Let $H$ be a proper subgroup of $G$, a finite group. Show that there exists $x\in G$ which is not conjugate to an element of $H$.
Attempt: let $G$ act on $X$, the set of all proper groups of $G$, by conjugation. We have: $$|G|=|\text{Orb}(H)||\text{Stab}(H)|\geq |\text{Orb}(H)||H|\quad (h\in H\Rightarrow h\in \text{Stab}(H))$$ (ignore from here) Since $H$ is a proper subgroup:
$$2\leq |G/H|\leq |\text{Orb}(H)|$$ We can write $\text{Orb}(H)=\{H, g_1Hg_1^{-1},\;\dots ,\;g_nHg_n^{-1}\}$ where $n\geq 1$. I feel like this is close but need a little push to see what to do next, or maybe a hint if I am going in the wrong direction. Any help appreciated.
This is true for finite groups.
proof: Since number of the the conjugate of $H$ equal to $|G:N_G(H)|\leq|G:H|$ and every conjugate of $H$ has same order with $H$ and $H^g\cap H $ is nonempty set for all $g\in G$.
Thus, their union can not equal to $G$ as $|G|=|H||G:H|\geq |H||G:N_G(H)|$. We are done.