I may be heavily mistaken but these are my partial arguments in favour of the following statement
" No subgroup of order 8 in $S_5$ is abelian"
If not, let there be an abelian subgroup of order 8 in $S_5$ .Then it is isomorphic to either $\mathbb{Z}_8$, $\mathbb{Z}_4 \oplus \mathbb{Z}_2$ or $\mathbb{Z}_2\oplus \mathbb{Z}_2\oplus \mathbb{Z}_2$
case 1) The subgroup $H$ (say) is isomorphic to $\mathbb{Z}_8$.Then there should be a corresponding element of order 8 in $H$ which implies $S_5$ has an element of order 8 which is impossible by considering the cycle type of different elements.
case 2) The subgroup $H$ is isomorphic to $\mathbb{Z}_4\oplus\mathbb{Z}_2$ .The latter group has $(1,0)$ and $(0,1)$ of order $4$ and $2$ respectively which commute and hence there are corresponding elements of order $4$ and $2$ in $H$. Now, The elements of order 4 in $S_5$ are of the cycle type $(a,b,c,d)$ and that of order 2 are of cycle type $(a,b)$ or $(a,b)(c,d)$. In any case , permutations of order 4 and 2 in $S_5$ will not commute since they do not have disjoint cycle types.Thus contradiction.
Now I am puzzled with the last case. Will there be some contradiction ?Are the above two cases handled properly? Is there rather a smart way to arrive at the result??
Sorry for a long list of questions but I need your help. Thanks for your ideas.
Let $S$ be an arbitrary subgroup of $S_5$, Since $|| = 23×3×5$ and $2^3|(G)$, $2^4\nmid o(G)$, therefore the number of $2$-sylow subgroups of $G$ of order $8$ is of the form $1+2k$ and $1+2k|o(G)$, that is, $1+2k|2^3×3× 5$. Therefore $1+2k|5×3$, the possible values for $k = 0, 1, 2$ and $7$. Therefore the maximum number of $2$-sylow subgroup of order $8$ is $15$ when $k = 7$. Since there is no element of order $8$, so every element of $S$ must have order $1, 2 $and $4$. Can you finish from here?