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 $n$.

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.

• – lhf Oct 2 '15 at 15:00

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 $\langle g^{n/k} \rangle$).