$G$ is a group of order $60$. Will there be a subgroup of order $ 6$?
Alternating group $A_5$ has a subgroup of order $6$. That is the group generated by this set $\{(123), (23) (45)\}$.
Will we be able to prove that there always exists a subgroup of order $6$ in a group of order $ 60$?
Can anyone help me to understand by giving a hint?
No, that's not always true. Take for example $G = C_5 \times A_4$. This group has order $60$ and no subgroups of order $6$. If you know that $A_4$ has no subgroups of order $6$ (it is the smallest group which fails to satisfy the converse of Lagrange's theorem), it is easy to find this example.
However, can you prove that this is the only exception?