Let $G=\langle a\rangle$ be a cyclic group of order $n$. Show that, for every divisor $d$ of $n$, there exists a subgroup of $G$ whose order is $d$.

Question

Let $G=\langle a\rangle$ be a cyclic group of order $n$. Show that, for every divisor $d$ of $n$, there exists a subgroup of $G$ whose order is $d$.

This time I have no approach. I haven't found any relation between the divisors of $n$ and the order of the subgroups of $G$. How would approach this?

Answer 1

If you realize $G$ as the set $\{0,1,\ldots, n-1\}$ with addition modulo $n=de$, then $\{0,e,2e,\ldots, (d-1)e\}$ is a subgroup of order $d$.

Answer 2

‎ ‎$ d‎ ‎\vert n ‎‎‎‎‎‎ ‎\Rightarrow ‎‎‎‎\exists k‎ ‎\in ‎‎\mathbb{N} ‎‎\qquad‎ st ‎‎\qquad‎ n=kd ‎‎$‎‎‎‎

‎$ 1=‎ ‎‎a^n=a^kd=(a^k)^{d} ‎\Rightarrow ‎o(<a^k>)=d ‎‎$‎‎‎

then ‎$<a^k> \leq G ‎$ ‎and ‎‎$‎\vert ‎<a^k‎> \vert = d ‎$‎‎

Answer 3

For any divisor d of n,we shall get an integer k such that dk=n. Now consider the element $a^k$ and show that the subgroup generated by it is the required one.