Normality of a quotient group

  • Thread starter Hjensen
  • Start date
23
0

Main Question or Discussion Point

I have a question I need to resolve before my exam on thursday. It relates to the following result:

Let [itex]N[/itex] be a normal subgroup of [itex]G[/itex], and let [itex]K[/itex] be any subgroup of [itex]G[/itex] containing [itex]N[/itex]. Then [itex]K/N[/itex] is a subgroup of [itex]G/N[/itex]. Furthermore, [itex]K/N[/itex] is normal in [itex]G/N[/itex] if and only if [itex]K[/itex] is normal in [itex]G[/itex].

The first part is rather simple, and half of the other statement is just from the third isomorphism theorem. What I want to prove is, that [itex]K/N[/itex] normal in [itex]G/N[/itex] implies that [itex]K[/itex] is normal in [itex]G[/itex]. I suppose I could define a homomorphism like

[itex]G\rightarrow G/N\rightarrow (G/N)/(K/N)[/itex]

with kernel [itex]K[/itex]. That just seems like a lot of work to prove something which is probably rather simple. If I have to go through this at my exam, I'd prefer not to spend much time on this particular result. Does anyone have an idea for a short proof?
 

Answers and Replies

21,992
3,274
You need to prove for every [itex]k\in K[/itex] that [itex]gkg^{-1}\in K[/itex].
But if we do the calculation in G/N, then we get by normality of K/N that

[tex][gkg^{-1}]=[g][k][g]^{-1}\in K/N[/tex]

By definition, this means that there is a k' in K and a n in N such that [itex]gkg^{-1}=k^\prime n[/itex]. But N is a subset of K, thus [itex]gkg^{-1}\in K[/itex].
 
Last edited by a moderator:
77
0
Using the isomorphism theorems here seems like "killing a fly with a nuke" or whatever the saying is (although your idea of that homomorphism does seem interesting.) micromass seems to have covered the rest.
 

Related Threads for: Normality of a quotient group

Replies
5
Views
1K
Replies
1
Views
539
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
8
Views
753
Top