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.
Show that if a group $G$ of finite order has elements of order $d,$ then $G$ has subgroups of order of every divisor of $d.$
abstract-algebracyclic-groupsfinite-groupsgroup-theory
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Best Answer
If there's an element of order $d$, say $g\in G$, then $\langle g\rangle$ is cyclic of order $d$.