To determine the stable homotopy groups of spheres $\pi_*(S)$ is one of the central problems in homotopy theory. One of the main tools to reach it is the Adams spectral sequence(ASS).

Let $A$ be the mod $p$ Steenrod algebra, and $S$ be the sphere spectrum localized at an odd prime $p$. For connected finite type spectra $X$, $Y$, there exists the ASS $\{E_r^{s,t}, d_r\}$ such that

(1) $d_r: E_r^{s,t}\rightarrow E_r^{s+r,t+r-1}$ is the differential;

(2) $E_2^{s,t}\cong \text{Ext}_A^{s,t}(H^*(X),H^*(Y))\Rightarrow [\Sigma^{t-s}Y,X]_p$, where $E_2^{s,t}$ is the cohomology of $A$. When $X$ is sphere spectrum $S$, Toda-Smith spectrum $V(n)(n=1,2,3)$, respectively, $(\pi_{t-s}(X))_p$ is the stable homotopy groups of $S$, $V(n)$. So, for computing the stable homotopy groups of spheres with the ASS, we must compute the $E_2$-term of the ASS, $\text{Ext}_A^{*,*}(\mathbb{Z}_p, \mathbb{Z}_p)$.

From [1], $\text{Ext}_A^{1,*}(\mathbb{Z}_p, \mathbb{Z}_p)$ has the $\mathbb{Z}_p$-base consisting of

$a_0\in \text{Ext}_A^{1,1}(\mathbb{Z}_p, \mathbb{Z}_p), h_i\in\text{Ext}_A^{1,p^i q}(\mathbb{Z}_p, \mathbb{Z}_p)$

for all $i\geqslant0$ and $\text{Ext}_A^{2,*}(\mathbb{Z}_p, \mathbb{Z}_p)$ has the $\mathbb{Z}_p$-base consisting of $\alpha_2$, $a_0^2$, $a_0h_i(i>0)$, $g_i(i\geqslant0)$, $k_i(i\geqslant0)$, $b_i(i\geqslant0)$ and $h_ih_j(j\geqslant i+2,i\geqslant0)$ whose internal degrees are $2q+1$, 2, $p^iq+1$, $p^{i+1}q+2p^iq$, $2p^{i+1}q+p^iq$, $p^{i+1}q$ and $p^i q+p^j q$, respectively. From [2,P.110, Table 8.1], the $\mathbb{Z}_p$-base of $\text{Ext}_A^{3,*}(\mathbb{Z}_p, \mathbb{Z}_p)$ has been completely listed and there is a generator $\stackrel {\sim}{\gamma_t}\in \text{Ext}_A^{t,tp^2q+(t-1)pq+(t-2)q+t-3}(\mathbb{Z}_p, \mathbb{Z}_p)$ which is described in [3].

Our main theorems of this paper are as follows.

Theorem 1.1  Let $p\geqslant 11$, $3\leqslant t<p-5$, then

$ \stackrel{\thicksim}{\gamma_t} h_0 b_1^5\in \text{Ext}_A^{t+11,(t+5)p^2q+(t-1)(p+1)q+t-3}(\mathbb{Z}_p,\mathbb{Z}_p) $

is a permanent cycle in the ASS and converges to a non-trivial element in $\pi_*(S)$.

Based on a new homotopy element in $\pi_*(V(2))$, the above homotopy element in $\pi_*(S)$ will be constructed .

For the reader's convenience, let us firstly give some preliminaries on Toda-Smith spectrum $V(n)$.

The $\mathbb{Z}_p$ cohomology group of Toda-Smith spectrum $V(n)$ is $H^*V(n)\cong E[Q_0,Q_1,$ $\cdots,Q_n]\\ \cong Q(2^{n+1})$, where $Q_i(i\geqslant 0)$ is the Milnor's elements of Steenrod algebra $A$, and E[ ] is the exterior algebra. From [4], when $n=1,2,3$ and $p>2n$, we know that $V(n)$ is realized, and there exists a cofibre sequence $(V(-1)=S)$:

$ \Sigma^{2(p^n-1)}V(n-1) \xrightarrow{{{\alpha }_{n}}} V(n-1) \xrightarrow{{{i}_{n}}} V(n) \xrightarrow{{{j}_{n}}} \Sigma^{2p^n-1}V(n-1), $

where $\alpha_n(n=0,1,2,3)$ are $p$, $\alpha$, $\beta$, $\gamma$, respectively. The cofibre sequence can induce a short exact sequence of $\mathbb{Z}_p$ cohomology groups. Thus, we get the following long exact sequence of Ext groups:

$ \cdots\xrightarrow{{(j_n)_*}} \text{Ext}_A^{s-1,t-(2p^n-1)}(H^*V(n-1),\mathbb{Z}_p) \xrightarrow{{(\alpha_n)_*}} \text{Ext}_A^{s,t}(H^*V(n-1),\mathbb{Z}_p) $

$ \xrightarrow{{(i_n)_*}} \text{Ext}_A^{s,t}(H^*V(n),\mathbb{Z}_p) \xrightarrow{{(j_n)_*}} \text{Ext}_A^{s,t-(2p^n-1)}(H^*V(n-1),\mathbb{Z}_p)\xrightarrow{{(\alpha_n)_*}} \cdots. $

The following theorem is a key step to prove Theorem 1.1.

Theorem 1.2 Let $p\geqslant 11$, then $h_0b_1^5\in \text{Ext}_A^{11,5p^2q+q}(H^*V(2),\mathbb{Z}_p)$ is a permanent cycle in the ASS and converges to a non-trivial element in $\pi_*V(2)$.

It is very difficult to determine the stable homotopy groups of spheres. So far, not so many nontrivial elements in the stable homotopy groups of spheres were detected. See, for example [1,5,6].

The detection of the element $\stackrel{\sim}{\gamma_t}h_0b_1^5$ is parallel to that of the element $\stackrel{\sim}{\gamma_t}h_0b_1^2$ given in [7]. Actually, our results are more complicated, especially to Proposition 3.3 and Proposition 3.4.

This paper is organized as follows: after giving some preliminaries on the May spectral sequence (MSS) in Section 2, the proofs of the main theorems will be given in Section 3.

The most successful tool for computing $\text{Ext}_A^{*,*}(\mathbb{Z}_p,\mathbb{Z}_p)$ is the MSS. From [8,Theorem 3.2.5] {Ravenel}, there exists the MSS $\{E_r^{s,t,*},d_r\}$ which converges to $\text{Ext}_A^{s,t}(\mathbb{Z}_p,\mathbb{Z}_p)$. The $E_1$-term and differential of the MSS are

$E_1^{*,*,*}=E(h_{i,j} |i>0,j\geqslant 0)\otimes P(b_{i,j} |i>0,j\geqslant 0)\otimes P(a_i |i\geqslant 0),$

$d_r:E_r^{s,t,u}\rightarrow E_r^{s+1,t,u-r}, r\geqslant 1,$

where $E$ is the exterior algebra, $P$ is the polynomial algebra and

$h_{i,j}\in E_1^{1,2(p^i -1)p^j,2i-1},\hspace{0.1cm} b_{i,j}\in E_1^{2,2(p^i-1)p^{j+1}, p(2i-1)},\hspace{0.1cm} a_i\in E_1^{1,2p^i-1,2i+1}.$

Lemma 2.1 (see [9]) Let $t=(c_np^n+c_{n-1}p^{n-1}+\cdots +c_1p+c_0)q+c_{-1}$, $c_i\in\mathbb{Z}$, and $p-1\geqslant c_n\geqslant c_{n-1}\geqslant\cdots \geqslant c_1\geqslant c_0\geqslant c_{-1}\geqslant 0$, then the number of $h_{n+1-i,i}$ in the generator of $E_1^{c_n,t,*}$ will be $(c_i-c_{i-1})$ $(0\leqslant i\leqslant n)$.

Corollary 2.2 (see [9]) If $p>a\geqslant b\geqslant c\geqslant d\geqslant 0$, then the number of $h_{1,2}$, $h_{2,1}$ and $h_{3,0}$ in the generator of $E_1^{a,ap^2q+bpq+cq+d,*}$ will be $(a-b)$, $(b-c)$ and $(c-d)$, respectively.

Corollary 2.3 (see [9]) Let $t\geqslant 3$, then $E_1^{t,tp^2 q+(t-1)pq+(t-2)q+t-3}=\mathbb{Z}_p\{h_{2,1}h_{1,2}h_{3,0}a_3^{t-3}\}$.

Lemma 2.4 (see [9]) Let

$t=(c_np^n+c_{n-1}p^{n-1}+\cdots +c_1 p+c_0)q+c_{-1},$

$c_i\in \mathbb{Z}_p (-1\leqslant i\leqslant n)$, for $c_i<c_{i-1}$, $0\leqslant i\leqslant n-1$, then $E_1^{c_n,t,*}=0$.

Lemma 2.5 (see [9]) Let $u>0$, $p>c_2,c_1,c_0,c_{-1}\geqslant 0$ and $c_2-c_{-1}\geqslant 4$, there don't exist $u$ factors in the generator of $E_1^{u,c_2 p^2 q+c_1 pq+c_0 q+c_{-1}}$.

Let $P$ be the subalgebra of $A$ generated by the reduced power operations $P^i(i>0),$ then we have the following results.

Proposition 3.1 (see [9]) $\text{Ext}_A^{s,t}(H^*V(3),\mathbb{Z}_p)\cong \text{Ext}_P^{s,t}(\mathbb{Z}_p,\mathbb{Z}_p)$, $t-s<2p^4-1$.

Corollary 3.2 Let $s\geqslant 2$, then

$\text{Ext}_A^{s+11,5p^2q+q+s-1}(H^*V(3),\mathbb{Z}_p)\cong \text{Ext}_P^{s+11,5p^2q+q+s-1}(\mathbb{Z}_p,\mathbb{Z}_p).$

Proposition 3.3 Let $3\leqslant t<p-5, p\geqslant 11$, then

$0\neq \stackrel{\thicksim}{\gamma_t} h_0 b_1^5 \in \text{Ext}_A^{t+11,(t+5)p^2q+(t-1)(p+1)q+t-3} (\mathbb{Z}_p ,\mathbb{Z}_p).$

The generators of $E_1^{s,t,u}$ and their first, second degrees satisfying $t<p^3 q$ are listed in Table 1.

To compare the degrees, $h_0$, $b_1\in \text{Ext}_A^{*,*}(\mathbb{Z}_p,\mathbb{Z}_p)$ are represented by $h_{1,0}\in E_1^{1,q,*}$, $b_{1,1}\in E_1^{2,p^2 q,*}$ in the MSS. From Corollary 2.3, we conclude that $\stackrel{\thicksim}{\gamma_t}\in \text{Ext}_A^{t,tp^2 q+(t-1)pq+(t-2)q+t-3}$ $(\mathbb{Z}_p,\mathbb{Z}_p)$ is represented by $h_{2,1}h_{1,2}h_{3,0}a_3^{t-3}\in E_1^{t,tp^2 q+(t-1)pq+(t-2)q+t-3,*}(t\geqslant 3)$ in the MSS.

Thus, $\stackrel{\thicksim}{\gamma_t} h_0 b_1^5$ is represented by $h_{1,0}b_{1,1}^5h_{2,1}h_{1,2}h_{3,0}a_3^{t-3}\in E_1^{t+11,(t+5)p^2 q+(t-1)pq+(t-1)q+t-3,*}$ in the MSS. If we want to prove that $0\neq \stackrel{\thicksim}{\gamma_t} h_0 b_1^5 \in \text{Ext}_A^{t+11,(t+5)p^2q+(t-1)(p+1)q+t-3}(\mathbb{Z}_p ,\mathbb{Z}_p), $ we must prove that $E_1^{t+10,(t+5)p^2q+(t-1)(p+1)q+t-3,*}=0$. For any $\sigma\in E_1^{t+10,(t+5)p^2q+(t-1)(p+1)q+t-3,*}$, we have the following discussions.

Case 1 When $t\geqslant 6$, from Lemma 2.5, the number of the factors in $\sigma$ will be $t+9$, $t+8$, $t+7$, $t+6$ or $t+5$.

Subcase 1.1 If $\sigma$ has $t+9$ factors, there exists a factor $b_{i,j}(b_{1,0},b_{1,1},b_{2,0})$. Due to the commutativity, the possible forms will be $\sigma=\sigma_{1.1} b_{1,0}$, $\sigma=\sigma_{1.2} b_{1,1}$, $\sigma=\sigma_{1.3} b_{2,0}$, where

$\begin{eqnarray*} &&\sigma_{1.1}\in E_1^{t+8,(t+5)p^2q+(t-2)pq+(t-1)q+t-3,*}, \sigma_{1.2}\in E_1^{t+8,(t+4)p^2q+(t-1)pq+(t-1)q+t-3,*},\nonumber\\ &&\sigma_{1.3}\in E_1^{t+8,(t+4)p^2q+(t-2)pq+(t-1)q+t-3,*}. \end{eqnarray*}$

By Lemma 2.5, the number of the factors in $\sigma_{1.1}$ is $t+7, t+6$ or $t+5,$ thus the number of the factors in $\sigma$ will be $t+8, t+7$ or $ t+6$. It is in contradiction with that $\sigma$ has $t+9$ factors, so $\sigma_{1.1}=0$. Similarly, we conclude that $\sigma_{1.2}=0$, $\sigma_{1.3}=0$, so $\sigma=0$.

Subcase 1.2 If $\sigma$ has $t+8$ factors, there exist two factors $b_{i,j}(b_{1,0},b_{1,1},b_{2,0})$. Due to the commutativity, the possible forms will be $\sigma=\sigma_{2.1} b_{1,0}^2$, $\sigma=\sigma_{2.2} b_{1,1}^2$, $\sigma=\sigma_{2.3} b_{2,0}^2$, $\sigma=\sigma_{2.4} b_{1,0}b_{1,1}$, $\sigma=\sigma_{2.5} b_{1,0}b_{2,0}$, $\sigma=\sigma_{2.6} b_{1,1}b_{2,0}$, where

$\begin{eqnarray*} &&\sigma_{2.1}\in E_1^{t+6,(t+5)p^2q+(t-3)pq+(t-1)q+t-3,*}, \sigma_{2.2}\in E_1^{t+6,(t+3)p^2q+(t-1)pq+(t-1)q+t-3,*},\nonumber\\ &&\sigma_{2.3}\in E_1^{t+6,(t+3)p^2q+(t-3)pq+(t-1)q+t-3,*}, \sigma_{2.4}\in E_1^{t+6,(t+4)p^2q+(t-2)pq+(t-1)q+t-3,*},\nonumber\\ &&\sigma_{2.5}\in E_1^{t+6,(t+4)p^2q+(t-3)pq+(t-1)q+t-3,*}, \sigma_{2.6}\in E_1^{t+6,(t+3)p^2q+(t-2)pq+(t-1)q+t-3,*}. \end{eqnarray*}$

By the similar argument in Subcase1.1, we can get that $\sigma_{2.i}=0(i=1,2\cdots 6)$, thus $\sigma=0$.

Subcase 1.3 If $\sigma$ has $t+7$ factors, there exist three factors $b_{i,j}(b_{1,0},b_{1,1},b_{2,0})$. Due to the commutativity, the possible forms will be $\sigma=\sigma_{3.1} b_{1,0}^3$, $\sigma=\sigma_{3.2} b_{1,0}^2 b_{1,1}$, $\sigma=\sigma_{3.3} b_{1,0}^2 b_{2,0}$, $\sigma=\sigma_{3.4}b_{1,1}^3 $, $\sigma=\sigma_{3.5} b_{2,0}^3$, $\sigma=\sigma_{3.6} b_{1,1}^2 b_{2,0}$, $\sigma=\sigma_{3.7} b_{1,1}^2 b_{1,0}$, $\sigma=\sigma_{3.8} b_{2,0}^2 b_{1,0}$, $\sigma=\sigma_{3.9} b_{2,0}^2 b_{1,1}$, $\sigma=\sigma_{3.10} b_{1,0} b_{1,1} b_{2,0}$, where

$\begin{eqnarray*} &&\sigma_{3.1}\in E_1^{t+4,(t+5)p^2q+(t-4)pq+(t-1)q+t-3,*}, \sigma_{3.2}\in E_1^{t+4,(t+4)p^2q+(t-3)pq+(t-1)q+t-3,*}, \nonumber\\ &&\sigma_{3.3}\in E_1^{t+4,(t+4)p^2q+(t-4)pq+(t-1)q+t-3,*}, \sigma_{3.4}\in E_1^{t+4,(t+2)p^2q+(t-1)pq+(t-1)q+t-3,*},\nonumber\\ &&\sigma_{3.5}\in E_1^{t+4,(t+2)p^2q+(t-4)pq+(t-1)q+t-3,*}, \sigma_{3.6}\in E_1^{t+4,(t+2)p^2q+(t-2)pq+(t-1)q+t-3,*},\nonumber\\ &&\sigma_{3.7}\in E_1^{t+4,(t+3)p^2q+(t-2)pq+(t-1)q+t-3,*}, \sigma_{3.8}\in E_1^{t+4,(t+3)p^2q+(t-4)pq+(t-1)q+t-3,*},\nonumber\\ &&\sigma_{3.9}\in E_1^{t+4,(t+2)p^2q+(t-3)pq+(t-1)q+t-3,*}, \sigma_{3.10}\in E_1^{t+4,(t+3)p^2q+(t-3)pq+(t-1)q+t-3,*}. \end{eqnarray*}$

It is obvious that $\sigma_{3.1}=0$. By the similar argument in Subcase1.1, we can get that $\sigma_{3.i}=0$ $(i=4,5,\cdots, 10).$ From Lemma 2.4, note that $t-3<t-1$, $t-4<t-1$, thus the remainder are all zero. Therefore, we can get $\sigma=0$.

Subcase 1.4 If $\sigma$ has $t+6$ factors, there exist four factors $b_{i,j}(b_{1,0},b_{1,1},b_{2,0})$. Due to the commutativity, $\sigma=\sigma^\prime b_{1,0}^x b_{1,1}^yb_{2,0}^z$, where $x+y+z=4,x,y,z\geqslant 0$ and $\sigma^\prime\in E_1^{t+2,T},$ $T=(t+5-y-z)p^2q+(t-1-x-z)pq+(t-1)q+(t-3).$ If $x\geqslant 2,$ we have that $y+z<3$ and $t+5-y-z>t+2$. It is obvious that $\sigma^\prime =0$. Thus, the possible nontrivial forms will be $\sigma=\sigma_{4.1} b_{1,1}^4$, $\sigma=\sigma_{4.2} b_{2,0}^4$, $\sigma=\sigma_{4.3} b_{1,1}^3 b_{1,0}$, $\sigma=\sigma_{4.4} b_{1,1}^3 b_{2,0}$, $\sigma=\sigma_{4.5} b_{2,0}^3 b_{1,0}$, $\sigma=\sigma_{4.6} b_{2,0}^3 b_{1,1}$, $\sigma=\sigma_{4.7} b_{1,1}^2 b_{2,0}^2$, $\sigma=\sigma_{4.8} b_{1,0} b_{1,1}^2b_{2,0}$, $\sigma=\sigma_{4.9} b_{1,0} b_{1,1}b_{2,0}^2$, where the first degrees of $\sigma_{4.i}(i=1,2,3,\cdot\cdot\cdot,9)$ are all $t+2$ and the second degrees of them are listed in Table 2 ($M=t-1$, and $N=t-3$).

By the argument similar to Subcase 1.3, we get that $\sigma_{4.i}=0(i=1,2\cdots 9)$, thus $\sigma=0$.

Subcase 1.5 If $\sigma$ has $t+5$ factors, there exist five factors $b_{i,j}(b_{1,0},b_{1,1},b_{2,0})$. Due to the commutativity, the possible nontrivial forms will be $\sigma=\sigma_{5.1} b_{1,1}^3b_{2,0}^2$, $\sigma=\sigma_{5.2} b_{1,1}^2b_{2,0}^3$, $\sigma=\sigma_{5.3} b_{2,0}^4 b_{1,1}$, $\sigma=\sigma_{5.4} b_{1,1}^4 b_{2,0}$, $\sigma=\sigma_{5.5} b_{2,0}^5$, $\sigma=\sigma_{5.6} b_{1,1}^5$, where the first degrees of $\sigma_{5.i}(i=1,2,3,\cdot\cdot\cdot,6)$ are all $t$ and the second degrees of them are listed in Table 3 ($M=t-1$, and $N=t-3$).

Similarly to $\sigma_{3.2}$, we can get that $\sigma_{5.i}=0(i=1,2\cdots 5)$. As for $\sigma_{5.6}$, from the Corollary 2.2, there exist two factors $h_{3,0}$, so $\sigma_{5.6}=0$. Thus, we can get $\sigma=0$.

Case 2 When $t=5$, $E_1^{t+10,(t+5)p^2q+(t-1)(p+1)q+t-3,*}=E_1^{15,10p^2q+4pq+3q+q+2,*},$ the generator contains $(q+2)$ factors $a_i$. Therefore, the first degree $\geqslant q+2>15,$ it's a contradiction. So, we get $\sigma=0$. When $t=4$, $ t=3$, the proofs are the similar to $t=5$. Summarize the above Case 1 and Case 2, $E_1^{t+10,(t+5)p^2q+(t-1)(p+1)q+t-3,*}=0.$ That is

$0\neq \stackrel{\thicksim}{\gamma_t} h_0 b_1^5 \in \text{Ext}_A^{t+11,(t+5)p^2q+(t-1)(p+1)q+t-3} (\mathbb{Z}_p ,\mathbb{Z}_p)(3\leqslant t<p-5).$

Proposition 3.4 Let $r\geqslant 2$, $3\leqslant t<p-5$, then

$\text{Ext}_A^{t+11-r,(t+5)p^2q+(t-1)pq+(t-1)q+t-r-2,*}(\mathbb{Z}_p,\mathbb{Z}_p)=0.$

It is sufficient if we can show that $E_1^{t+11-r,(t+5)p^2q+(t-1)pq+(t-1)q+t-r-2,*}=0$.

Case 1 If $r>6$, $t+11-r<t+5$, so we have

$E_1^{t+11-r,(t+5)p^2q+(t-1)pq+(t-1)q+t-r-2,*}=0.$

Case 2 If $r=6$, then

$E_1^{t+11-r,(t+5)p^2q+(t-1)pq+(t-1)q+t-r-2,*}=E_1^{t+5,(t+5)p^2q+(t-1)pq+(t-1)q+t-8,*}.$

Subcase 2.1 When $t\geqslant 8$, from the Corollary 2.2, there exist six factors $h_{1,2}$, so $\sigma=0$.

Subcase 2.2 When $t=7$, $E_1^{t+5,(t+5)p^2q+(t-1)pq+(t-1)q+t-8,*}=E_1^{12,12p^2q+6pq+5q+q-1,*}$. The generator contains $(q-1)$ factors $a_i$. Therefore, the first degree$\geqslant q-1>12$, it's a contradiction. So, the generator is impossible to exist.

Subcase 2.3 When $3\leqslant t\leqslant 6$, by the similar argument in Subcase 2.2, the generator is impossible to exist.

Case 3 If $r=5$, then

$E_1^{t+11-r,(t+5)p^2q+(t-1)pq+(t-1)q+t-r-2,*}=E_1^{t+6,(t+5)p^2q+(t-1)pq+(t-1)q+t-7,*}.$

Subcase 3.1 When $t\geqslant 7$, from the Lemma 2.5, we know that the generator contains $t+5$ factors, one of which must be the factor $b_{i,j}$. Thus, the possible nontrivial forms will be $\sigma=\sigma_{3.1} b_{1,1}$, $\sigma=\sigma_{3.2} b_{2,0}$, where $\sigma_{3.1}\in E_1^{t+4,(t+4)p^2q+(t-1)pq+(t-1)q+t-7,*}$, $\sigma_{3.2}\in E_1^{t+4,(t+4)p^2q+(t-2)pq+(t-1)q+t-7,*}$. By the similar argument in Subcase2.1, we can get that $\sigma_{3.1}=0$, $\sigma_{3.2}=0$.

Subcase 3.2 When $3\leqslant t\leqslant 6$, by the similar argument in Subcase 2.2, the generator is impossible to exist.

Case 4 If $r=4$, then

$E_1^{t+11-r,(t+5)p^2q+(t-1)pq+(t-1)q+t-r-2,*}=E_1^{t+7,(t+5)p^2q+(t-1)pq+(t-1)q+t-6,*}.$

Subcase 4.1 When $t\geqslant 6$, from Lemma 2.5, we know that the number of the factors in $\sigma$ will be $t+5$ or $t+6$.

Subcase 4.1.1 If $\sigma$ contains $t+6$ factors, then there exists a factor $b_{i,j}(b_{1,0},b_{1,1},$ $b_{2,0})$. Due to the commutativity, the possible nontrivial forms will be $\sigma=\sigma_{4.1} b_{1,0}, \sigma=\sigma_{4.2} b_{1,1},$ $ \sigma=\sigma_{4.3}b_{2,0}, $ where

$\begin{eqnarray*} &&\sigma_{4.1}\in E_1^{t+5,(t+5)p^2q+(t-2)pq+(t-1)q+t-6,*}, \sigma_{4.2}\in E_1^{t+5,(t+4)p^2q+(t-1)pq+(t-1)q+t-6,*},\nonumber\\ &&\sigma_{4.3}\in E_1^{t+5,(t+4)p^2q+(t-2)pq+(t-1)q+t-6,*}. \end{eqnarray*}$

By the similar argument in Subcase 2.1, we know that $\sigma_{4.1}=0$. As for $\sigma_{4.2}$, from Lemma 2.5, $\sigma_{4.2}$ must contain $t+4$ factors, thus $\sigma=\sigma_{4.2}b_{1,1}$ contains $t+5$ factors. It is a contradiction with that $\sigma$ contains $t+6$ factors, then $\sigma_{4.2}=0$. Similarly, we can get $\sigma_{4.3}=0$.

Subcase 4.1.2 If $\sigma$ contains $t+5$ factors, then there exist two factors $b_{i,j}(b_{1,0},b_{1,1},$ $b_{2,0})$. Due to the commutativity, the possible nontrivial forms will be $\sigma=\sigma_{4.4} b_{1,1}^2, $ $ \sigma=\sigma_{4.5}b_{2,0}^2,$ $\sigma=\sigma_{4.6} b_{1,1}b_{2,0}, $ where

$\begin{eqnarray*} &&\sigma_{4.4}\in E_1^{t+3,(t+3)p^2q+(t-1)pq+(t-1)q+t-6,*}, \sigma_{4.5}\in E_1^{t+3,(t+3)p^2q+(t-3)pq+(t-1)q+t-6,*}, \nonumber\\ &&\sigma_{4.6}\in E_1^{t+3,(t+3)p^2q+(t-2)pq+(t-1)q+t-6,*}. \end{eqnarray*}$

By the similar argument in Subcase 2.1, we can get $\sigma_{4.4}=0$. As for $\sigma_{4.5}$, from the Lemma 2.4 and $t-3<t-1$, so $\sigma_{4.5}=0$. Similarly, we can get $\sigma_{4.6}=0$.

Subcase 4.2 When $3\leqslant t\leqslant 5$, by the similar argument in Subcase 2.2, we know that the generator is impossible to exist. Thus, we have $\sigma=0$.

Case 5 If $r=3$, then

$E_1^{t+11-r,(t+5)p^2q+(t-1)pq+(t-1)q+t-r-2,*}=E_1^{t+8,(t+5)p^2q+(t-1)pq+(t-1)q+t-5,*}.$

Subcase 5.1 When $t\geqslant 5$, from Lemma 2.5, the number of the factors in $\sigma$ will be $t+5$, $t+6$ or $t+7$.

Subcase 5.1.1 If $\sigma$ contains $t+7$ factors, then there exists a factor $b_{i,j}(b_{1,0},b_{1,1},$ $b_{2,0})$. Due to the commutativity, the possible nontrivial forms will be $\sigma=\sigma_{5.1} b_{1,0}, \sigma=\sigma_{5.2} b_{1,1},$ $\sigma=\sigma_{5.3}b_{2,0}, $ where

$\begin{eqnarray*} &&\sigma_{5.1}\in E_1^{t+6,(t+5)p^2q+(t-2)pq+(t-1)q+t-5,*}, \sigma_{5.2}\in E_1^{t+6,(t+4)p^2q+(t-1)pq+(t-1)q+t-5,*},\nonumber\\ &&\sigma_{5.3}\in E_1^{t+6,(t+4)p^2q+(t-2)pq+(t-1)q+t-5,*}. \end{eqnarray*}$

Similarly to $\sigma_{4.2}$, we can get $\sigma_{5.i}=0(i=1,2,3)$.

Subcase 5.1.2 If $\sigma$ contains $t+6$ factors, then there exist two factors $b_{i,j}(b_{1,0},b_{1,1},b_{2,0})$. Due to the commutativity, the possible nontrivial forms will be $\sigma=\sigma_{5.4} b_{1,1}^2$, $\sigma=\sigma_{5.5}b_{2,0}^2$, $\sigma=\sigma_{5.6} b_{1,0} b_{1,1} $, $\sigma=\sigma_{5.7}b_{1,0}b_{2,0}$, $\sigma=\sigma_{5.8} b_{1,1}b_{2,0}$, where

$\begin{eqnarray*} &&\sigma_{5.4}\in E_1^{t+4,(t+3)p^2q+(t-1)pq+(t-1)q+t-5,*}, \sigma_{5.5}\in E_1^{t+4,(t+3)p^2q+(t-3)pq+(t-1)q+t-5,*},\nonumber\\ &&\sigma_{5.6}\in E_1^{t+4,(t+4)p^2q+(t-2)pq+(t-1)q+t-5,*}, \sigma_{5.7}\in E_1^{t+4,(t+4)p^2q+(t-3)pq+(t-1)q+t-5,*},\nonumber\\ &&\sigma_{5.8}\in E_1^{t+4,(t+3)p^2q+(t-2)pq+(t-1)q+t-5,*}. \end{eqnarray*}$

Similarly to $\sigma_{4.2}$, we can get that $\sigma_{5.4}=0$, $\sigma_{5.5}=0$, $\sigma_{5.8}=0$. Similarly to $\sigma_{4.5}$, we can get that $\sigma_{5.6}=0$, $\sigma_{5.7}=0$.

Subcase 5.1.3 If $\sigma$ contains $t+5$ factors, then there exist three factors $b_{i,j}(b_{1,0},b_{1,1},b_{2,0})$. Due to the commutativity, the possible nontrivial forms will be $\sigma=\sigma_{5.9} b_{1,1}^3$, $\sigma=\sigma_{5.10}b_{2,0}^3$, $\sigma=\sigma_{5.11} b_{1,1}^2b_{2,0}$, $\sigma=\sigma_{5.12} b_{1,1}b_{2,0}^2$, where

$\begin{eqnarray*} && \sigma_{5.9}\in E_1^{t+2,(t+2)p^2q+(t-1)pq+(t-1)q+t-5,*}, \sigma_{5.10}\in E_1^{t+2,(t+2)p^2q+(t-4)pq+(t-1)q+t-5,*},\nonumber\\ &&\sigma_{5.11}\in E_1^{t+2,(t+2)p^2q+(t-2)pq+(t-1)q+t-5,*}, \sigma_{5.12}\in E_1^{t+2,(t+2)p^2q+(t-3)pq+(t-1)q+t-5,*}, \end{eqnarray*}$

By the similar argument in Subcase2.1, we can know that $\sigma_{5.9}=0$. Similarly to $\sigma_{4.5}$, we can get that $\sigma_{5.10}=0$, $\sigma_{5.11}=0$, $\sigma_{5.12}=0$.

Subcase 5.2 When $t=4$ or $t=3$, by the similar argument in Subcase2.2, we know that the generator is impossible to exist. Thus, in this case, we can get $\sigma=0$.

Case 6 If $r=2$, then

$E_1^{t+11-r,(t+5)p^2q+(t-1)pq+(t-1)q+t-r-2,*}=E_1^{t+9,(t+5)p^2q+(t-1)pq+(t-1)q+t-4,*}.$

Subcase 6.1 When $t\geqslant 5$, from Lemma 2.5, the number of the factors in $\sigma$ will be $t+5$, $t+6$, $t+7$ or $t+8$.

Subcase 6.1.1 If $\sigma$ contains $t+8$ factors, then there exists a factor $b_{i,j}(b_{1,0},b_{1,1},b_{2,0})$. Due to the commutativity, the possible nontrivial forms will be $\sigma=\sigma_{6.1} b_{1,0}, \sigma=\sigma_{6.2} b_{1,1},$ $\sigma=\sigma_{6.3}b_{2,0}, $ where

$\begin{eqnarray*} &&\sigma_{6.1}\in E_1^{t+7,(t+5)p^2q+(t-2)pq+(t-1)q+t-4,*}, \sigma_{6.2}\in E_1^{t+7,(t+4)p^2q+(t-1)pq+(t-1)q+t-4,*}, \nonumber\\ &&\sigma_{6.3}\in E_1^{t+7,(t+4)p^2q+(t-2)pq+(t-1)q+t-4,*}. \end{eqnarray*}$

Similarly to $\sigma_{4.2}$, we can get $\sigma_{6.1}=0$, $\sigma_{6.2}=0$, $\sigma_{6.3}=0$.

Subcase 6.1.2 If $\sigma$ contains $t+7$ factors, then there exist two factors $b_{i,j}(b_{1,0},b_{1,1},b_{2,0})$. Due to the commutativity, the possible nontrivial forms will be $\sigma=\sigma_{6.4} b_{1,0}^2$, $\sigma=\sigma_{6.5} b_{1,1}^2$, $\sigma=\sigma_{6.6}b_{2,0}^2$, $\sigma=\sigma_{6.7} b_{1,0} b_{1,1} $, $\sigma=\sigma_{6.8}b_{1,0}b_{2,0}$, $\sigma=\sigma_{6.9} b_{1,1}b_{2,0}$, where

$\begin{eqnarray*} &&\sigma_{6.4}\in E_1^{t+5,(t+5)p^2q+(t-3)pq+(t-1)q+t-4,*}, \sigma_{6.5}\in E_1^{t+5,(t+3)p^2q+(t-1)pq+(t-1)q+t-4,*},\nonumber\\ &&\sigma_{6.6}\in E_1^{t+5,(t+3)p^2q+(t-3)pq+(t-1)q+t-4,*}, \sigma_{6.7}\in E_1^{t+5,(t+4)p^2q+(t-2)pq+(t-1)q+t-4,*},\nonumber\\ &&\sigma_{6.8}\in E_1^{t+5,(t+4)p^2q+(t-3)pq+(t-1)q+t-4,*}, \sigma_{6.9}\in E_1^{t+5,(t+3)p^2q+(t-2)pq+(t-1)q+t-4,*}. \end{eqnarray*}$

Similarly to $\sigma_{4.5}$, we can get that $\sigma_{6.4}=0$. Similarly to $\sigma_{4.2}$, we can get that $\sigma_{6.i}=0$ $(i=5,6\cdots 9)$.

Subcase 6.1.3 If $\sigma$ contains $t+6$ factors, then there exist three factors $b_{i,j}(b_{1,0},b_{1,1},b_{2,0})$. Due to the commutativity, the possible nontrivial forms will be $\sigma=\sigma_{6.10} b_{1,1}^3$, $\sigma=\sigma_{6.11}b_{2,0}^3$, $\sigma=\sigma_{6.12}b_{1,1}^2b_{1,0}$, $\sigma=\sigma_{6.13} b_{1,1}^2b_{2,0}$, $\sigma=\sigma_{6.14} b_{1,0}b_{2,0}^2$, $\sigma=\sigma_{6.15} b_{1,1}b_{2,0}^2$, $\sigma=\sigma_{6.16} b_{1,0}b_{1,1}b_{2,0}$, where the first degrees of $\sigma_{6.i}(i=10,11,\cdot\cdot\cdot,16)$ are all $t+3$ and the second degrees of them are listed in Table 4 ($M=t-1$, and $N=t-4$).

Similarly to $\sigma_{4.2}$, we can get that $\sigma_{6.10}=0$, $\sigma_{6.11}=0$, $\sigma_{6.13}=0$. Similarly to $\sigma_{4.5}$, we can get that $\sigma_{6.12}=0$, $\sigma_{6.14}=0$, $\sigma_{6.15}=0$, $\sigma_{6.16}=0$.

Subcase 6.1.4 If $\sigma$ contains $t+5$ factors, then there exist four factors $b_{i,j}(b_{1,0},b_{1,1},b_{2,0})$. Due to the commutativity, the possible nontrivial forms will be $\sigma=\sigma_{6.17} b_{1,1}^4$, $\sigma=\sigma_{6.18}b_{2,0}^4$, $\sigma=\sigma_{6.19}b_{1,1}^3b_{2,0}$, $\sigma=\sigma_{6.20} b_{1,1}b_{2,0}^3$, $\sigma=\sigma_{6.21} b_{1,2}^2b_{2,0}^2$, where the first degrees of $\sigma_{6.i}(i=17,18,\cdot\cdot\cdot,21)$ are all $t+1$ and the second degrees of them are listed in Table 5 ($M=t-1$, and $N=t-4$).

Similarly to Subcase2.1, we can get that $\sigma_{6.17}=0$. Similarly to $\sigma_{3.2}$, we can get that $\sigma_{6.i}=0(i=18,19,20,21)$.

Subcase 6.2 When $t=3, t=4$, by the similar argument in Subcase 2.2, we know that the generator is impossible to exist. Thus, we have $\sigma=0$.

Therefore, we can get that $E_1^{t+11-r,(5+t)p^2q+(t-1)pq+(t-1)q+t-r-2,*}=0$.

That is $\text{Ext}_A^{t+11-r,(5+t)p^2q+(t-1)pq+(t-1)q+t-r-2,*}(\mathbb{Z}_p,\mathbb{Z}_p)=0.$

Proposition 3.5 Let $s\geqslant 2$, $p\geqslant 11$, then $\text{Ext}_A^{s+11,5p^2q+q+s-1}(H^*V(2),\mathbb{Z}_p)=0.$

From Corollary 3.2, we have

$\text{Ext}_A^{s+11,5p^2q+q+s-1}(H^* V(3),\mathbb{Z}_p)\cong\text{Ext}_P^{s+11,5p^2q+q+s-1}(\mathbb{Z}_p,\mathbb{Z}_p).$

From [4,Lemma 2.2]{Toda}, we know that the rank of $\text{Ext}_P^{s+11,5p^2q+q+s-1}(\mathbb{Z}_p,\mathbb{Z}_p)$ is less than or equal to that of $[P(b_j^i)\otimes H^{*,*}(U(L))]^{s+11,5p^2q+q+s-1}$ , and $[P(b_j^i)\otimes H^{*,*}(U(L))]^{s,t}$ is the $E_2-$term of the MSS, where $P()$ is the polynomial algebra. Up to the total degree $t-s<(p^3+3p^2+2p+1)q-4$, $H^{s,t}(U(L))$ is multiplicative by the following cohomology classes

$\begin{alignat}{5} h_i & =\{R_1^i\}, & g_i & =\{R_2^i R_1^i\}, & k_i & =\{R_2^i R_1^{i+1}\}(i\geqslant 0),\nonumber\\ l_1 & =\{R_3^0 R_2^0 R_1^0 \}, & l_2 & =\{R_2^1 R_2^0 R_1^1 \}, & l_3 & =\{R_3^0 R_1^2 R_1^0 \},\nonumber\\ l_4 & =\{R_3^0 R_2^1 R_1^2 \}, & l_5 & =\{R_3^1 R_2^1 R_1^1 \}, & l_6 & =\{R_2^2 R_2^1 R_1^2 \},\nonumber\\ m_1 & =\{R_3^0 R_2^1 R_2^0 R_1^1 \}, & m_2 & =\{R_4^0 R_3^0 R_2^0 R_1^0 \} , & & \nonumber\\ m_3 & =\{R_3^1 R_2^1 R_2^0 R_1^1 \}, & m_4 & =\{R_2^2 R_3^0 R_1^2 R_1^0 \}. & & \nonumber \end{alignat}$

Moreover, we have additively

$\begin{eqnarray*} &&H^{*,*}(U(L))\cong \{1,l_4,h_3 \}\otimes \{1,h_0,h_1,g_0,k_0,k_0h_0\}\nonumber\\ &&+\{h_2,h_2h_0,g_1,l_1,l_2,l_1h_1,k_1,l_3,k_1h_1,l_1h_2,m_1,m_1h_0,g_2,g_2h_0,l_5,m_2,m_3,l_6,m_4\}, \end{eqnarray*}$

and the bidegrees of $R_j^i,b_j^i$ are $(1,2(p^{i+j}-p^i))$, $(2,2(p^{i+j-1}-p^{i+1}))$, respectively. In the MSS, $b_1^0$ converges to $b_0\in \text{Ext}_P^{2,pq}(\mathbb{Z}_p,\mathbb{Z}_p)$, and the total degree of $b_1^0$ is $|b_1^0|=pq-2$. The generators whose total degrees are less than or equal to $5p^2q+q-12$ in $[P(b_j^i)\otimes H^{*,*}(U(L))]^{s,t}$ and the total degrees $|\lambda|$ mod $pq-2$ are listed in Table 6 $(t=1,2,3,4,5,t'=1,2,3,4).$

Let $x$ be a generator of $ \text{Ext}_A^{s+11,5p^2q+q+s-1}(H^* V(2),\mathbb{Z}_p)$, then we have

$(i_3)_*(x)\in \text{Ext}_A^{s+11,5p^2q+q+s-1}(H^* V(3),\mathbb{Z}_p). $

The total degree of $(i_3)_*(x)$ is $5p^2q+q+s-1-(s+11)=5p^2q+q-12\equiv 6q-2$ (mod $ pq-2)$. From the above Table, we know that the generator $\lambda$ with total degree mod $pq-2$ being equal to $6q-2$ in $[P(b_j^i)\otimes H^{*,*}(U(L))]^{s,t}$ doesn't exist. So, we can get that $(i_3)_*(x)=0$. Consider the following exact sequence:

$ \cdots \xrightarrow{{(j_3)_*}} \text{Ext}_A^{s+10,5p^2q+q+s-1-(2p^3-1)}(H^*V(2),\mathbb{Z}_p) \xrightarrow{{(\alpha_3)_*}} \text{Ext}_A^{s+11,5p^2q+q+s-1}(H^*V(2),\mathbb{Z}_p) $

$ \xrightarrow{{(i_3)_*}} \text{Ext}_A^{s+11,5p^2q+q+s-1}(H^*V(3),\mathbb{Z}_p)\xrightarrow{{(j_3)_*}} \cdots, $

there exists an element $x_1\in \text{Ext}_A^{s+10,5p^2q+q+s-1-(2p^3-1)}(H^*V(2),\mathbb{Z}_p)$ satisfying $(\alpha_3)_*(x_1)=x$. The total degree of $(i_3)_*(x_1)$ is $5p^2q+q+s-1-(2p^3-1)-(s+10)\equiv 4q-6$(mod $pq-2)$. From the above Table, we know that

$0=(i_3)_*(x_1)\in \text{Ext}_A^{s+10,5p^2q+q+s-1-(2p^3-1)}(H^*V(3),\mathbb{Z}_p).$

Using the exactness repeatedly, there exists an element $x_k\in \text{Ext}_A^{s+11-k,5p^2q+q+s-1-k(2p^3-1)}$ $(H^*V(2),\mathbb{Z}_p)$ satisfying $(\alpha_3)_*(x_k)=x_{k-1}$. But the total degree of $(i_3)_*(x_k)$ mod $pq-2$ is different from that in the above table, so we know that

$0=(i_3)_*(x_k)\in \text{Ext}_A^{s+11-k,5p^2q+q+s-1-k(2p^3-1)}(H^*V(3),\mathbb{Z}_p).$

Let $k=5,$ then

$x_5\in \text{Ext}_A^{s+6,5p^2q+q+s-1-5(2p^3-1)}(H^*V(2),\mathbb{Z}_p)=\text{Ext}_A^{s+6,-10p^2+q+s+4}(H^*V(2),\mathbb{Z}_p)=0.$

Therefore, we have $x=\underbrace{(\alpha_3)_* \cdots (\alpha_3)_*}\limits_5(x_5)=0$, that is

$\text{Ext}_A^{s+11,5p^2q+q+s-1}(H^*V(2),\mathbb{Z}_p)=0 (s\geqslant2, p\geqslant 11).$

Proposition 3.6 Let $r\geqslant 2$, $p\geqslant 11$, then

$\text{Ext}_A^{11-r,5p^2q+q-r+1}(H^*V(2),\mathbb{Z}_p)=0.$

The proposition is evident for $r\geqslant 11$. Thus, we need only to consider the case of $2\leqslant r<11$.

For any $y\in \text{Ext}_A^{11-r,5p^2q+q-r+1}(H^*V(2),\mathbb{Z}_p)$, we can know $0<q-r+1<q$ from $2\leqslant r<11$. Since $Sdim((i_3)_*(y))=5p^2q+q-r+1=q-r+1\neq 0$(mod $ q)$ and $\text{Ext}_P^{s,t}(\mathbb{Z}_p,\mathbb{Z}_p)=0$ $ (t\neq 0$ mod $q )$, from Proposition 3.1, we can get that

$0=(i_3)_*(y)\in \text{Ext}_A^{11-r,5p^2q+q-r+1}(H^*V(3),\mathbb{Z}_p).$

According to the exactness, there exists an element

$y_1\in \text{Ext}_A^{10-r,5p^2q+q-r+1-(2p^3-1)}(H^*V(2),\mathbb{Z}_p)$

such that $(\alpha_3)_*(y_1)=y$, and

$Sdim((i_3)_*(y_1))=5p^2q+q-r+1-(2p^3-1)=4p^2q-pq-r=q-r\neq 0 ({\hbox{mod}} q),$

so $0=(i_3)_*(y_1)\in \text{Ext}_A^{10-r,5p^2q+q-r+1-(2p^3-1)}(H^*V(3),\mathbb{Z}_p)$.

Similarly, there exists an element $y_k\in \text{Ext}_A^{11-k-r,5p^2q+q-r+1-k(2p^3-1)}(H^*V(2),\mathbb{Z}_p)$ satisfying $(\alpha_3)_*(y_k)=y_{k-1}$, and $Sdim((i_3)_*(y_k))\neq 0$ (mod $q).$ Let $k=5,$ then

$y_5\in \text{Ext}_A^{6-r,5p^2q+q-r+1-5(2p^3-1)}(H^*V(2),\mathbb{Z}_p)=\text{Ext}_A^{6-r,-10p^2+q-r+6}(H^*V(2),\mathbb{Z}_p)=0.$

Thus, we get that $y=\underbrace{(\alpha_3)_* \cdots (\alpha_3)_*}\limits_5(y_5)=0$, that is

$\text{Ext}_A^{11-r,5p^2q+q-r+1}(H^*V(2),\mathbb{Z}_p)=0(r\geqslant 2,p\geqslant 11). $

The Proof of Theorem 1.2  First, we consider the ASS with $E_2$-term:

$\text{Ext}_A^{s,t}(H^*V(2),\mathbb{Z}_p)\Rightarrow \pi_{t-s}V(2),$

and its differential is $d_r:E_r^{s,t}\rightarrow E_r^{s+r,t+r-1}.$

From Proposition 3.5,

$E_2^{r+11,5p^2q+q+r-1}=\text{Ext}_A^{r+11,5p^2q+q+r-1}(H^*V(2),\mathbb{Z}_p), $

we can get that $E_r^{r+11,5p^2q+q+r-1}=0 (r\geqslant 2).$ Let $h_0b_1^5$ be the image of $h_0b_1^5\in \text{Ext}_A^{11,5p^2q+q}(\mathbb{Z}_p,\mathbb{Z}_p)$ under the map

$(i_2)_*(i_1)_*(i_0)_*:\text{Ext}_A^{11,5p^2q+q}(\mathbb{Z}_p,\mathbb{Z}_p)\rightarrow \text{Ext}_A^{11,5p^2q+q}(H^*V(2),\mathbb{Z}_p),$

then $d_r(h_0b_1^5)\in E_r^{r+11,5p^2q+q+r-1}=0(r\geqslant 2). $ Furthermore, we can get that

$h_0b_1^5\in E_2^{11,5p^2q+q}=\text{Ext}_A^{11,5p^2q+q}(H^*V(2),\mathbb{Z}_p)$

is a permanent cycle in the ASS. Moreover, from Proposition 3.6,

$E_2^{11-r,5p^2q+q-r+1}=\text{Ext}_A^{11-r,5p^2q+q-r+1}(H^*V(2),\mathbb{Z}_p)=0(r\geqslant 2),$

we have that $E_r^{11-r,5p^2q+q-r+1}=0(r\geqslant 2).$ So, $h_0b_1^5$ is impossible to be the $d_r$-boundary in the ASS, and $h_0b_1^5\in \text{Ext}_A^{11,5p^2q+q}(H^*V(2),\mathbb{Z}_p)$ converges to a nontrivial element in $\pi_*V(2)$.

The Proof of Theorem 1.1 From Theorem 1.2, we know that there exists a nontrivial element $f$ in $\pi_*V(2)$, which is represented by $h_0b_1^5\in \text{Ext}_A^{11,5p^2q+q} (H^*V(2),\mathbb{Z}_p)$, where $h_0b_1^5$ denotes the image of $h_0b_1^5\in \text{Ext}_A^{11,5p^2q+q}(\mathbb{Z}_p,\mathbb{Z}_p)$ under the homomorphism

$(i_2)_*(i_1)_*(i_0)_*:\text{Ext}_A^{11,5p^2q+q}(\mathbb{Z}_p,\mathbb{Z}_p)\rightarrow \text{Ext}_A^{11,5p^2q+q}(H^*V(2),\mathbb{Z}_p).$

Consider the following composition of maps

$ \tilde{f}:\Sigma^{5p^2q+q-11}S\xrightarrow{{f}} V(2)\xrightarrow{{\gamma^t}} \Sigma^{t(p^2+p+1)q} V(2)\xrightarrow{{j_0j_1j_2}} \Sigma^{-t(p^2+p+1)q+(p+1)q+q+3}S, $

the composed map $\tilde{f}= j_0j_1j_2\gamma^t f$ is represented by

$(j_0j_1j_2)_*(\gamma^t)_*(i_2i_1i_0)_*(h_0b_1^5)\in \text{Ext}_A^{11+t,(5+t)p^2q+(t-1)(p+1)q+t-3}(\mathbb{Z}_p,\mathbb{Z}_p).$

From [3], we have known that $\gamma_t=j_0j_1j_2\in \pi_*(S)$ is represented by $\stackrel{\thicksim}{\gamma_t}$ in the ASS. By the knowledge of Yoneda products, we know that the following composition:

$ \text{Ext}_A^{0,0}(\mathbb{Z}_p,\mathbb{Z}_p)\xrightarrow{{(i_2i_1i_0)_*}} \text{Ext}_A^{0,0}(H^*V(2),\mathbb{Z}_p)\xrightarrow{{(\gamma^t)_*}} \text{Ext}_A^{t,t(p^2+p+1)q+t}(H^*V(2),\mathbb{Z}_p) $

$ \xrightarrow{{(j_0j_1j_2)_*}} \text{Ext}_A^{t,tp^2q+(t-1)pq+(t-2)q+t-3}(\mathbb{Z}_p,\mathbb{Z}_p) $

is a homomorphism which is multiplied by $\stackrel{\thicksim}{\gamma_t}$. Hence, $\stackrel{\thicksim}{f}\in \pi_*(S)$ is represented by

$\stackrel{\thicksim}{\gamma_t} h_0 b_1^5\in \text{Ext}_A^{11+t,(5+t)p^2q+(t-1)(p+1)q+t-3}(\mathbb{Z}_p,\mathbb{Z}_p)$

in the ASS.

Moreover, from the Proposition 3.4 we know that $\stackrel{\thicksim}{\gamma_t}h_0 b_1^5$ can't be hit by the differentials in the ASS, then we get that $\stackrel{\thicksim}{\gamma_t}h_0 b_1^5$ converges to a nontrivial element $\stackrel{\thicksim}{f}$ in $\pi_*(S)$.