a group with specific orders of elements I want to find a group with elements of order $1,2,3,4$ and $5$  (at least one of each order). 
All I can say is that the order of the group is $60$ itself, but cannot find the correct one.
Please let me know about your solutions too.
 A: One example is the alternating group $A_6$.
All groups with this property have been classified in the following paper.

R. Brandl and W.J. Shi, Finite groups whose element orders are consecutive integers, J. Algebra 143, (1991), 388-400. DOI

In the above article Brandl and Shi classify all finite groups which have at least one element of order $1,\ 2,\ \ldots,\ n$, and none of any other order. It turns out that such groups exist only for $n \leq 8$. Another interesting result is that a finite group $G$ has $\{1,2,3,4,5,6,7\}$ as its set of element orders if and only if $G \cong A_7$.
A: If you really insist that your group does a element of order $4,$ then there is no such group of order $60$ satisfying the conditions: for if $G$ were such a group, it would have a cyclic Sylow $2$-subgroup, and then a normal subgroup $N$ of order $15,$ and furthermore $N$ would be cyclic, so would have an element of order $15$ (thanks to the comment of Hagen von Eitzen for shortening the argument). Clearly a group of order $60$ has to contain elements of order $1,2,3$ and $5,$ and there is only one group of order $60$ up to isomorphism which contains elements of just these orders.
