Problem about finite group and conjugation Let $G$ be a finite group with order $n\geq 7$ and H be a proper subgroup of $G$ then prove that there exists at least $3$ elements of $G$ that do not conjugate to any elements of $H$
Any idea? I still have no idea how to prove this.For example when using contradiction method.We assume that there is at most 2 element $g_1,g_2 \in G\backslash H$ that do not conjugate to $h ,\forall{h}\in H$ then how to deal with it?
Thank for your help.

Let us consider the permutation action of $G$ on the set of cosets $G/H$. The stabilizer of the coset $xH$ consists of the elements of $xHx^{-1}$, so the element of $G$ that are not conjugate to any elements of $H$ are precisely those that act without fixed points on $G/H$. This action also gives us a group homomorphism
$\phi: G\to \operatorname{Sym}(G/H)$.
If $H\lhd G$, then all those stabilizers are equal to $H$. The elements in $G\setminus H$ then fit the bill, and this set contains at least half the elements of $G$, so we are done.
If $H$ is not normal, then $\phi(G)$ is a transitive subgroup of $S_n$, where
$n=[G:H]\ge 3$. It is known (can be proven by a counting argument) that the average number of fixed points of an element of a transitive group is $1$. As the identity has $n$ fixed points, there are at least $n-1$ elements of
$\phi(G)$ that have no fixed points. So we are done unless $n=3$. But in that case $\ker\phi$ is a subgroup of index six, so it is non-trivial. As the fixed point free elements of $\phi(G)$ correspond to cosets $\ker\phi$, we are done in this case as well.
