Say we have the Group $(\Bbb Z_3 \times \Bbb Z_4,+)$, this group has the following elements;
$\{(0,0)(0,1)(0,2)(0,3)(1,0)(1,1)(1,2)(1,3)(2,0)(2,1)(2,2)(2,3)\}$ and so is a group of order 12. Therefore ( I believe) that the subgroups are found by looking for groups of order 1,2,3,4,6,12.
We can also note that $|\langle g \rangle|=o(g)$(i.e the order of the element g)
$\therefore \langle(0,1)\rangle:=\{(0,1)(0,2)(0,3),(0,0)\}$ is a subgroup as $o((0,1))=4$ and $|\langle(0,1)\rangle|=4$
$\langle(2,3)\rangle:=\{(2,3),(1,2),(0,1),(2,0),(1,3),(0,2),(2,1),(1,0),(0,3),(2,2)(1,1)(0,0)\}$ has order 12 and so is also a subgroup.
When I was going though the list of elements in $\Bbb Z_3 \times \Bbb Z_4$ It seemed like they all generated a group of order 1,2,3,4,6 or 12, Is this true ?
Also if my method for computing subgroups is correct, is there any other ways to find subgroups apart from using this generator technique ?
Since $\gcd(3,4)=1$, we have $\Bbb Z_3\times \Bbb Z_4\cong \Bbb Z_{12}$ (by the Chinese Remainder Theorem), so the group is cyclic and so, hence, all its subgroups are cyclic; it is then a matter of finding which elements of $\Bbb Z_{12}$ generate a proper subgroup then seeing which of these is distinct from any others, a significantly easier task (in my opinion).