Existence of $g$ whose conjugacy class is disjoint from a proper subgroup Let $H$ be a proper subgroup of a finite group $G$. Prove that there exists a $g\in{G}$ whose conjugacy class is disjoint from $H$
I'm stuck with this problem. What I only know is that if every conjugacy of $g$ is not disjoint from $H$, then there at most $\vert{H}\vert$ conjugacy classes..
I think this exercise is beautiful ..So any hint will be helpful!
Thank you !
 A: Oops... Here's the proof: Since $H<G$, $G$ is not the union of $g$-conjugates of $H$. There exists some $a$ such that $a$ is not contained in any $g$-conjugate of $H$. i.e. Any $g$-conjugates of $a$ is not contained in $H$. Thus the conjugacy class of $a$ and $H$ are disjoint.
A: Hint: use the following two facts.
Lemma 1 If $K$ is a proper subgroup of $G$, then $\bigcup_{g \in G}K^g \subsetneq G$.
Lemma 2 Let $\{g_1, \cdots, g_k\}$ be a set of representatives of the different conjugacy classes of $G$, then $\langle g_1, \cdots, g_k \rangle=G$.
And note that Lemma 1 implies Lemma 2. The proof of Lemma 1 has appeared many times on StackExchange and can be found here for example.
A: We assume Jordan's theorem which states that if $G$ acts transitively on $X$ with $|X| \geq 2$ then there exists some $g$ such that $X^g = \varnothing$.  To prove the fact that there exists some $C_g$ disjoint from $H$, we want $X^g$ to somehow relate to $H \cap C_g$. Consider the $G$ action on left cosets $\{gH\}$ given by $g' \cdot gH = g'gH$. Note that this action is transitive since $gg'^{-1}$ sends $g'H$ to $gH$. Also, since $H$ is a proper subgroup, there are $\geq 2$ cosets so $|X| = |\{gH\}| \geq 2$. Therefore, by Jordan's theorem, there is some $g \in G$ such that $X^g = \varnothing$. Suppose for the sake of contradiction that $h \in H \cap C_g$. Thus, there exist $g' \in G$ such that $g'h = gg'$. So then $g$ sends the coset $g'H$ to $gg'H = g'hH = g'H$. In other words, $g$ fixes $g'H$ so $g'H \in X^g$ — a contradiction. Therefore, $H \cap C_g$ must be empty, and $C_g$ is the conjugacy class disjoint from $H$.
