# Normal subgroups

1. Apr 26, 2008

### titaniumx3

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

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

Show that $$N_{G}(H)$$ is a subgroup of G.

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

2. Apr 26, 2008

### d_leet

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.

3. Apr 27, 2008

### HallsofIvy

Staff Emeritus
I think d leet mean "and the element g-1xg".

What is the product (g-1xg)(g-1x-1g)?

4. Apr 27, 2008

### titaniumx3

It's the identity.

But, aren't we supposed to show that (g-1)-1x(g-1) = gxg-1 is in H?

5. Apr 27, 2008

### ircdan

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?

6. Apr 27, 2008

### titaniumx3

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?).

7. Apr 27, 2008

### HallsofIvy

Staff Emeritus
?? 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).

8. Apr 27, 2008

### titaniumx3

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?

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