Let $G$ which is a cyclic group with $n$ elements. Find:
a) all possible subgroups
b) all elements which generate whole group.

I know that number of subgroups is the number of total divisors of $n$. Moreover if $l|(|G|=n)$ then exist exactly one subgroup $H$ such that $|H|=l$.

However I know how to do this task when I know $n$.

For example for $n=6$ we have divisors: $1,2,3,6$ so we have $4$ supgroups.

Meanwhile I have a problem when I don't know $n$:

Let $G= \left\{ e,g,g^{2},\dots ,g^{n-1} \right\}$ then we have: $$\langle e\rangle=\left\{ e \right\}$$ $$\langle g\rangle =\left\{ e,g,g^{2},\dots ,g^{n-1} \right\}$$ But I already have a problem with $\langle g^2\rangle $ because firstly I have: $\langle g^2\rangle =\left\{ e,g^{2},g^{4},\dots ,g^{n-1}\right\}$ but I don't know if $n=2k+1$ and this is whole $\langle g^2\rangle $ or $n=2k$ and $g^2$ is not a generator of any subgroups.

How can I do this task for every $n$?

  • 2
    $\begingroup$ About b). If $g$ is a generator, than any $g^k$ is a generator iff $(k, n)=1$. As a corollary, if $n$ is prime, than any non-unit element of $G$ is a generator. $\endgroup$
    – user615081
    Oct 19, 2019 at 15:01

1 Answer 1


If $\langle g\rangle =G$, then $\mid g^k\mid=\frac n{\operatorname {gcd}(n,k)}$.

For $b)$, using this fact we get that whenever $\operatorname {gcd}(n,k)=1$, $\langle g^k\rangle =G$.


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