Subgroups of $G=(\mathbb{Z}_{12},+)$ 
  
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*Draw the Hasse Diagram for the subgroup lattice of $G=(\mathbb{Z}_{12},+)$
  
*Identify the subgroups $K=\langle [6]\rangle$ and $L=\langle [9]\rangle$ on your diagram. What are $K\cap L$ and $\langle K,L\rangle$?
  

In solving the above problems from a practice test, I've noticed a couple of things that I'd like to have clarified:


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*All of the subgroups generated by $\langle[x]\rangle$ such that $gcd(x,12)=1$ have an order of 12 (they generate the whole set $\mathbb{Z}_{12}$). Is it generally true that any group $G=(\mathbb{Z}_i,+)$ is generated by $\langle [x]\rangle$ if $gcd(x,i)=1$? If so, under what theorem?

*It also seems that for all $x\leq 12$, $\langle[x]\rangle=\langle[12-x]\rangle$. For example, $\langle[4]\rangle=\langle[8]\rangle$ and $\langle[3]\rangle=\langle[9]\rangle$. Is this also true in general (that is, for any group $G=(\mathbb{Z}_i,+)$), or is this just serendipitous?

*Since $\langle[6]\rangle\subset\langle[9]\rangle=\langle[3]\rangle$, I'm surmising that $K\cap L=\langle[6]\rangle$. Is this correct?

*Does $\langle K,L\rangle=K\cup L$ in the case where $K\subseteq L$ or $L\subseteq K$? In this case, it would appear so, since $K=\{[0],[6]\}$ and $L=\{[0],[3],[6],[9]\}$, and $\langle K,L\rangle=\{[3]^0,[3]^1,[3]^2,[3]^3\}$

 A: *

*Yes, it is generally true. Assume that GCD$(x,i)=1$. Then there exist $a,b\in\mathbb{Z}$ such that $$ax+yi=1.$$ If we multiplay above equation with $k\in\mathbb{Z_i}$, we get $$axk+kiy=k,$$ so
$$axk\equiv k(mod i).$$
From there you conclude that there exists $n\in\mathbb{N} (n=xk)$ such that $an=k, \forall k\in \mathbb{Z_i}$ which means $\langle x\rangle =\mathbb{Z_i}$.

*Let us take a look at $\mathbb{Z_i}$. From $0\in \langle x\rangle $ and $-x\in \langle x\rangle $ we get $i-x\in \langle x\rangle $, so $\langle i- x\rangle \subseteq \langle x\rangle $.
On the other side $(i-1)(i-x)=(i-1)(-x)=x\in \langle x\rangle $, so $\langle x\rangle \subseteq \langle i-x\rangle$. From these two follows equality.

*Yes, it is correct.

*If $K\subseteq L$, then $\langle K, L\rangle =\langle L\rangle$.
A: The order of an element in this subgroup depends upon the divisors is shares with 12. If the greatest common divisor of x and 12 is 1, then you're going to have to add x to itself 12 times ($12x$) to get a multiple of 12 i.e. the identity in the group. This is precisely what it means to have order 12 i.e. the order of the subgroup it generates is 12. 
This argument works in general for $\mathbb{Z}/n\mathbb{Z}$. 
What you are doing by saying $\gcd(x,12)=1$ is saying the order of $x$ does not (properly) divide $12$ (the order of the group). So by Lagrange's theorem, $x$ cannot generate a proper subgroup. 
Your second point follows from the definition of a cyclic subgroup. The subgroup $\langle x \rangle$ contains elements of the form $ax\mod 12$, where $a \in \mathbb{N}$. While elements in $\langle 12 - x \rangle$ contains elements of the form $a(12-x) = 12a -ax \equiv -ax$ mod 12. However, it should be clear that a subgroup generated by the inverse of an element is the same as the subgroup generated by the element. Hence $x$ and $12-x$ will always generate the same subgroup. 
Your third point is correct. 
