# Prove $\langle a^s \rangle=\langle a^t \rangle \iff \gcd(n,s)=\gcd(n,t)$

Problem

Let $G$ be a cyclic group with $n$ elements and with $a$ as the generator let $b \in G$ and let $b=a^s$. Then $b$ generates a cyclic subgroup of $H$ of $G$ containing $\dfrac{n}{d}$ elements, where $d=\gcd(n,s)$. Also show that $\langle a^s \rangle=\langle a^t \rangle \iff \gcd(n,s)=\gcd(n,t)$

I could prove all the remaining parts of the problem but I can't figure out the way to prove the second part of the problem i.e. $\langle a^s \rangle=\langle a^t \rangle \impliedby \gcd(n,s)=\gcd(n,t)$ without using the concept of Isomorphism.

The only things that can be used are,

1. Definition of cyclic group

2. The theorem that every cyclic group is abelian.

3. The theorem that every subgroup of a cyclic group is cyclic.

Any help will be appreciated.

• – lhf Nov 11 '14 at 8:31

Any cyclic group has a $unique$ subgroup of the orders which are the divisors of its order. Since $gcd\ (n,s)=gcd\ (n,t)||G|$, hence there is a uniques subgroup in $G$ of the order $\frac {n}{gcd(n,s)}=\frac {n}{gcd(n,t)}$ and hence $\langle{a^s}\rangle$=$\langle{a^t}\rangle$.