Prove N_{G}(H) is a Subgroup of G

[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 $N_{G}(H)$ ?

 

[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 N_G(H) since x^-1 is in H.

 

[HallsofIvy]

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

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

 

[titaniumx3]

It's the identity.

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

 

[ircdan]

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?

 

[titaniumx3]

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 N_G(H)? (i.e. do all elements of N_G(H) have inverses?).

 

[HallsofIvy]

?? Was that a typo. You are not trying to prove that g^-1 is in N_G(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 N_G(H).

 

[titaniumx3]

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

Am I correct in saying this?