Prove that the total number of subgroups of a finite cyclic group $G=\langle a \rangle$ such that $o(G)=n$ is the number of divisors of $n$.

If $G=\langle a \rangle $ is a finite cyclic group such that $o(G)=n$, then any subgroup of $G$ has the form $S=\langle g^d\rangle$ for a divisor $d\mid n$. Different $d$ give different subgroup. But how to give a concrete proof of the problem.

Please help me to solve the problem using the elementary tools specifically using Lagrange's theorem.


I'm not sure how you could prove this without characterizing what the subgroups of a cyclic group look like.

Your result follows directly from the fact that a (finite) cyclic group (of order $n$) has a unique subgroup of order $k$ for each divisor $k$ of $n$. [so the number of divisors = the number of distinct subgroups.]

To prove this fact about uniqueness, typically you'll need to show:

(1) Every subgroup of a cyclic group is cyclic. [Sketch of proof: Suppose $G$ is generated by $g$. Let $H$ be a subgroup. For $H$ non-trivial, there is a unique smallest positive power $k$ of $g$ such that $g^k \in H$. Then show $H=\langle g^k \rangle$.]

(2) Given any subgroup, $H=\langle g^k \rangle$, show that $H= \langle g^k \rangle = \langle g^d \rangle$ where $d=\mathrm{gcd}(k,n)$. [Use the fact that $d=kx+ny$ for some integers $x,y$ so get this.]

(3) Note that if $k$ divides $n$, $\langle g^{n/k} \rangle$ is a subgroup of order $k$. So there is a subgroup for each divisor.

(4) Let $H$ be a subgroup of order $k$. Well, $H=\langle g^d \rangle$ for some divisor $d$ (by (1) and (2) put together). But $|H|=|\langle g^d \rangle|=n/d = k$ so $d=n/k$. This means that the order of $H$ uniquely determines $d$. Thus any two subgroups of order $k$ must be equal to the same $\langle g^d \rangle$. Thus there is only one subgroup for each divisor $k$ (namely $\langle g^{n/k} \rangle$).


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