In the case of solvable groups, this may not say much about the structure of the group. For example, if $G$ is a finite $\{p,q\}$-group, where $p,q$ are distinct primes (hence $G$ is solvable by Burnside's $p^{a}q^{b}$-theorem), then it is quite unusual (though certianly not impossible) for $G$ not to have an element of order $pq.$ If $G$ contains an elementary Abelian $p$-group of order $p^{2}$ and an elementary Abelian $q$-group of order $q^{2}$, then $G$ will contain an element of order $pq.$ If, for example, $G$ contains no elementary Abelian $q$-sugroup of order $q^{2},$ then the Sylow $q$-subgroups of $G$ are cyclic or generalized quaternion

In the case of solvable groups, this may not say much about the structure of the group. For example, if $G$ is a finite $\{p,q\}$-group, where $p,q$ are distinct primes (hence $G$ is solvable by Burnside's $p^{a}q^{b}$-theorem), then it is quite unusual (though certianly not impossible) for $G$ not to have an element of order $pq.$ If $G$ contains an elementary Abelian $p$-group of order $p^{2}$ and an elementary Abelian $q$-group of order $q^{2}$, then $G$ will contain an element of order $pq.$ If, for example, $G$ contains no elementary Abelian $q$-subgroup of order $q^{2},$ then the Sylow $q$-subgroups of $G$ are cyclic or generalized quaternion