# Equality of subgroups of finite cyclic groups

My abstract algebra text has the following proof of the statement, "Let $G$ be a cyclic group with $n$ elements and generated by $a.$ Then $\langle a^s \rangle = \langle a^t \rangle \iff \gcd(s,n) = \gcd(t,n).$"

Proof: Taking for the moment $\mathbb Z_n$ as a model for a cyclic group of order $n,$ we see that if $d$ is a divisor of $n,$ then the cyclic subgroup $\langle d \rangle$ of $\mathbb Z_n$ has $n/d$ elements, and contains all the positive integers $m$ less than $n$ such that $\gcd(m,n) = d.$ Thus there is only one subgroup of $\mathbb Z_n$ of order $n/d$. Combined with our above result regarding the order of subgroups of a cyclic group, this shows at once that if $a$ is a generator of the cyclic group $G,$ then $\langle a^s \rangle = \langle a^t \rangle$ iff $\gcd(s,n) = \gcd(t,n).$

What is confusing me about this proof is the assertion that if $d$ is a divisor of $n,$ then the cyclic subgroup $\langle d \rangle$ of $\mathbb Z_n$ contains all the positive integers $m$ less than $n$ such that $\gcd(m,n) = d.$ For example, let $n = 24$ and $d = 6.$ Then $\langle 6 \rangle = \{0,6,12,18\}$, but $\gcd(12,24) = 12 \neq 6.$ Isn't this a contradiction of the assertion in the book?

No, since $\langle 12 \rangle \neq \langle 6 \rangle$.
On the other hand $\langle 18 \rangle = \langle 6 \rangle$ because $(6,24)=(18,24)$
• What would the elements of $\langle 6 \rangle$ be? Was I correct in listing those? Sep 26, 2014 at 18:04
• Okay, but then according to the book's statement that "if $d$ is a divisor of $n,$ then the cyclic subgroup $\langle d \rangle$ of $\mathbb Z_n$ contains all the positive integers $m$ less than $n$ such that $\gcd(m,n) = d,$" I think it would follow that $\langle 6 \rangle = \{0=24, 6, 18\}.$ Did they mean to write simply that $\langle d \rangle$ contains all the positive integers $m$ less than $n$ such that $d|m?$ Sep 26, 2014 at 18:13
• Well, it says it contains, but doesn't say it is equal to it. That is, $$gcd(m,n)=d \implies m\in \langle d\rangle$$ Saying it contains all $m$ with $d|m$ is more or less definition of $\langle d\rangle$ Sep 26, 2014 at 18:19