Showing that the conjugates of a proper subgroup do not cover the group. I am trying to figure out the following.
Suppose $G$ is a group and is finite; let $H$ be a proper subgroup. Show that the conjugates of $H$ do not cover $G$ (that is, there is some $g \in G$ which is not conjugate to an element of $H$).
As I understand it, we say that given $a,b \in G$, $a$ is conjugate with $b$ if $\exists g' \in G (a = g'bg'^{-1})$. Since $H$ is a proper subgroup of $G$, then there $\exists g \in G$ such that $g \notin H$. Since $g \notin H$, then $\forall h \in H (g \neq h)$. Suppose then that $\exists q \in G, \exists r \in H (g = qrq^{-1})$ $-$ i.e., that $g$ is conjugate with with some element of $H$.
Then $q^{-1}gq = r$. However, I'm not sure where to go from here. I'm guessing that I have to somehow use the assumption that $G$ is finite however I'm not sure how to.

For $h,g\in G$ let $h^g=ghg^{-1}$; for a fixed $g\in G$, the map $h\mapsto h^g$ is an automorphism of $G$. Let $N=\{g\in G:H^g=H\}$, where $H^g=\{h^g:h\in H\}$; this is the normalizer of $H$, and it’s not hard to check that $N$ is a subgroup of $G$. 


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*Show that $H\subseteq N$.  

*Show that the number of conjugates of $H$, including $H$ itself, is $|G/N|$.  

*Show that each conjugate of $H$ is a subgroup of $G$.


Since the identity element belongs to each conjugate of $H$, we have
$$\left|\bigcup_{g\in G}H^g\right|<|G/N|\cdot|H|=\frac{|G|}{|N|}\cdot|H|\le\frac{|G|}{|H|}\cdot|H|=|G|\;.$$
