Find all subgroups of $\mathbb{Z_7}\times\mathbb{Z_5}$ Find all subgroups of $\mathbb{Z_7}\times\mathbb{Z_5}$ without repeating the same subgroup. You should specify a generator of each subgroup.
I understand the question. However the only way I know how to find all the subgroups is to individually calculate the subgroup generated by each element of $\mathbb{Z_7}\times\mathbb{Z_5}$. That process is long, annoying and will definitely turn up some identical subgroups. Is there some method for doing this that is quicker/more concise? I suspect it has something to do with prime numbers but we've never actually covered a quicker method in class.
 A: The method of writing down the subgroup generated by each element is not as long a process as you anticipate, because it is fairly easy to recognize repeated subgroups. For that we use a consequence of Lagrange theorem wherein any non-identity element of $\mathbb{Z}_7$ generates the whole of $\mathbb{Z}_7$. Similarly for $\mathbb{Z}_5$.
Thus, we see that $(0, 0)$ generates the trivial subgroup. Next, $(0, k)$  for $k\in\{1, \dots, 4\}$ generates $\{0\}\times\mathbb{Z}_5$. Similarly, $(k, 0)$ for $k\in\{1, \dots, 6\}$ generates $\mathbb{Z}_7\times\{0\}$. Finally, $(i, j)$ for $i\in\{1, \dots, 6\}$ and $i\in\{1, \dots, 4\}$ generates $\mathbb{Z}_7\times \mathbb{Z}_5$, because $5$ and $7$ are coprime.
We conclude that $\mathbb{Z}_7\times \mathbb{Z}_5$ has four subgroups: $\{(0, 0)\}$, $\{0\}\times\mathbb{Z}_5$, $\mathbb{Z}_7\times\{0\}$ and $\mathbb{Z}_7\times \mathbb{Z}_5$.
A: $\mathbb Z_7×\mathbb Z_5\cong\mathbb Z_{35} $.
And for a cyclic group,  the subgroups are the cyclic groups of all orders dividing the order of the group.
That gives us the trivial group,  $\mathbb Z_5,\mathbb Z_7$, and the whole group.
For generators, use a generator, say $(1,1) $ for $\Bbb Z_{35} $.  Then $(0,0),(7,7),(5,5) $ and $(1,1) $ will do.
