Is the Injective Hull of an Irreducible Module in Group K?

[TMO]

Let's suppose that I have an element $e$ of order $p$ in the group of complex numbers whose elements all have order $p^n$ for some $n\in\mathbb{N}$ (henceforth called $K$), and the module generated by $(e)$ is irreducible.

How do I show that the injective hull of the module generated by $(e)$ is in fact, equal to $K$?

Attempted Work. I was told that this submodule generated by $(e)$ is isomorphic to $\mathbb{Z}[p^{-1}]$. I don't know how to proceed from there... but I think it involves showing that two maps are surjective.

 

[fresh_42]

Let's suppose that I have an element $e$ of order $p$ in the group of complex numbers whose elements all have order $p^n$ for some $n\in\mathbb{N}$ (henceforth called $K$), and the module generated by $(e)$ is irreducible.

How do I show that the injective hull of the module generated by $(e)$ is in fact, equal to $K$?

Attempted Work. I was told that this submodule generated by $(e)$ is isomorphic to $\mathbb{Z}[p^{-1}]$. I don't know how to proceed from there... but I think it involves showing that two maps are surjective.

What is the module operation? Module above what?

 

[mathwonk]

do you know what a divisible group is? and how it relates to injectivity? it seems obvious in your example that to make your subgroup divisible, you would need to include everything in K.

https://en.wikipedia.org/wiki/Divisible_group

 

[TMO]

do you know what a divisible group is? and how it relates to injectivity?

I think there is a theorem that for a principle ideal domain X, X is divisible iff. X is injective. Tell me, is it iff. or if?

it seems obvious in your example that to make your subgroup divisible, you would need to include everything in K.

https://en.wikipedia.org/wiki/Divisible_group

It isn't obvious to me. Could you explain why?

 

[mathwonk]

divisibility involves the presence of roots, and in your example everything is a root of your basic elements. i.e. everything in K becomes an element of your small subgroup after raising to a high power. note that your group is written multiplicatively and the definition for abelian groups is usually stated additively so that means when they say a group is divisible if na belonging to the goup forces a to also belong, means that if a^n belongs then a also belongs.
try to relax over this stuff. just try to actually learn what some of these things are, if you don't finish everything in a certain time, that matters less than actually learning some of it so you don't have to repeat that part.

 

[mathwonk]

i also read that for an abelian group divisible is equivalent to injective. but we should try to understand this. i.e. try proving that injective implies divisible say.
recall that injectivity means that all maps into it extend. so take e.g. the map of the subgroup nZ into an abelian group. If we want to extend this to all of Z, we are going to need to send 1 to 1/n times whatever n went to, so we have to be able to divide by n in the target group, or in multiplicative notation, to take nth roots.

so that direction looks easy. the other direction is probably harder and may even use zorns lemma or some such nonsense. try it for finite abelian groups, i.e. direct sums of cyclic groups. or first just for cyclic groups like Z, as I just used. just try to get a feel for it.

have you had the basic courses on abstract algebra? e.g. have you done the classification of finite abelian groups?