Show that if a group $G$ of finite order has elements of order $d,$ then $G$ has subgroups of order of every divisor of $d.$

Question

My attempt: Let $G$ be cyclic and let $a\in G$ has order $d.$ Since order of an element of a group divides order of that group, and since converse of Lagrange theorem is true for cyclic groups, so we can have cyclic subgroups of order of every divisors of $d.$ However, I am stuck to prove for an arbitrary group of finite order.

Answer 1

If there's an element of order $d$, say $g\in G$, then $\langle g\rangle$ is cyclic of order $d$.

Answer 2

If $c$ divides $d$, the order of $a^{d\over c}$ is $c$.