If $H \leq G$ has finite index, then $G$ has a normal subgroup of finite index From a practice test for a Masters Qual. Exam:  
Let $H$ be a subgroup of finite index in a group $G$. Prove that $G$ has a normal subgroup $K$ of finite index with $K \subseteq H$.
It makes perfect sense intuitively that $K$ should exist, I just don't know how to get ahold of it.
 A: Let $x_1, ... , x_n$ be a set of distinct left coset representatives for $H$ in $G$, and let 
$$K = \bigcap\limits_{i=1}^n x_iHx_i^{-1}$$
This is a subgroup of finite index in $G$, because the intersection of two subgroups of finite index is still of finite index.
To show that $K$ is normal in $G$, note that if $g \in G$, then $gx_1, ... , gx_n$ is another set of left coset representations for $H$ in $G$.  Hence there is a permutation $\sigma$ such that $gx_iH = x_{\sigma(i)}H$.  Then also $Hx_i^{-1}g^{-1} = Hx_{\sigma(i)}^{-1}$, and so
$$gx_iHx_i^{-1}g^{-1} = x_{\sigma(i)}Hx_i^{-1}g^{-1} = x_{\sigma(i)}Hx_{\sigma(i)}^{-1}$$
Thus $$gKg^{-1} = \bigcap\limits_{i=1}^n gx_iHx_i^{-1}g^{-1} = \bigcap\limits_{i=1}^n x_{\sigma(i)}Hx_{\sigma(i)}^{-1} = K$$
You can obtain the same subgroup $K$ less tediously by defining it to be the kernel of the homorphism $\phi: G \rightarrow \textrm{Sym}(G/H)$ of $G$ into the group of bijections of the set of left cosets of $H$ in $G$, where $\phi(g)[xH] = gxH$ for all $g \in G$ and left cosets $xH$.
