Player A can win for $n=2$ (by selecting the identity) and for $n=3$ (selecting a 3-cycle). We prove that B has a winning strategy for $n \geq 4$. Consider the moment when all permitted moves lose immediately, and let $H$ be the subgroup generated by the elements selected by the players. Choosing another element from $H$ would not lose immediately, so all elements of $H$ must have been selected. Since $H$ and any other element generate $S_{n}, H$ must be a maximal subgroup in $S_{n}$. If $|H|$ is even, then the next player is A, so B wins. Denote by $n_{i}$ the order of the subgroup generated by the first $i$ selected elements; then $n_{1}\left|n_{2}\right| n_{3} \mid \ldots$ We show that B can achieve that $n_{2}$ is even and $n_{2}<n!$; then $|H|$ will be even and A will be forced to make the final - losing - move. Denote by $g$ the element chosen by A on his first move. If the order $n_{1}$ of $g$ is even, then B may choose the identical permutation $id$ and he will have $n_{2}=n_{1}$ even and $n_{2}=n_{1}<n$!. If $n_{1}$ is odd, then $g$ is a product of disjoint odd cycles, so it is an even permutation. Then B can chose the permutation $h=(1,2)(3,4)$ which is another even permutation. Since $g$ and $h$ are elements of the alternating group $A_{n}$, they cannot generate the whole $S_{n}$. Since the order of $h$ is 2, B achieves $2 \mid n_{2}$. Remark. If $n \geq 4$, all subgrups of odd order are subgroups of $A_{n}$ which has even order. Hence, all maximal subgroups have even order and B is never forced to lose.