If $G$ is a group and $S$ is a subset of $G$, $\langle S\rangle$, the subgroup generated by $S$, is simply the smallest subgroup of $G$ that contains every element of $S$. It can be defined as $\bigcap\{H\leqslant G:S\subseteq H\}$, the intersection of all subgroups of $G$ containing $S$ as a subset. However, it’s probably more instructive to look at it a bit differently.
Consider the subset $S=\{4,6\}$ of $\Bbb Z_{12}$. $S$ is not a subgroup of $\Bbb Z_{12}$: it’s not closed under addition, it doesn’t contain the identity element, and it doesn’t contain the additive inverse of $4$. What, at a bare minimum, do we have to add to $S$ to turn it into a subgroup of $\Bbb Z_{12}$? We certainly have to throw in $-4=8$, though $-6=6$ is already in $S$; that expands $S$ to $\{4,6,8\}$. We also have to close the set under addition, so we have to throw in $4+6=10$, $4+8=0$, and $6+8=2$, bringing the set up to $\{0,2,4,6,8,10\}$. This is a subgroup of $\Bbb Z_{12}$, so nothing more need be added: $\big\langle\{4,6\}\big\rangle=\{0,2,4,6,8,10\}\;.$
$S=\{4,6\}$ generates the subgroup $\{0,2,4,6,8,10\}$ in the sense that we can work out from $S$ as I did above, adding elements as we discover that they’re needed if we’re to have a group, until we’ve added just enough to get a group, and the process ends.
Finding $\langle S\rangle$ when $S$ is a subset of one of the cyclic groups $Z_n$ turns out to be especially easy. That’s where the greatest common divisor comes in. Notice that in my example the subgroup of $\Bbb Z_{12}$ generated by $\{4,6\}$ turned out to be the set of multiples of $2$ in $\Bbb Z_{12}$, which is easily seen to be the subgroup of $\Bbb Z_{12}$ generated by the single element $2$: once you have $2$, you must have $2+2=4$, $4+2=6$, $6+2=8$, $8+2=10$, and $10+2=0$, and these six elements do indeed form a subgroup of $\Bbb Z_{12}$. It’s not an accident that $2=\gcd\{4,6,12\}$; in general is you have a subset $S\subseteq Z_n$, you’ll find that $\langle S\rangle$ is just the set of multiples (in $\Bbb Z_n$) of the gcd of $n$ and the members of $S$. This is a matter of elementary number theory; I suspect that it’s been proved either in class or in your text.