# Prove that in $(\mathbb{Z}_n,+): grp\{(x)\} = grp\{y\} \iff gcd(n,x)=gcd(n,y)$ [closed]

I am trying to prove that in $$(\mathbb{Z}_n,+): grp\{(x)\} = grp\{y\} \iff gcd(n,x)=gcd(n,y)$$

I came up with the following, so far:

$$\implies$$ $$grp\{x\}=grp{y} \iff y \in grp\{x\}\implies \exists p\in \mathbb{Z}: y=px$$ And also the $$ord(x)=ord(y)$$ Then is $$gcd(y,n)=gcd(px,n)=gcd(x,n)$$ because $$p \nmid n$$

$$\impliedby$$ $$ggd(x,n)=d=ggd(y,n)\implies d|x,d|y,d|n$$ $$\implies \exists c\in \mathbb{Z}: cd=n \implies grp\{d\} \subseteq (\mathbb{Z}_n,+)$$

$$\subseteq$$

$$\forall x_1 \in grp\{x\}: \exists e,l \in \mathbb{Z}: x_1=ex=eld\subseteq grp\{d\}$$ (Because $$d|x\implies \exists l\in\mathbb{Z}: x=ld\implies x\in grp\{d\}$$) Analogue for $$grp\{y\}$$

$$\supseteq$$

I thought at defining an bijection $$f_x: grp\{x\} \to grp\{d\}$$ And analogously for $$f-y:grp\{y\}\to grp\{d\}$$ And prove like that that $$grp\{x\}=grp\{d\}=grp\{y\}$$

I know my proof is flawed and I would appreciate if someone could help me further with it. Many thanks in advance.

• What do you mean by, for example, $grp\{ (x)\}$? Commented Jan 19, 2023 at 11:50
• @Shaun I meant with $grp\{x\}$ the cyclic group which is generated by $x$ I saw just now, that I wrote sometimes $grp\{x\}$ and sometimes $grp\{(x)\}$ although I meant the same.
– T_B
Commented Jan 19, 2023 at 12:03

The first step is wrong. $$\langle x\rangle =\langle y\rangle$$ is not equivalent to $$y\in \langle x\rangle .$$
Here's a hint: $$\langle x\rangle =\langle y\rangle \iff\lvert x\rvert =\lvert y\rvert.$$ This is because cyclic groups have a unique subgroup of each allowable (Lagrange) order.
But, $$\lvert x\rvert =n/(x,n).$$
• thank you for your correction. What dies $|x|$ mean? And is there a proof available of $\<x\>=\<y\> \iff |x|=|y|$? Thanks again
• $\lvert x\rvert$ is the order of $x.$. I'm using the fact that there's a unique subgroup of each order dividing the order of a cyclic group. It isn't that hard to prove. Commented Jan 19, 2023 at 17:28