A group of order $120$ has a subgroup of index $3$ or $5$ (or both) What I have tried that number of $2$-sylow subgroup can be $1,3,5$ or $15$.I have solved the problem when the number of $2$-sylow subgroup is $1,3,5$. But I am not able to solve it for $15$.
Any help will be appreciated..
 A: Suppose there are $15$ Sylow $2$-subgroups, and let $S$ be one of them. Note that $N_G(S)=S$. If $S$ is not maximal, then it is contained in a subgroup of index $3$ or $5$, so assume it is maximal.
Consider the permutation representation of $G$ by left multiplication on the left cosets of $S$. Then $S$ is a point stabilizer. If $S$ fixes more than one point in this representation, then the normalizer $N_G(S)$ permutes the fixed points and strictly contains $S$, contradicting $N_G(S)=S$.
So $S$ fixes a unique point. Since its orbits have length a power of $2$ and their lengths add up to $15$, there must be an orbit of length $2$. The point stabilizer $T$ of that orbit has index $2$ in $S$ and (since it stabilizes more than one point) it lies in more than one Sylow subgroup. So $N_G(T)$ properly contains $S$. If $N_G(T) = G$ then $T \lhd G$ and $|G/T| = 30$. Now all groups of order $30$ have subgroups of index $3$ and $5$ (exercise), and hence so does $G$. Otherwise $S < N_G(T) < G$, contradicting maximality of $S$.
