What are the subgroups of $C_2\times C_{202}$?

Question

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?

Answer 1

$\{0\}\times C_{202}$ 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 $\{(0,0),(1,101)\}$

Answer 2

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?

Answer 3

Please by all means if I'm missing any subgroups do correct me. Since I'm actually doing this myself for homework. Without necessarily defining an operation. Let's call "g" the generator for Cyclic group C₂, and the same way let's call "h" the generator for cyclic group C₂₀₂.

We have:
C₂ = {g⁰, g¹} and C₂₀₂ = {h⁰, h¹, h², h³, ... , h²⁰¹}, with g⁰ and h⁰ being the identities.
The subgroups for C₂ are {g⁰}, which is the trivial subgroup, and {g⁰, g¹}, which is the entire group.
The subgroups for C₂₀₂ are {h⁰}, {h⁰, h¹⁰¹}, {h⁰, h², h⁴, h⁶, ..., h²⁰⁰} and {h⁰, h¹, h², h³, ... , h²⁰¹}, of orders 1, 2, 101, 202 respectively, once again {h⁰} being the trivial subgroup and {h⁰, h¹, h², h³, ... , h²⁰¹} the entire group.

So we have the group C₂xC₂₀₂, which we can write as {(g^a, h^b)|a ∈ Z[0, 1] and b ∈ Z[0, 201]} or {(g⁰, h⁰), (g⁰, h¹), (g⁰, h²), ..., (g⁰, h²⁰¹), (g¹, h⁰), (g¹, h¹), (g¹, h²), ..., (g¹, h²⁰¹)} (All of this is not really necessary, but it helped me visualize the whole thing better). The subgroups of C₂xC₂₀₂ can be of orders 1, 2, 4, 101, or 202. We technically also have the entire group which is subgroup of itself and that would be of order 404.

Of order 1 we have the Trivial Subgroup which is the identity, so {(g⁰, h⁰)}
Of order 2 we have 3 subgroups {(g⁰, h⁰), (g⁰, h¹⁰¹)}, {(g⁰, h⁰), (g¹, h¹⁰¹)}, {(g⁰, h⁰), (g¹, h⁰)} which are the cyclic groups <(g⁰, h¹⁰¹)>, <(g¹, h¹⁰¹)> and <(g¹, h⁰)> respectively.
Of order 4 we have the non-cyclic subgroup {(g⁰, h⁰), (g⁰, h¹⁰¹), (g¹, h¹⁰¹), (g¹, h⁰)}
Of order 101 we have 1 cyclic subgroup <(g⁰, h²)> = {(g⁰, h⁰), (g⁰, h²), (g⁰, h⁴), ... , (g⁰, h²⁰⁰)}
Of order 202 we have 2 cyclic subgroups <(g⁰, h¹)> = {(g⁰, h⁰), (g⁰, h¹), (g⁰, h²), ..., (g⁰, h²⁰¹)} and <(g¹, h¹)> = {(g⁰, h⁰), (g¹, h¹), (g⁰, h²), (g¹, h³), (g⁰, h⁴), ..., (g¹, h²⁰¹)}
and finally we have the entire group which is a subgroup of itself (this is a non-cyclic group of order 404), C₂xC₂₀₂ = {(g^a, h^b)|a ∈ Z[0, 1] and b ∈ Z[0, 201]} = {(g⁰, h⁰), (g⁰, h¹), (g⁰, h²), ..., (g⁰, h²⁰¹), (g¹, h⁰), (g¹, h¹), (g¹, h²), ..., (g¹, h²⁰¹)}

Once again, if I did miss any subgroups please don't hesitate to correct me. I hope I was helpful.