - 3}} \in E_1^{s - 2, {t^\prime }, * }, \end{array}$

where $t^\prime=sp^{2}q+(s-1)pq+(s-1)q+s-3$.

We list all the possibilities of $h_{i}^{\prime}$ in the following table $(i=1, 2, \cdots, 8)$, thus $h$ doesn't exist in this case.

Case 2 If $ s+2=p$, then $E_1^{s+2, t^{\prime\prime}, \ast}=E_1^{p, t^{\prime\prime}, \ast}$, where $t^{\prime\prime}=p^{n+1}q+2p^{n}q+(p-2)p^{2}q+(p-3)pq+(p-3)q+p-5$. Let $h=x_{1}x_{2}\cdots x_{r}$ be the generator of $E_1^{p, t^{\prime\prime}, \ast}$, and assume that

$\deg x_{i}=q(c_{i, n+1}p^{n+1}+c_{i, n}p^{n}+\cdots+c_{i, 1}p+c_{i, 0})+e_{i}, $

where $c_{i, j}=0$ or 1, $e_{i}=1$ if $x_{i}=a_{k_{i}}$ or $e_{i}=0$, then

$\deg h =\sum\limits_{i=1}^{r}\deg x_{i} =q((\sum\limits_{i=1}^{r}c_{i, n+1})p^{n+1}+(\sum\limits_{i=1}^{r}c_{i, n})p^{n}+\cdots +(\sum\limits_{i=1}^{r}c_{i, 0}))+(\sum\limits_{i=1}^{r}e_{i})\\ \quad \quad =q(p^{n+1}+2p^{n}+(p-2)p^{2}+(p-3)p+(p-3))+p-5, \\ \dim h =\sum\limits_{i=1}^{r}\dim x_{i}=p. $

We claim that $\sum\limits_{i=1}^{r}c_{i, 0}$, $\sum\limits_{i=1}^{r}c_{i, 1}$ and $\sum\limits_{i=1}^{r}c_{i, 2}$ are impossible to constitute $p$. The reason is the following: if $\sum\limits_{i=1}^{r}c_{i, 0}=p$, because of $\sum\limits_{i=1}^{r}e_{i}=p-5$, then

$q\Big((\sum\limits_{i=1}^{r}c_{i, n+1})p^{n+1}+(\sum\limits_{i=1}^{r}c_{i, n})p^{n}+\cdot \cdot\cdot+(\sum\limits_{i=1}^{r}c_{i, 0})\Big)+(\sum\limits_{i=1}^{r}e_{i})= \sum\limits_{i=1}^{r}e_{i} ({\rm{mod}} p), $

this contradicts to $q(p^{n+1}+2p^{n}+(p-2)p^{2}+(p-3)p+(p-3))+p-5=(p-3)q+p-5 ({\rm{mod}} p)$. For the same reason, $\sum\limits_{i=1}^{r}c_{i, 1}$ and $\sum\limits_{i=1}^{r}c_{i, 2}$ are impossible to constitute $p$.

From $\dim x_{i}=1$ or 2 and $\sum\limits_{i=1}^{r}\dim x_{i}=p$, we can see that $r\leqslant p$. By Remark 2.2 and $r\leqslant p$, $c_{i, j}=0$ or 1, $e_i=0$ or 1, we have

$\sum\limits_{i=1}^{r}e_{i}=p-5, \sum\limits_{i=1}^{r}c_{i, 0}=p-3, \sum\limits_{i=1}^{r}c_{i, 1}=p-3, \sum\limits_{i=1}^{r}c_{i, 2}=p-2, \nonumber\\ (\sum\limits_{i=1}^{r}c_{i, 3})p^{3}+\cdots+(\sum\limits_{i=1}^{r}c_{i, n})p^{n}+ (\sum\limits_{i=1}^{r}c_{i, n+1})p^{n+1}=p^{n+1}+2p^{n}, $

(3.1)

so

$(\sum\limits_{i=1}^{r}c_{i, 3})+\cdots+(\sum\limits_{i=1}^{r}c_{i, n})p^{n-3}+ (\sum\limits_{i=1}^{r}c_{i, n+1})p^{n-2}=p^{n-2}+2p^{n-3}. $

(3.2)

Thus $p\mid \sum\limits_{i=1}^{r}c_{i, 3}$. Note that $c_{i, 3}=0$ or 1, $r\leqslant p$, it is known that $\sum\limits_{i=1}^{r}c_{i, 3}=0$ or $p$.

Case 2.1 When $\sum\limits_{i=1}^{r}c_{i, 3}=0$, we have

$(\sum\limits_{i=1}^{r}c_{i, 4})p+\cdots+(\sum\limits_{i=1}^{r}c_{i, n})p^{n-3}+ (\sum\limits_{i=1}^{r}c_{i, n+1})p^{n-2}=p^{n-2}+2p^{n-3}. $

Case 2.1.1  When $n>4$, we claim that $\sum\limits_{i=1}^{r}c_{i, 4}=0$. Otherwise, if $\sum\limits_{i=1}^{r}c_{i, 4}=p$, then $r=p$. So $\dim x_{i}=1(1\leqslant i\leqslant p)$ and $\deg x_{i}= (\textrm {higher terms}) +p^{4}q+(\textrm {lower terms})$. Because of $\sum\limits_{i=1}^{r}e_{i}=p-5$, $\deg a_{k}\equiv1({\rm mod} q)$, $\dim h_{l, j}\equiv0({\rm mod} q)$ and $\dim b_{u, z}\equiv0({\rm mod} q)$, there exist factors $a_{j_{1}}a_{j_{2}}\cdots a_{j_{p-5}}$ among the generators $x_{i} ( j_{i}\geqslant 5, 1\leqslant i\leqslant p-5)$. Thus, $\sum\limits_{i=1}^{r}c_{i, 3}\geqslant p-5$, which contradicts to $\sum\limits_{i=1}^{r}c_{i, 3}=0$, so $\sum\limits_{i=1}^{r}c_{i, 4}=0$. By induction on $j$, we can get $\sum\limits_{i=1}^{r}c_{i, j}=0 (5\leqslant j\leqslant n-1)$, $\sum\limits_{i=1}^{r}c_{i, n}=2$, $\sum\limits_{i=1}^{r}c_{i, n+1}=1$.

Case 2.1.2  When $n=4$, it is easy to get $\sum\limits_{i=1}^{r}c_{i, 4}=2$ and $\sum\limits_{i=1}^{r}c_{i, 5}=1$.

From the above discussion of Case 2.1.1 and Case 2.1.2, similarly to Case 1, we can see that $b_{1, n}b_{1, n-1}h_{1, n}$, $h_{2, n}h_{1, n}$, $h_{2, n}b_{1, n-1}$, $ b_{2, n-1}h_{1, n}$, $b_{1, n-1}b_{2, n-1}$, $b_{1, n}b_{1, n-1}^{2}, $ $h_{1, n}b_{1, n-1}h_{1, n+1}$ and $h_{1, n+1}b_{1, n-1}^{2} $ are contained in the $x_{i}$, so $h$ is impossible to exist.

Case 2.2  When $\sum\limits_{i=1}^{r}c_{i, 3}=p$, then $r=p$. We get ${\rm dim} x_{i}=1$ from ${\rm dim} h=p$, then $h=x_{1}x_{2}\cdots x_{p}$, $x_{i}\in E(h_{i, j}\mid i>0, j\geqslant 0)\otimes P(a_{k}\mid k\geqslant0)$.

Case 2.2.1  When $n>4$, we get

$p\cdot p^{3}+(\sum\limits_{i=1}^{r}c_{i, 4})p^{4}+\cdots+(\sum\limits_{i=1}^{r}c_{i, n})p^{n}+ (\sum\limits_{i=1}^{r}c_{i, n+1})p^{n+1}=p^{n+1}+2p^{n}, $

that is $(1 + \sum\limits_{i = 1}^r {{c_{i,4}}} ) + (\sum\limits_{i = 1}^r {{c_{i,5}}} )p + \cdots + (\sum\limits_{i = 1}^r {{c_{i,n + 1}}} ){p^{n - 3}} = {p^{n - 3}} + 2{p^{n - 4}},$ thus $p\mid(1+\sum\limits_{i=1}^{r}c_{i, 4})$, so $\sum\limits_{i=1}^{r}c_{i, 4}=p-1$ from $c_{i, 4}=0$ or 1 and $r=p$. By induction on $j$, we can get

$\sum\limits_{i=1}^{r}c_{i, j}=p-1(4\leqslant j\leqslant n-1), \sum\limits_{i=1}^{r}c_{i, n}=1, \sum\limits_{i=1}^{r}c_{i, n+1}=1. $

By the reason of degree and the Proposition 2.3, $h$ is impossible to exist.

Case 2.2.2  When $n=4$, we know that $\sum\limits_{i=1}^{r}c_{i, 4}=1$, $\sum\limits_{i=1}^{r}c_{i, 5}=1$ from (3.2), then

$\deg h=q(p^{5}+2p^{4}+(p-2)p^{2}+(p-3)p+p-3)+(p-5). $

By the reason of degree and the Proposition 2.3, $h$ is impossible to exist.

From the above discussion, for $5\leqslant s+2<p+1$, $E_1^{s+2, t, *}=0$, so $E_r^{s+2, t, *}=0 (r\geqslant 2)$. It is known that $h_{2, n}h_{1, n}$, $h_{1, n}, a_3^{s-3}h_{3, 0}h_{2, 1}h_{1, 2}\in E_1^{\ast, \ast, \ast}$ are permanent cycles in the MSS and converge nontrivially to $g_{n}$, $h_{n}$, $\widetilde{\gamma}_{s}\in{\rm{Ext}}_A^{\ast, \ast}(\mathbb{Z}_p, \mathbb{Z}_p) (n\geqslant0)$, respectively, so $a_3^{s-3}h_{3, 0}h_{2, 1}h_{1, 2}h_{1, n}h_{2, n}h_{1, 0}\in E_1^{s+3, t, \ast}$ is a permanent cycle in the MSS and converges nontrivially to $\widetilde{\gamma}_{s}h_{0}g_{n}\in{\rm{Ext}}_A^{s+3, t}(\mathbb{Z}_p, \mathbb{Z}_p)$. Note that $E_r^{s+2, t, \ast}=0 (r\geqslant1)$, thus the permanent cycle is not $d_{r}$-boundary and converges nontrivially to $\widetilde{\gamma}_{s}h_{0}g_{n}\in{\rm{Ext}}_A^{s+3, t}(\mathbb{Z}_p, \mathbb{Z}_p)$, that is, when $5\leqslant s+2<p+1$, $0\neq \widetilde{\gamma_{s}}h_{0}g_{n}\in{\rm{Ext}}_A^{s+3, t}(\mathbb{Z}_p, \mathbb{Z}_p)$.

Proposition 3.2  Let $3\leqslant s<p-1, n>3, p\geqslant7$, $2\leqslant r< s+3$, then

${\rm{Ext}}_A^{s+3-r, {p^{n+1}}q+2{p^{n}}q+sp^{2}q+(s-1)pq+(s-1)q+s-r-2}(\mathbb{Z}_p, \mathbb{Z}_p)=0. $

Proof  We only need to prove that $E_1^{s+3-r, t, \ast}=0$ in the MSS, where $t=p^{n+1}q+2{p^{n}}q+sp^{2}q+(s-1)pq+(s-1)q+s-r-2$. Let $h=x_{1}x_{2}\cdots x_{m}$ be the generator of $E_1^{s+3-r, t, \ast}$, where $x_{i}$ is $a_{k}$, $h_{i, j}$ or $b_{u, z}$, $0\leqslant k\leqslant n+2$, $0< i+j\leqslant n+2$, $0< u+z\leqslant n+1$, $i>0$, $j\geqslant 0$, $u>0$, $z\geqslant 0$.

Assume that $\deg x_{i}=q(c_{i, n+1}p^{n+1}+c_{i, n}p^{n}+\cdots+c_{i, 0})+e_{i}$, where $c_{i, j}=0$ or 1, $e_{i}=1$ if $x_{i}=a_{k_{i}}$ or $e_{i}=0$, then

$\deg h =\sum\limits_{i=1}^{m}\deg x_{i} =q((\sum\limits_{i=1}^{m}c_{i, n+1})p^{n+1}+(\sum\limits_{i=1}^{m}c_{i, n})p^{n}+\cdots+ (\sum\limits_{i=1}^{m}c_{i, 0}))+(\sum\limits_{i=1}^{m}e_{i})\\ \quad \quad =q(p^{n+1}+2p^{n}+sp^{2}+(s-1)p+(s-1))+s-r-2, \\ \dim h =\sum\limits_{i=1}^{m}\dim x_{i}=s+3-r. $

Note that $\dim x_{i}=1$ or 2, we can see that $m\leqslant s+3-r\leqslant s+1<p$. We claim that $s-r-2\geqslant0$, otherwise, $p>\sum\limits_{i=1}^{m}e_{i}=q+(s-r-2)\geqslant q-5>p$. Because of $c_{i, j}=0$ or 1, $e_i=0$ or 1 and $r<p$, we have

$ \sum\limits_{i=1}^{m}e_{i}=s-r-2, \sum\limits_{i=1}^{m}c_{i, 0}=s-1, \sum\limits_{i=1}^{m}c_{i, 1}=s-1, \sum\limits_{i=1}^{m}c_{i, 2}=s, \\ \sum\limits_{i=1}^{m}c_{i, 3}=\cdots=\sum\limits_{i=1}^{m}c_{i, n-1}=0, \sum\limits_{i=1}^{m}c_{i, n}=2, \sum\limits_{i=1}^{m}c_{i, n+1}=1. $

From the above results, we can see that $b_{1, n}b_{1, n-1}h_{1, n}$, $h_{2, n}h_{1, n}$, $h_{2, n}b_{1, n-1}$, $ b_{2, n-1}h_{1, n}$, $b_{1, n-1}b_{2, n-1}$, $b_{1, n}b_{1, n-1}^{2}, $ $h_{1, n}b_{1, n-1}h_{1, n+1}$ and $h_{1, n+1}b_{1, n-1}^{2}$ are contained in the $x_{i}$. By the commutativity of $E_1^{\ast, \ast, \ast}$, we can denote

$\begin{array}{l} {h_1} = {x_1}{x_2} \cdots {x_{m - 3}}{b_{1, n}}{h_{1, n}}{b_{1, n - 1}}, \quad \quad h_1^\prime = {x_1}{x_2} \cdots {x_{m - 3}} \in E_1^{s - 2 - r, t(r), * };\\ {h_2} = {x_1}{x_2} \cdots {x_{m - 3}}{b_{1, n}}b_{1, n - 1}^2, \quad \quad \quad h_2^\prime = {x_1}{x_2} \cdots {x_{m - 3}} \in E_1^{s - 3 - r, t(r), * };\\ {h_3} = {x_1}{x_2} \cdots {x_{m - 3}}{h_{1, n + 1}}b_{1, n - 1}^2, \quad \quad \;\;h_3^\prime = {x_1}{x_2} \cdots {x_{m - 3}} \in E_1^{s - 2 - r, t(r), * };\\ {h_4} = {x_1}{x_2} \cdots {x_{m - 2}}{h_{2, n}}{h_{1, n}}, \quad \quad \;\;\;\;\;h_4^\prime = {x_1}{x_2} \cdots {x_{m - 2}} \in E_1^{s + 1 - r, t(r), * };\\ {h_5} = {x_1}{x_2} \cdots {x_{m - 2}}{h_{2, n}}{b_{1, n - 1}}, \quad \quad \;\;\;h_5^\prime = {x_1}{x_2} \cdots {x_{m - 2}} \in E_1^{s - r, t(r), * };\\ {h_6} = {x_1}{x_2} \cdots {x_{m - 2}}{b_{2, n - 1}}{h_{1, n}}, \quad \quad \;\;\;h_6^\prime = {x_1}{x_2} \cdots {x_{m - 2}} \in E_1^{s - r, t(r), * };\\ {h_7} = {x_1}{x_2} \cdots {x_{m - 2}}{b_{2, n - 1}}{b_{1, n - 1}}, \quad \quad \;h_7^\prime = {x_1}{x_2} \cdots {x_{m - 2}} \in E_1^{s - 1 - r, t(r), * };\\ {h_8} = {x_1}{x_2} \cdots {x_{m - 3}}{h_{1, n + 1}}{h_{1, n}}{b_{1, n - 1}}, \quad \;h_8^\prime = {x_1}{x_2} \cdots {x_{m - 3}} \in E_1^{s - 1 - r, t(r), * }, \end{array}$

where $t(r)=sp^{2}q+(s-1)pq+(s-1)q+s-2-r$.

For $h_{1}^{\prime}$, $s-2-r< \sum\limits_{i=1}^{m-3}c_{i, 2}=s$, by Proposition $2.3$, we get that $h_{1}^\prime$ is impossible to exist. For the same reason, $h_{i}^\prime (i=2, 3, \cdots 8)$ are impossible to exist. So we have $E_1^{s+3-r, t, \ast}=0, $ that is ${\rm{Ext}}_A^{s+3-r, t}(\mathbb{Z}_p, \mathbb{Z}_p)=0$.

Proposition 3.3  Let $p\geqslant7$, $tq=p^{n+1}q+2p^{n}q$, $n>3$, then

(1)

${\rm{Ext}}_A^{4, {tq+rq+u}}(\mathbb{Z}_p, \mathbb{Z}_p)=0 (r=2, 3, 4, u=-1, 0 {\rm or} r=3, 4, u=1);\\ {\rm{Ext}}_A^{4, {tq+q}}(\mathbb{Z}_p, \mathbb{Z}_p)\cong\mathbb{Z}_p\{h_{0}l_{n}\}; {\rm{Ext}}_A^{4, {tq}}(\mathbb{Z}_p, \mathbb{Z}_p)=0;\\ {\rm{Ext}}_A^{4, {tq+2q+1}}(\mathbb{Z}_p, \mathbb{Z}_p)\cong\mathbb{Z}_p\{\tilde\alpha_2g_n\}, a_0^2g_n\neq 0.$

(2)

${\rm{Ext}}_A^{5, {tq+rq+1}}(\mathbb{Z}_p, \mathbb{Z}_p)=0 (r=1, 3, 4); {\rm{Ext}}_A^{5, {tq+rq}}(\mathbb{Z}_p, \mathbb{Z}_p)=0 (r=2, 3);\\ {\rm{Ext}}_A^{5, {tq+2q+1}}(\mathbb{Z}_p, \mathbb{Z}_p)\cong\mathbb{Z}_p\{\tilde\alpha_2\}; {\rm{Ext}}_A^{5, {tq+2}}(\mathbb{Z}_p, \mathbb{Z}_p)\cong\mathbb{Z}_p\{a_{0}^{2}l_{n}\};\\ {\rm{Ext}}_A^{5, {tq+1}}(\mathbb{Z}_p, \mathbb{Z}_p)=0.$

Proof  (1) Consider the second degrees $({\rm{mod}} p^{n+1}q)$ of the generators in the $E_{1}$-terms of the MSS, where $0\leqslant j\leqslant n+1$,

${\rm{deg}} h_{s, j} =(p^{s+j-1}+\cdots+p^{j})q ({\rm{mod}} p^{n+1}q), 0\leqslant j<s+j-1<n+1, \\ \quad \quad \quad =(p^{n}+\cdots+p^{j})q ({\rm{mod}} p^{n+1}q), 0\leqslant j<s+j-1=n+1;\\ {\rm{deg}} b_{s, j-1} =(p^{s+j-1}+\cdots+p^{j})q ({\rm{mod}} p^{n+1}q), 1\leqslant j<s+j-1<n+1, \\ \quad \quad \quad =(p^{n}+\cdots+p^{j})q ({\rm{mod}} p^{n+1}q), 1\leqslant j<s+j-1=n+1;\\ {\rm{deg}} a_{j+1} =(p^{j}+\cdots+1)q+1 ({\rm{mod}} p^{n+1}q), 0\leqslant j<n+1, \\ \quad \quad \quad =(p^{n}+\cdots+1)q+1 ({\rm{mod}} p^{n+1}q), j=n+1.$

For the second degree $k=tq+rq+u (0\leqslant r\leqslant 4, -1\leqslant u\leqslant2)=2p^{n}q+rq+u ({\rm{mod}} p^{n+1}q)$, and excluding the factor which has second degree $\geqslant tq+pq$, we can get that the possibility of the factor of the generators in $E_{1}^{w, tq+rq+u, \ast}$ $ (4\leqslant w\leqslant 5)$ are $a_{0}, a_{1}, h_{1, 0}, h_{1, n+1}, h_{1, n}, h_{2, n}, $ $b_{1, n}, $ $b_{1, n-1}$ and $b_{2, n-1}$. Thus from the degree we know that

$E_1^{4, {tq+rq+u, \ast}}=0 (r=3, 4, u=1);\\ E_1^{4, {tq+rq+u, \ast}}=0 (r=2, 3, 4, u=-1, 0);\\ E_1^{4, {tq, \ast}}\cong\mathbb{Z}_p\{b_{1, n-1}b_{2, n-1}\};\\ E_1^{4, {tq+q, \ast}}\cong\mathbb{Z}_p\{h_{1, 0}h_{1, n}b_{2, n-1}, h_{2, n}b_{1, n-1}h_{1, 0}\};\\ E_1^{4, {tq+2q+1, \ast}}\cong\mathbb{Z}_p\{a_{1}h_{1, 0}h_{1, n}h_{2, n}\}.$

In the MSS, note that $d_{r}(xy)=d_{r}(x)y+(-1)^{s}xd_{r}(y)$ $(x\in E_1^{s, t, \ast}, y\in E_1^{s^\prime, t^\prime, \ast})$. Since $d_{1}(b_{1, n-1}h_{2, n}h_{1, 0})\neq 0$, then $E_{r}^{4, tq+q}=\mathbb{Z}_p\{b_{2, n-1}h_{1, n}h_{1, 0}\}$ $(r\geqslant 2)$. Moreover, $h_{2, n}h_{1, n}h_{1, 0}$ is permanent cycle in the MSS which converges to $h_{0}g_{n}\in{\rm{Ext}}_A^{3, \ast}(\mathbb{Z}_p, \mathbb{Z}_p)$, then $d_{r}(E_{r}^{3, tq+q, \ast})=0$ for $r\geqslant 1$, so that $b_{2, n-1}h_{1, n}h_{1, 0}$ is not $d_{r}$-boundary and it converges nontrivially to $h_{0}l_{n}$.

In addition, we say that ${\rm{Ext}}_A^{4, {tq}}(\mathbb{Z}_p, \mathbb{Z}_p)=0$, since $E_1^{4, tq, \ast}\cong\mathbb{Z}_p\{b_{1, n-1}$ $b_{2, n-1}\}$, where $b_{1, n-1}$ converges to $b_{n-1}$, while in the ${\rm{Ext}}_A^{2, {\ast}}(\mathbb{Z}_p, \mathbb{Z}_p)$, there is no element in relation to $b_{2, n-1}\in E_1^{2, p^{_n+1}q+p^{n}q, \ast}$.

(2) Similarly, due to the reason of the degree, we can get the following results

$E_1^{5, {tq+q+1, \ast}}\cong\mathbb{Z}_p\{a_{1}b_{1, n-1}b_{2, n-1}, a_{0}h_{1, 0}h_{2, n}b_{1, n-1}, a_{0}h_{1, 0}b_{2, n-1}h_{1, n}\};\\ E_1^{5, {tq+rq+1, \ast}}=0 (r=3, 4);\\ E_1^{5, {tq+rq, \ast}}=0 (r=2, 3);\\ E_1^{5, {tq+2q+1, \ast}}\cong\mathbb{Z}_p\{a_{1}h_{1, 0}b_{2, n-1}h_{1, n}, h_{2, n}b_{1, n-1}h_{1, 0}a_{1}\};\\ E_1^{5, {tq+2, \ast}}\cong\mathbb{Z}_p\{a_0^{2}h_{1, n}b_{2, n-1}, a_{0}^{2}h_{2, n}b_{1, n-1}\};\\ E_1^{5, {tq+1, \ast}}\cong\mathbb{Z}_p\{a_{0}b_{1, n-1}b_{2, n-1}, a_{0}h_{1, n-1}h_{1, n}h_{1, n+1}\}.$

The generators of $E_1^{5, {tq+q+1, \ast}}$ in the MSS all die, this is because that

$d_{1}(a_{1}b_{1, n-1}b_{2, n-1}) =-a_{0}h_{1, 0}b_{1, n-1}b_{2, n-1}\neq 0, \\ a_{0}h_{1, 0}h_{2, n}b_{1, n-1} =-d_{1}(a_{1} h_{2, n}b_{1, n-1}) $

and

$a_{0}h_{1, 0}b_{2, n-1}h_{1, n}=-d_{1}(a_{1}h_{2, n-1}h_{1, n}). $

So we have ${\rm{Ext}}_A^{5, {tq+q+1}}(\mathbb{Z}_p, \mathbb{Z}_p)=0$. In addition, with a similar proof of (1), we know that $d_{r}E_r^{4, tq+2, \ast}=0$. So the generator of $E_1^{5, {tq+2, \ast}}$ in the MSS converges to $a_{0}^{2}l_{n}$.

Since $d_{1}(h_{2, n}b_{1, n-1}h_{1, 0}a_{1})\neq 0$, then $E_{r}^{5, tq+2q+1}=\mathbb{Z}_p\{b_{2, n-1}h_{1, n}h_{1, 0}a_{1}\}$ for $r\geqslant 2$. Moreover, $h_{2, n}h_{1, n}h_{1, 0}a_{1}$ is a permanent cycle in the MSS which converges to $\tilde{\alpha}_2g_{n}$($a_{1}h_{1, 0}$ is a permanent cycle in the MSS and converges to $\widetilde{\alpha}_{2}\in{\rm{Ext}}_A^{\ast, {\ast}}(\mathbb{Z}_p, \mathbb{Z}_p)$), then $d_{r}(E_{r}^{4, tq+2q+1})=0$ for $r\geqslant 1$. Thus $b_{2, n-1}h_{1, n}h_{1, 0}a_{1}$ is not $d_{r}$-boundary and converges nontrivially to $\widetilde{\alpha}_{2}l_{n}$.

Since $a_{0}$, $b_{1, n-1}$, $h_{1, n}$ and $h_{1, n+1}$ are all permanent cycles in the MSS and converge to $a_{0}$, $b_{n-1}$, $h_{n}$ and $h_{n+1}$, respectively, it is easy to get that $a_{0}b_{1, n-1}h_{1, n}h_{1, n+1}$ is a permanent cycle in the MSS and converges to $a_{0}b_{n-1}h_{n}h_{n+1}$ which equals $0\in {\rm{Ext}}_A^{5, {tq+1}}(\mathbb{Z}_p, \mathbb{Z}_p)$ by $h_{n}h_{n+1}=0$. Furthermore, we have $d_{2p-1}(b_{2, n-1})=b_{1, n}h_{1, n}-b_{1, n-1}h_{1, n+1}$ from [10], then $d_{2p-1}(a_{0}b_{2, n-1}b_{1, n-1})\neq 0$ and so ${\rm{Ext}}_A^{5, {tq+1}}$ $(\mathbb{Z}_p, \mathbb{Z}_p)=0$.

Theorem 3.4  Let $p\geqslant7$, $n>3$, then

$h_{0}g_{n}\in{\rm{Ext}}_A^{3, {p^{n+1}q+2p^{n}q+q}}(\mathbb{Z}_p, \mathbb{Z}_p) $

is a permanent cycle in the ASS, and converges to a nontrivial element in $\pi_{p^{n+1}q+2p^{n}q+q-3}S.$

Proof  From [[20], Theorem 1.1], there is a nontrivial differential $d_{2}(g_{n})=a_{0}l_{n} (n\geqslant1)$ in the ASS, the elements $g_{n}$ and $l_{n}$ are called a pair of $a_{0}$-related elements. The condition of Theorem A in [7] can be established by the $\mathbb{Z}_p$-bases of ${\rm{Ext}}_A^{s, \ast}(\mathbb{Z}_p, \mathbb{Z}_p)$ $(s\leqslant 3)$ in [10] and Proposition 3.3 in the above. Furthermore, we have $\kappa\cdot (\alpha_{1})_{L}=(1_{E_{4}}\wedge p)f$ with $f\in [\sum^{tq+q}L, E_{4}]$ (see [7], 9.2.34), then $(1_{E_{4}}\wedge i)\kappa\cdot(\alpha_{1})_{L}=0$. Thus

$(1_{E_{4}}\wedge 1_{L}\wedge i)(\kappa\wedge 1_{L})\phi=(1_{E_{4}}\wedge 1_{L}\wedge i)(\kappa\wedge1_{L})((\alpha_{1})_{L}\wedge 1_{L})\widetilde{i}^{\prime\prime}=0, $

where $\widetilde{i}^{\prime\prime}\in\pi_{q}L\wedge L$ such that $((\alpha_{1})_{L}\wedge 1_{L})\widetilde{i}^{\prime\prime}=\phi$. It can be easily proved that $(\kappa\wedge1_{L})\phi=(\overline{c}_{3}\wedge1_{L})\sigma\phi, $ where $\sigma\phi\in\pi_{tq+2q}(KG_{3}\wedge L)$ is a $d_{1}$-cycle which represents $(\phi)_{\ast}(\sigma)\in {\rm{Ext}}_A^{3, tq+2q}(H^{\ast}L, \mathbb{Z}_p)$. Thus

$(\overline{c}_{3}\wedge 1_{L\wedge M})(1_{KG_{3}}\wedge i)\sigma\phi=0.$

So we can get that $(1_{L}\wedge i)_{\ast}\phi_{\ast}(g_{n})\in{\rm{Ext}}_A^{3, p^{n+1}q+2p^{n}q+q}(H^{\ast}L\wedge M, \mathbb{Z}_p)$ is a permanent cycle in the ASS. Then Theorem $3.4$ will be concluded by Theorem C in [7], here $\phi\in [\sum^{2q-1}S, L]$, $\kappa\in\pi_{tq+1}E_{4}$.

The Proof of Theorem 1.1  From Theorem 3.4, $h_{0}g_{n}\in {\rm{Ext}}_A^{3, {p^{n+1}q+2p^{n}q+q}}(\mathbb{Z}_p, \mathbb{Z}_p)$ is a permanent cycle in the ASS and converges to a nontrivial element $\varphi\in\pi_{p^{n+1}q+2p^{n}q+q-3}S$ for $n>3$.

Consider the following composition of mappings

$\tilde{f}{{\Sigma }^{{{p}^{n+1}}+2{{p}^{n}}q+q-3}}S\xrightarrow{\varphi }S\text{ }\xrightarrow{{{i}^{\prime \prime }}{{i}^{\prime }}i}V(2)\xrightarrow{{{\gamma }^{s}}}$

${{\Sigma }^{-s({{p}^{2}}+p+1)q}}V(2)\text{ }\xrightarrow{j{{j}^{\prime }}{{j}^{\prime \prime }}}{{\Sigma }^{-s({{p}^{2}}+p+1)q+(p+1)q+q}}S, $

because $\varphi$ is represented by $h_{0}g_{n}$ in the ASS, then the above $\widetilde{f}$ is represented by

$\widetilde{g}=(jj^{\prime}j^{\prime\prime})_{\ast}(\gamma^{s})_{\ast}(i^{\prime\prime}i^{\prime}i)_{\ast} (h_{0}g_{n})=(jj^{\prime}j^{\prime\prime}\gamma^{s}i^{\prime\prime}i^{\prime}i)_{\ast}(h_{0}g_{n})$

in the ASS. Furthermore, we know that $\gamma_{s}=jj^{'}j^{''}\gamma^{s}i^{\prime\prime}i^{\prime}i\in\pi_{\ast}S $ is represented by $\widetilde{\gamma}_{s}$ in the ASS. By using the Yoneda products, we know that the composition

$ \text{Ext}_{A}^{\text{0}, \text{0}}({{\mathbb{Z}}_{p}}, {{\mathbb{Z}}_{p}})\xrightarrow{{{({{i}^{\prime \prime }}{{i}^{\prime }}i)}_{\text{*}}}}\text{Ext}_{A}^{\text{0}, \text{0}}({{H}^{\text{*}}}V(\text{2}), {{\mathbb{Z}}_{p}})$

$\xrightarrow{{{(j{{j}^{\prime }}{{j}^{\prime \prime }})}_{*}}{{({{\gamma }^{s}})}_{*}}}\text{Ext}_{A}^{s, (s{{p}^{2}}+(s-1)p+(s-2))q+s-3}({{\mathbb{Z}}_{p}}, {{\mathbb{Z}}_{p}})$

is a multiplication (up to nonzero scalar) by

${\widetilde{\gamma}_{s}}\in{\rm{Ext}}_A^{s, sp^{2}q+(s-1)pq+(s-2)q+s-3}(\mathbb{Z}_p, \mathbb{Z}_p).$

Hence, the composite map $\widetilde{f}$ is represented (up to nonzero scalar) by

$\widetilde{\gamma}_{s}h_{0}g_{n}\in {\rm{Ext}}_A^{s+3, p^{n+1}q+2p^{n}q+sp^{2}q+(s-1)pq+(s-1)q+s-3}({\mathbb{Z}_p}, {\mathbb{Z}_p}) $

in the ASS.

From Proposition 3.1, we see that $\widetilde{\gamma}_{s}h_{0}g_{n}\neq 0$. Moreover, from Proposition 3.2, it follows that $\widetilde{\gamma}_{s}h_{0}g_{n}$ can not be hit by any differential in the ASS. Thus $\widetilde{\gamma}_{s}h_{0}g_{n}$ survives nontrivially to a homotopy element in $\pi_{\ast}S$.