What are the subgroups of $C_2\times C_{202}$? Basically, if $C_2$ is a cyclic group of order two and $C_{202}$ is the same with order $202$, what are the subgroups of the product of the two?
The farthest I've gotten is that if $C_2$ is addition mod 2 and $C_{202}$ is addition mod 202, one subgroup is (0, c) and (1, c) where c is in $C_{202}$. With Lagrange's theorem, $C_{202}$ would also have subgroups sized 2 and 101, so would the same apply to them?
 A: Here is a suggested systematic way of thinking about the problem:
Think about $C_2$ and $C_{202}$ as additive groups, each having identity $0$ and generator $1$ (there may be other generators of course).
Then the subgroups include the whole group, generated by $(0,1)$ and $(1,0)$ - it doesn't have a single generator because it isn't cyclic. We can tell this because $202\cdot (a,b)=(202a, 202b)=(0,0)$ - so every element of the group has an order which is a factor of $202$ and none can therefore have order $404$.
Then there is the trivial subgroup generated by $(0,0)$ which has order $1$.
Now the order of a subgroup is a factor of the order of the original group, and the possibilities for proper subgroups are therefore $202, 101, 4, 2$
Now check each of these orders (I'll leave some bits for you to check):
A group of order $202$ is cyclic (can you show this?), so you need to find a generator. Either $(0,n)$ will serve if $n$ is coprime to $202$ or $(1,m)$ will have order $202$ provided $m$ has order $101$ or $202$ in $C_{202}$ - how many different subgroups does this give you, bearing in mind that the same group can have a number of different generators?
Since $101$ is prime, any subgroup of this order will be cyclic. Can you see that the elements of the original group which have order $101$ have form $(0,2n)$? How many groups does this give you?
A subgroup of order $2$ is cyclic, and will be generated by an element of order $2$ in the original group - how many of these can you find?
A subgroup of order $4$ cannot be cyclic, because the order of any element must be a factor of $202$, so there can be no element of order $4$ - so any subgroup of order $4$ must be of type $C_2\times C_2$. Since the original group is $C_2\times C_{202}$ and we can see $C_2$ inside $C_{202}$, there ought to be a subgroup of this form (also by Sylow's Theorem, when you reach it). Any such subgroup will consist of the identity, plus three elements of order two - how many can you find, and is this consistent with your identification of subgroups of order 2?
A: $\{0\}\times C_{202}$ is a subgroup, but not $\{1\}\times C_{202}$.
$C_2\times{0}$ is also a subgroup, and $C_2\times 2.C_{202}$ and ${0}\times2.C_{202}$ too.
More generally, if $H$ and $K$ are subgroups of $C_2$ and $C_{202}$, $H\times K$ is a subgroup of $C_2\times C_{202}$
But all the subgroups aren't of this form. Look for example at $\{(0,0),(1,101)\}$
