Normal subgroups

  • Thread starter titaniumx3
  • Start date
53
0

Main Question or Discussion Point

Let G be a group and H a subgroup of G. We define the following:

[tex]N_{G}(H) = \{g \in G \,\,|\,\, g^{-1}hg \in H,\, for\, all\,\, h\in H\}[/tex]

Show that [tex]N_{G}(H)[/tex] is a subgroup of G.

_______________________

I've shown that for all [itex]x,\, y[/itex] of [itex]N_{G}(H)[/itex], [itex]xy[/itex] is an element of [itex]N_{G}(H)[/itex], but how do I show that [itex]x^{-1}[/itex] is an element of [itex]N_{G}(H)[/itex] ?
 

Answers and Replies

1,073
1
H is a subgroup, hence H has inverses so for the element x in H, and the element g-1hg consider the element g-1x-1g which is in NG(H) since x-1 is in H.
 
HallsofIvy
Science Advisor
Homework Helper
41,732
893
I think d leet mean "and the element g-1xg".

What is the product (g-1xg)(g-1x-1g)?
 
53
0
It's the identity.

But, aren't we supposed to show that (g-1)-1x(g-1) = gxg-1 is in H?
 
229
0
It's the identity.

But, aren't we supposed to show that (g-1)-1x(g-1) = gxg-1 is in H?
yup that's right, think about what halls just said though and it answers your question, you know (g^-1xg)(g^-1x^-1g) = e. This tells you what?
 
53
0
yup that's right, think about what halls just said though and it answers your question, you know (g^-1xg)(g^-1x^-1g) = e. This tells you what?
It tells me the inverse of g-1xg is g-1x-1g (which must also be contained in H), but how does that show that g-1 is in NG(H)? (i.e. do all elements of NG(H) have inverses?).
 
HallsofIvy
Science Advisor
Homework Helper
41,732
893
?? Was that a typo. You are not trying to prove that g-1 is in NG(H), you are trying to prove x-1 is.

And the fact that g-1x-1g is in H shows that x-1 is in NG(H).
 
53
0
Now I'm getting confused lol. According to the definition of NG(H) from my original post, we want to show that for all g in NG(H), g-1 is also in NG(H). In other words we have to show that for any g in NG(H) and all x in H, (g-1)-1x(g-1) is H.

Am I correct in saying this?
 

Related Threads for: Normal subgroups

  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
1
Views
672
  • Last Post
Replies
3
Views
626
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
2
Views
426
  • Last Post
Replies
6
Views
1K
Top