I don't know to find subgroups. $G=\mathbb Z_3\times\mathbb Z_5$
I don't know to find a subgroup.
Give me an example a subgroup of $G$ how to find that ?
 A: There are exactly 4 subgroups


*

*$G$ itself is a subgroup of $G$

*$\{(0,0)\}$ is a subgroup of $G$.

*$\mathbb{Z}_3\times\{0\}$ is a subgroup of $G$

*$\{0\}\times\mathbb{Z}_5$ is a subgroup of $G$.
There are no more subgroups of $G$.
How we find those? The group $G$ is a cyclic group because it is generated by $(1,1)$, hence every subgroup is a cyclic group as well. So you just take an element in $a\in G$ and you have that $\left<a\right>$ is a subgroup of $G$ (and every subgroup is of that form).
$G$ has 15 elements, but some of the elements define the same group (for example $\left<(1,0)\right> = \left<(2,0)\right> = \mathbb{Z}_3\times\{0\}$.
You can go over all elements and confirm that those 4 subgroups I mentioned before are the only subgroups of $G$.
A: Because $(3,5)=1$, $G$ is cuclic and $([1]_3,[1]_5)$ is a generator of $G$.
Now as I commented for every positive $n| |G|=15$ there exists a subgroup of $G$ the one that is generated by $([1]_3,[1]_5)^n$.So we have:
$1)$ for $n=1$:$<([1]_3,[1]_5)>=G$
$2)$ for $n=3$ :$<([0]_3,[3]_5)>=\{0\}\times\mathbb{Z_5}$
$3)$for $n=5$:$<([2]_3,[0]_5)>=\mathbb{Z_3}\times\{0\}$
$4)$for $n=15$:$<([0]_3,[0]_5)>=\{0,0\}$
