Proving N(H) is a Subgroup of G Containing H

[Jen917]

Let G be a group and let H be a subgroup.

Define N(H)={x∈G|xhx^-1 ∈H for all h∈H}. Show that N(H) is a subgroup of G which contains H.

To be a subgroup I know N(H) must close over the operations and the inverse, but I am not sure hot to show that in this case.

 

[I like Serena]

Let G be a group and let H be a subgroup.

Define N(H)={x∈G|xhx^-1 ∈H for all h∈H}. Show that N(H) is a subgroup of G which contains H.

To be a subgroup I know N(H) must close over the operations and the inverse, but I am not sure hot to show that in this case.

Hi Jen917! Welcome to MHB! (Smile)

To prove that the operation is closed, we need to prove that for all $x, y \in N(H)$:
$$\forall h \in H: (xy)h(xy)^{-1} \overset ?\in H$$
Can we find that using $xhx^{-1} \in H$ and $yhy^{-1} \in H$? (Wondering)To prove that the inverse belongs to the set, we need a sub step first.
Either way, it requires that $H$ is finite. Is it?

Suppose $x$ is an element of $N(H)$.
Now consider the function $H\to H$ given by $h \mapsto xhx^{-1}$.
Is it a bijection? (Wondering)