1. ## Abstract Algebra

Let G be a finite group. and let H be a subgroup of G, and call N(H) the normalizer of H in G defined as, {x|xHx^-1 = H}.
Prove that N(H) is a subgroup of G
and prove that |{xHx^-1|x in G}| = |G/N(H)|

2. To show N(H) is a subgroup take $\displaystyle a,b \in N(H)$ and show $\displaystyle ab^{-1}\in N(H)$. This is fairy simple if you first note that if $\displaystyle xHx^{-1}=H$ then $\displaystyle x^{-1}Hx=H$.

3. ## Group Actions

The second part is a result of the counting formula which states that the size of the orbit is equal to the index of the stabilizer . $\displaystyle N(H)\unlhd G$ and $\displaystyle N(H)=stab(H)$ under the group action conjugation. The LHS is the orbit of H under conjugation, so these sets are equal in cardinality.

Plus there is a really easy bijection to establish this fact. $\displaystyle \phi : xHx^{-1} \rightarrow G/N(H)$ by $\displaystyle \phi (xHx^{-1})=xN(H)$. Surjectivity is clear.
Injectivity and well-defined is also pretty obvious.
$\displaystyle xHx^{-1}=yHy^{-1} \Leftrightarrow y^{-1}xHx^{-1}y=H\Leftrightarrow (y^{-1}x)H(y^{-1}x)^{-1}=H$$\displaystyle \Leftrightarrow y^{-1}x \in N(H)\Leftrightarrow xN(H)=yN(H)$