Prove that N is a normal subgroup

[AdrianZ]

Homework Statement

If H is any subgroup of G and $N={\cap_{a\in G} a^{-1}Ha}$, prove that N is a normal subgroup of G.

The Attempt at a Solution

Is this statement true? $\forall n: n \in N \implies \exists h \in H : n=a^{-1}ha$
The theorem looks intuitively true, but I don't know how to write a formal argument knowing that $N={\cap_{a\in G} a^{-1}Ha}$

 

[micromass]

The statement you write is true. How would you find h?? Hint: rewrite the equation $n=a^{-1}ha$ to find an expression for h.

 

[AdrianZ]

The statement you write is true. How would you find h?? Hint: rewrite the equation $n=a^{-1}ha$ to find an expression for h.

well, then is this argument valid?

first I should show that N is a subgroup of G, it means I should prove that for any a and b in N, ab^-1 is also in N.

$\forall p,b \in N: \forall a \in G, \exists h_1,h_2 \in H : p=a^{-1}h_1a , b=a^{-1}h_2a$
$pb^{-1}= a^{-1}h_1a(a^{-1}h_2a)^{-1}=a^{-1}h_1a(a)^{-1}h_2^{-1}(a^{-1})^{-1}=a^{-1}hh^{-1}a \implies \exists h_3 \in H, \forall a \in G: a^{-1}h_3a=pb^{-1}$
hence, $pb^{-1} \in N$ and N is a subgroup of G.

If that argument is valid, how can I show that N is normal?

 

[micromass]

well, then is this argument valid?

first I should show that N is a subgroup of G, it means I should prove that for any a and b in N, ab^-1 is also in N.

$\forall p,b \in N: \forall a \in G, \exists h_1,h_2 \in H : p=a^{-1}h_1a , b=a^{-1}h_2a$
$pb^{-1}= a^{-1}h_1a(a^{-1}h_2a)^{-1}=a^{-1}h_1a(a)^{-1}h_2^{-1}(a^{-1})^{-1}=a^{-1}hh^{-1}a \implies \exists h_3 \in H, \forall a \in G: a^{-1}h_3a=pb^{-1}$
hence, $pb^{-1} \in N$ and N is a subgroup of G.

Looks good.

If that argument is valid, how can I show that N is normal?

Take p in N arbitrary and show that $a^{-1}pa$ is also in N.

 

[Deveno]

i think you're focusing too much on the h's. in the intersection it's the a's that serve as the index. and these range over every element of G. that's important.

so we are taking the intersection of H, aHa^-1,bHb^-1, etc.

now, suppose we have some fixed element of G, g.

as a ranges over the entire group G, doesn't ga as well?

that is, doesn't:

$$\bigcap_{a \in G} aHa^{-1} = \bigcap_{ga \in G} (ga)H(ga)^{-1}$$