Q:

Let $K<G$, where G is a group. Define $N_G(K)=(g\in G/gkg^{-1}=K)$ Prove the following:

Let $K<G$, where G is a group. Define $$N_G(K)= \{g\in G \mid g K g^{-1}=K\}$$ which is called the normalizer of subgroup K.
1.) Prove that $N_G(K)$ is a subgroup of G and K is a normal subgroup of $N_G(K)$
2.) Prove that $K$ is a normal subgroup if and only if $N_G(K)=G$
The term normalizer is new to me but I take it that it follows the same principles of a normal group. So, therefore, I'm thinking that to prove this, I must try and prove that $K$ is a subset of $N_G(K)$ for every $k \in K$ where $kKk^{-1}=K$ for $N_G(K)$ I'm assuming that if I can do this, then it will help with solving part 2) 

A:

Intuitively, you just need to show that if $h$ normalizes $K$, then its inverse $h^{-1}$ does as well, and that if $g$ also normalizes $K$, then the product $hg$ normalizes $K$. This will show that the set $N_G(K)$ is indeed a group; it contains inverses and products of all its elements.
Think about why it's clear that $K \subseteq N_G(K)$, I'm not sure it's too helpful here.
I'll also say that the definition of $N_G(K)$ basically tells you why $K$ is normal in $N_G(K)$, so the 1st part should be the most difficult.

