Hint for another basic group problem

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[gusty kiln]

Conceptually I understand that if G is cyclic, then G = {e, a, a^2, a^3}. In order for G to be isomorphic to D2 then e = a^2, which contradicts the fact that G is of order 4. But I'm not sure how to start an actual proof.

[gusty kiln]

@dense isle it's not for this specific problem but for two cyclic groups Cm and Cn, I want to show that Cm x Cn = Cmn is cyclic

[dense isle]

I mean in this case you can manually characterize all groups of order 4

[gusty kiln]

What do you mean?

Oh and m and n are coprime

[dense isle]

There's only 2 groups of order 4

One is cyclic and one is not

[gusty kiln]

C4 and D2 right

[dense isle]

Yea I had to google to check notation of D2 yes that's Klein 4

You can assume cyclic and see what structure u get assume non cyclic and see what structure u get

Cyclic means existence of element of order 4

Non cyclic means nonexistence of element of order 4 but order must divide grp so only 1 and 2 orders can exist

Go ahead with that

[dense isle]

This is true only if n and m are coprime that should be a hint for u to start

Think about he Cartesian product Cm X Cn =(a,b) a in Cm and b in Cn

[gusty kiln]

This is what I was confused about, I understand that (a,b) generates a cyclic group, but I'm not sure what (a,b)^{some exponent} means

And just for context this is the whole problem I'm trying to do

[dense isle]

(a,b)^x = (a^x,b^x)

That's essentially the defn but maybe you'll have to show that's true for all integers positive negative and 0 with induction

As (a,b)^2=(a,b) * (a,b)=a^2,b^2) for example

[gusty kiln]

I see ... I don't think I have to prove this definition

[gusty kiln]

I think in this part of the proof, the textbook recommends using chinese remainder theorem

But i'm not sure how to apply CRT here

[dense isle]

Hmm you really don't need crt here

Notice you need (a,b) to run across the entire group if they have a common factor then at that order it will cycle back so it won't generate the whole group

That's basically all it is

[dense isle]

It's instructive to do so but yea essentially trivial

[gusty kiln]

I see, but the textbook hints that I need CRT at some point, for reference this is what it says

[dense isle]

Hmm i mean that seems a bit redundant to me tbh

It says it "also" follows from crt

You can think of it as a special case of it

Crt itself is stronger tho

[dense isle]

For this you don't need crt you can prove it directly using this idea

[gusty kiln]

Ok, I’ll try to do that

[rancid raft]

The main approach here, I'll sketch out. Essentially, we can identify two generators of the group $C_n$ as $g_1$ and $C_m$ as $g_2$, which are given by $C_n = {e, g_1, g_1^2, \dots}$ and $C_n = {e, g_2, g_2^2, \dots}$ respectively. Consider the element of the group $(g_1, g_2) \in C_m \times C_n$. Since $\gcd(m,n) = 1$, we can apply Bezout's lemma, generalized, to state there are integers $xm + yn = d$ for any natural $d$. Consider the homomorphism $\varphi(g_1^x,g_2^y) = g_3^{mx + ny}$, where $g_3$ generates $C_{mn}$. Then one just proves this is an isomorphism.

[tepid tendon]

Magpie (N,^) #Magma4Member

[rancid raft]

Essentially the only difference between m,n being coprime or not is simply whether phi can actually be a surjection.

[gusty kiln]

In phi, shouldn't the exponents of g1 and g2 be n and m? Or is x and y right

[rancid raft]

x and y is right, else it couldn’t span the entire group C_mn

[gusty kiln]

Oh, idk why I always assumed they would be m and n

That helps a lot

[sturdy flint]

How come Bezout lemma is even here as wlel

[rancid raft]

Bezout was a chad

[sturdy flint]

I cannot accept that

[rancid raft]

you'll have to accept it

[gusty kiln]

For injectivity, do I need to use the fact that if $g_3^{mx+ny} = g_3^{mx'+ny'}$, then $x \equiv x' \mod m$

[tepid tendon]

UrgleMcPurfle

[rancid raft]

yes

[gusty kiln]

Do i need to prove that any further, or can I just leave it there?

It seems sort of self evident but idk what I can and can't assume

[rancid raft]

I mean, that's probably good enough to be frank

[gusty kiln]

I think I'm done w the problem and I understand in hindsight why the proof works, but (this is a dumb question) do I need to justify how we get phi in the first place

Or for future problems like this can I just define the function however I want

[rancid raft]

you can just choose any function as long as it satisfies isomorphism conditions

just choose a function and prove it does

that's the simplest route

usually when that gets laborious you switch to isomorphism theorems

where you know there is an isomorphism but you don't have to construct it

[gusty kiln]

Oh ok

[gusty kiln]

-close

+close