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.
 A: Here is an answer that only needs Jyrki's results for $A_3$ and $S_3$ (where they can be directly verified):
$$|G \setminus \cup_{g \in G} H^g| \geq |G| - 1 - \sum_{g \in G/H} | H^g \setminus \{1\}| =  |G|-1 -[G:H](|H|-1) = [G:H]-1$$ so unless the index is 1, 2, or 3, the statement is proven. For index 1, $H$ is not proper. For index 2, $H$ is normal, so in fact $|G \setminus H| = |G|/2 \geq 3$. For index 3, the kernel of the permutation action is non-trivial, so a very very simple version of Jyrki's answer applies.

If the index is 3 and $H$ is not normal, then there are three conjugates of $H$ in $G$, so $H$ contains a subgroup $K$ such that $K$ is normal in $G$, and $G/K$ is isomorphic to either the alternating group $A_3$ or the symmetric group $S_3$. In both cases, $G/K$ contains an element $gK$ of order 3, and $\{H,H^g,H^{g^2}\}$ are the three conjugates of $H$. Since $H\neq H^g$, $g \notin H$, but since $H^g \neq H^{g^2}$, $g \notin H^g$, and similarly $g\notin H^{g^2}$. However the same argument applies to every element $gk \in gK$ and $g^2K$. Since $[G:K] \leq 6$ and $|G| \geq 7$, we get that $|K|\geq 2$ and so $|gK \cup g^2K| \geq 4$, so that we in fact have 4 elements not contained in any conjugate of $H$.
A: 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.
