This paper considers flag-transitive $4$-$(q+1,k,\lambda)$ designs with $\lambda\geq5$ and $q+1>k>4$. Let the automorphism group of a design $\cal D$ be a simple group $G=PSL(2,q)$. Depend on the fact that the setwise stabilizer $G_B$ must be one of twelve kinds of subgroups, up to isomorphism we get the following two results. (i) If $10\geq \lambda \geq 5$, then except $(G,G_x,G_B,k,\lambda)=(PSL(2,761),{ E_{ 761}}\rtimes { C_{ 380}},S_4,24,7)$ or $(PSL(2,512),{ E_{ 512}}\rtimes { C_{ 511}},{ D_{ 18}},18,8)$ undecided, $\cal D$ is a $4$-$(24,8,5)$, $4$-$(9,8,5)$, $4$-$(8,6,6)$, $4$-$(10,9,6)$, $4$-$(9,6,10)$, $4$-$(9,7,10)$, $4$-$(12,11,8)$ or $4$-$(14,13,10)$ design with $G_B=D_8$, ${ E_8}\rtimes { C_7}$, $D_6$, ${ E_9}\rtimes { C_4}$, $PSL(2,2)$, $D_{ 14}$, ${ E_{ 11}}\rtimes { C_{ 5}}$ or ${ E_{ 13}}\rtimes { C_6}$ respectively. (ii) If $\lambda>10$, ${ G_B}=A_4$, $S_4$, $A_5$, $PGL(2,q_0)$($g>1$ even) or $PSL(2,q_0)$, where ${ q_0}^g=q$, then there is no such design.
Categories: math.CO