Finding $\langle a_1, a_2, ..., a_n \rangle$ in $\mathbb{Z}_{20}$ I realize a variation of this problem (probably in fact the same textbook problem from Gallian's text p.72 #40) has been asked here and here but my understanding is still a little cloudy. 
I am asked to find $\langle m,n\rangle$ in $\mathbb{Z}_{20}$ as well as $\langle l,m,n\rangle$. 
As a concrete example, let $m=6,n=15$, so we are trying to find $\langle 6,15\rangle$ in $\mathbb{Z}_{20}$ . My first thought was appealing to the definition: $$\langle h,k\rangle:= \{h^n,k^n|n\in\mathbb{Z}\}$$ where $h^n,k^n$ denotes repeated addition. So $$\langle6,15\rangle=\{6n,15n,9n,21n\space|n\in\mathbb{Z}\}$$
But I found in the first linked post that $\langle a_1, a_2, ..., a_n \rangle = \langle \operatorname{gcd}(a_1, a_2, ..., a_n) \rangle$ so instead $\langle 6,15\rangle=\langle3\rangle. $
Is there a relatively easy proof of the latter fact as well as intuition as to why it's true? 

In $\Bbb Z$, it is true that $\langle a_1, a_2, \ldots, a_n\rangle = \langle \gcd(a_1, a_2, \ldots, a_n)\rangle$. This follows from Bezout's identity in the case of $n = 2$, and induction for $n > 2$.
In $\Bbb Z_k$, you actually have $\langle a_1, a_2, \ldots, a_n\rangle = \langle \gcd(a_1, a_2, \ldots, a_n, k)\rangle$. For instance, in $\Bbb Z_{20}$ we have $\langle6, 15\rangle = \langle 1\rangle$. To see this explicitly, note that $6+15 = 1$, so $1\in \langle 6, 15\rangle$. And clearly $6, 15\in\langle 1\rangle$. Since the two subgroups contain eachother's generators, they must be the same.
One way to prove $\langle a_1, a_2, \ldots, a_n\rangle = \langle \gcd(a_1, a_2, \ldots, a_n, k)\rangle$ is to use the result from $\Bbb Z$, and use the fact that there is a natural bijective correspondence between subgroups of $\Bbb Z_k$ and subgroups of $\Bbb Z$ which contain $\langle k\rangle$.
