Question about normal subgroup

[Bruno Cavalcante]

Show that If N is a normal subgroup of G, a ∈ G and n ∈ N, so there is an element n' ∈ N such as an = n'a.

Can someone help me with this exercise, please? From the definition of normal subgroups, I know that aN = Na, for every a ∈ G. But I can't see how an = n'a (assuming n different from n') can happen.

 

[pka]

Show that If N is a normal subgroup of G, a ∈ G and n ∈ N, so there is an element n' ∈ N such as an = n'a. From the definition of normal subgroups, I know that aN = Na, for every a ∈ G.

If \((\forall a\in\mathcal{G})[a\mathcal{N}=\mathcal{N}a]\) then surely that means \(an\in\mathcal{N}a~?\)
What does it mean for \(x\in\mathcal{N}a~?\)

 

[Bruno Cavalcante]

If \((\forall a\in\mathcal{G})[a\mathcal{N}=\mathcal{N}a]\) then surely that means \(an\in\mathcal{N}a~?\)
What does it mean for \(x\in\mathcal{N}a~?\)

x = an, I supose.

 

[pka]

x = an, I supose.

Not formally. It means \((\exists n'\in\mathcal{N})[x=n'a]\)
So \(an\in a\mathcal{N}=\mathcal{N}a\Rightarrow (\exists n'\in\mathcal{N}[n'a=an]\)

 

[Singleton]

If I understand you correctly, you meant to say: "show there exists n' ...
not "so there exists n' ...

What can you say about ana^-1 ?

 

[Bruno Cavalcante]

If I understand you correctly, you meant to say: "show there exists n' ...
not "so there exists n' ...

What can you say about ana^-1 ?

I meant "then there is n' ...". Maybe I mistranslated the exercise, I'm not a native english speaker. My apologies.
I believe pka gave the formal answer, but I'm still trying to understand it.
Sure, n' = ana^-1 would be an answer, since ana^-1a = an, so n'a = an.
But I believe this only works for non-commutative grups.
For commutative groups, n' = ana^-1 implies n' = n.
I'm still a beginner at this topic and thank you both for your help.

 

[Bruno Cavalcante]

Not formally. It means \((\exists n'\in\mathcal{N})[x=n'a]\)
So \(an\in a\mathcal{N}=\mathcal{N}a\Rightarrow (\exists n'\in\mathcal{N}[n'a=an]\)

Thank you. I arranged it this way:

If [MATH]\mathcal{N}[/MATH] is a normal subgroup of [MATH]\mathcal{G}[/MATH], then [MATH](\forall a\in\mathcal{G})[a\mathcal{N}=\mathcal{N}a][/MATH].

So, [MATH]an\in\mathcal{N}a[/MATH]. But [MATH]\mathcal{N}a =[/MATH] {[MATH]na : n\in\mathcal{N}, a \in\mathcal{G}[/MATH]}.

Therefore, [MATH]\exists n'\in\mathcal{N}[/MATH] such as [MATH]n'a \in\mathcal{N}a[/MATH].

So, [MATH]an\in a\mathcal{N}=\mathcal{N}a\Rightarrow (\exists n'\in\mathcal{N})[n'a=an][/MATH].

 

[pka]

Thank you. I arranged it this way:

If [MATH]\mathcal{N}[/MATH] is a normal subgroup of [MATH]\mathcal{G}[/MATH], then [MATH](\forall a\in\mathcal{G})[a\mathcal{N}=\mathcal{N}a][/MATH].

So, [MATH]an\in\mathcal{N}a[/MATH]. But [MATH]\mathcal{N}a =[/MATH] {[MATH]na : n\in\mathcal{N}, a \in\mathcal{G}[/MATH]}.

Therefore, [MATH]\exists n'\in\mathcal{N}[/MATH] such as [MATH]n'a \in\mathcal{N}a[/MATH].

So, [MATH]an\in a\mathcal{N}=\mathcal{N}a\Rightarrow (\exists n'\in\mathcal{N})[n'a=an][/MATH].

Nice, spot on!

 

[Singleton]

I would think this exercise is asking you to prove
aN=Na from the definition of Normal subgroup. This part takes a little work.
What you did is read off the meaning of aN=Na which is rather immediate but can be a class exercise too, I guess.

 

[pka]

I would think this exercise is asking you to prove
aN=Na from the definition of Normal subgroup.

To: Singleton, Did you read post #1? In it the original poster says "I know that aN = Na, for every a ∈ G. "
Therefore it is highly unlikely that the question is asking for that to be reinvented.

 

[Singleton]

Hi pka, yes I read that. I think it is likely for students to assume the very thing they are asked to prove. This is sometimes a miscommunication from the instructor. In other words, I'm second guessing the poster. I know what he is asking and I know that you answered based on his assumptions. I am suggesting the exercise may not be what he thinks it is.

 

[pka]

Hi pka, yes I read that. I think it is likely for students to assume the very thing they are asked to prove. This is sometimes a miscommunication from the instructor. In other words, I'm second guessing the poster. I know what he is asking and I know that you answered based on his assumptions. I am suggesting the exercise may not be what he thinks it is.

Well O.K., I guess my having taught group theory many times (undergraduate & graduate) makes me assume that I know what is meant. Sorry.

 

[Singleton]

Well O.K., I guess my having taught group theory many times (undergraduate & graduate) makes me assume that I know what is meant. Sorry.

Yes I'm sure you know best. I only tossed out a possibility to the poster to help him. It wasn't directed at you, as I didn't read your posts before me. And yes, I had assumed we were all University mathematics professors.

 

[Bruno Cavalcante]

I would think this exercise is asking you to prove
aN=Na from the definition of Normal subgroup. This part takes a little work.
What you did is read off the meaning of aN=Na which is rather immediate but can be a class exercise too, I guess.

The original text of the exercise only consisted of the first line of post#1 (with that little correction: "then" instead of "so"), and it's on a chapter about normal subgroups, so I believe we can use the definition (aN = Na) without proving it (I didn't even know there was such a prove).

Definitly, I don't fully understand the implications of this exercise and need more thinking/reading about it. I'm going to discuss it further when my classes resume from the quarantine.

Anyways, I'm really thankful for the help in this forum.

Thank you. I arranged it this way:

If [MATH]\mathcal{N}[/MATH] is a normal subgroup of [MATH]\mathcal{G}[/MATH], then [MATH](\forall a\in\mathcal{G})[a\mathcal{N}=\mathcal{N}a][/MATH].

So, [MATH]an\in\mathcal{N}a[/MATH]. But [MATH]\mathcal{N}a =[/MATH] {[MATH]na : n\in\mathcal{N}, a \in\mathcal{G}[/MATH]}.

Therefore, [MATH]\exists n'\in\mathcal{N}[/MATH] such as [MATH]n'a \in\mathcal{N}a[/MATH].

So, [MATH]an\in a\mathcal{N}=\mathcal{N}a\Rightarrow (\exists n'\in\mathcal{N})[n'a=an][/MATH].

I want to correct the second line of this demonstration. I don't think [MATH]\mathcal{N}a[/MATH] is correctly described as

[MATH]\mathcal{N}a =[/MATH] {[MATH]na : n\in\mathcal{N}, a \in\mathcal{G}[/MATH]}.

A more precise description would be (rewriting the whole line):

So, [MATH]an\in\mathcal{N}a[/MATH]. But, for each [MATH]a\in\mathcal{G}[/MATH], [MATH]\mathcal{N}a =[/MATH] {[MATH]na : n\in\mathcal{N}[/MATH]}.

 

[Singleton]

The original text of the exercise only consisted of the first line of post#1 (with that little correction: "then" instead of "so"), and it's on a chapter about normal subgroups, so I believe we can use the definition (aN = Na) without proving it (I didn't even know there was such a prove).

If you took aN=Na to be the definition then you're done. You don't prove definitions.