Let $G=\langle a \rangle$. Show that the generators of $G$ are of the form $a^r$ where $\gcd(r,n)=1$

Question

Exercise :

Let $G=\langle a \rangle$ be a cyclic group of order $n \in \mathbb N$.

$(i) \space $ Show that $\langle a^s\rangle = \langle a^t \rangle$ if and only if $\gcd(s,n)=\gcd(t,n)$.

$(ii)$ Using $(i)$, show that the rest generators of $G$ are of the form $a^r$ where $\gcd(r,n)=1$

$(iii)$ Show that, for every divisor $d$ of $n$, the group $G=\langle a \rangle$ has a unique subgroup of order $d$ and that this is the only possible subgroup of $G$.

Discussion :

I have solved (correctly, I think) $(iii)$ as follows, but I'm totally stuck at $(i)$ and $(ii)$ as I don't seem to grasp how to start. There is a similar question for $(i)$ here, but do not rush to mark it as duplicate, as the solution discussed there uses the fact that is given to prove at $(iii)$ and since the exercise expects a different approach than taking advantage of the uniqueness.

For $(iii)$, I have showed the following :

We are given from the hypothesis of the exercise that the order of $G$ is $n \in \mathbb N \Rightarrow |G| = n$. Let $d$ be a divisor of $n$. Consider $H=\{ x \in G : x^d =1 \}$. Then $H$ is a subgroup of $G$ and $H$ contains all elements of $G$ that have order $d$.

If $K$ is a subgroup of $G$ of order $d$, then $K$ is cyclic, generated by an element of order $d$. Hence, $K\subseteq H$.

On the other hand, $x\in H$ iff $x=α^k$ with $0\le k < n$ and $α^{kd}=1$, where $α$ is a generator of $G$ as mentioned at the hypothesis. Hence, $kd=nt$ and so $k=(n/d) t$. The restriction $0\le k<n$ implies $0\le t<d$, and so $H$ has exactly $d$ elements. Therefore, $K=H$.

Question/Discussion : I would really appreciate if someone could lead me through $(i)$ and $(ii)$ since I don't seem to grasp how to approach them. Please correct me if I approached $(iii)$ wrongly though. I'm a beginner on our abstract algebra courses, so I'm sorry if this is considered an easy exercise.

Answer 1

Suppose that $\langle a^t\rangle = \langle a^s\rangle$. This means that for every $m\in\mathbb{N}$, there is a $k\in\mathbb{N}$ with $$a^{tm}=a^{sk}\implies tm\equiv sk\pmod n$$ and vice versa. Now, by Bezout's identity there exists an $m\in\mathbb{Z}$ such that: $$tm\equiv \gcd(t,n)\pmod n$$ Now, there must exist a $k$ such that: $$sk\equiv \gcd(t,n)\pmod n$$ However, since $\gcd(t,n)\mid n$, it follows that $\gcd(t,n)\mid sk$, so $\gcd(t,n)\mid\gcd(s,n)$. The same way, we can show that $\gcd(s,n)\mid \gcd(t,n)$ and we conclude that $\gcd(t,n)=\gcd(s,n)$.

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Now the other way around: Suppose that $\gcd(t,n)=\gcd(s,n)$. Define $d=\gcd(t,n)=\gcd(s,n)$. Write $$t=t'd,s=s'd$$ Clearly, $\gcd(n,t')=\gcd(n,s')=1$. Again, by Bezout's identity, there exists an $m\in\mathbb{N}$ such that: $$t'm\equiv 1\pmod n$$ So for all $l\in\mathbb{N}$, there exists an $m\in\mathbb{N}$ with $$t'm\equiv l\pmod n$$ multiplying both sides with $d$ shows that $tm$ can only be congruent to a multiple of $d$ modulo $n$ and that for each multiple of $d$, there exists an $m\in\mathbb{N}$ such that $tm$ is congruent to that multiple modulo $n$. The same goes for $sk$, so we have proven that for all $m\in\mathbb{N}$ there exists a $k\in\mathbb{N}$ such that $$tm\equiv sk\pmod n$$ and vice versa and that finishes the proof of (i).

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Using (i), (ii) is easy. Suppose $\langle a^r\rangle = G = \langle a^1\rangle$. By (i), we have: $$\gcd(r,n)=\gcd(1,n)=1$$

Answer 2

Let $|G| = n$. We'll prove the lemma $|a^d| = \frac{n}{\gcd(n,d)}$. To prove that we have:

$$\left(a^d\right)^{\frac{n}{\gcd(n,d)}} = \left(a^n\right)^{\frac{d} {\gcd(n,d)}}= \left(1\right)^{\frac{d} {\gcd(n,d)}} = 1 \implies |a^d| \le \frac{n}{\gcd(n,d)}$$

For the other direction we know that $\exists x,y \in \mathbb{Z}$ s.t. $nx + dy = \gcd(n,d)$. we have:

$$\left(a^{\gcd(n,d)}\right)^{|a^d|} = \left(a^n\right)^{x\cdot |a^d|} \cdot \left(a^d\right)^{|a^d| \cdot y} = 1 \implies \gcd(n,d)|a^d| \le n$$

This proves the lemma above.

For part $i)$ we have that if $\langle a^s \rangle = \langle a^t \rangle$, then both $a^s$ and $a^t$ are generators of the group and so $|a^s| = |a^t| \implies \frac{n}{\gcd(n,s)} = \frac{n}{\gcd(n,t)} \implies \gcd(n,s) = \text{gcd}(n,t)$. For the other direction note that:

$$\langle a^t \rangle = \langle a^{\gcd(n,t)} \rangle = \langle a^{\text{gcd}(n,t)} \rangle = \langle a^s \rangle $$

The part $ii)$ follows immediately from the lemma above, as $a^d$ is a generator of $\langle a \rangle$ if and only if $|a^d| = n \iff \frac{n}{\gcd(n,d)} =n \iff \gcd(n,d) = 1$