The union of all the conjugates of a subgroup is not the group. Let’s consider a finite group $G$ and a subgroup $H \subsetneq G$. I have to show that:
$$
\bigcup_{g\in G} gHg^{-1} \neq G
$$
In the solutions to the exercise this is done via explicit counting of elements. I had another idea but am stuck:
We can consider the action of $G$ on the set of its subgroups via conjugation. Then the number of different conjugacy classes of $H$ is the cardinality of the orbit of $H$ under the action i.e. $|O_G(H)|$. Now since $\forall g \in G : \ |gHg^{-1}|=|H|$ the number of elements in the union must be: $|O_G(H)| |H| $. Now we can use that $O_G(H)=G/N_G(H)$ by the orbit-stabilzier theorem. Thus $x:=|O_G(H)||H| = |G/N_G(H)||H|$. Finally using lagrange we get:
$$
x |N_G(H)|=|G||H|
$$
To get to the result I would need to show that $|N_G(H)|> |H|$. Since $H \subseteq N_G(H)$ I actually only need to find a single $g\in N_G(H)\setminus H$ but I somehow can’t find one. Any help is appreciated:)
 A: You don't actually need to show that $|N_{G}(H)| > |H|$, because you've overcounted. You write that the number of elements in the union
$$\bigcup_{g \in G} gHg^{-1}$$ is $|O_{G}(H)||H|,$ because each $gHg^{-1}$ has cardinality $|H|$ and there are $|O_{G}(H)|$ distinct $gHg^{-1}$'s. But this is wrong, because different $gHg^{-1}$'s are not necessarily disjoint. In fact, we know they are not disjoint: the identity element $e$ is contained in all the $gHg^{-1}$'s. Therefore, if we have distinct $gHg^{-1}$'s, $|O_{G}(H)||H|$ overcounts the identity element at least once.
So, we have two cases: in the first case, we have $|O_{G}(H)| =1$, in which case
$$\bigcup_{g \in G} gHg^{-1} = H \neq G.$$
In the second case, we have $|O_{G}(H)| \geq 2.$ In this case,
we have
$$\left|\bigcup_{g \in G} gHg^{-1}\right| \leq |O_{G}(H)||H| -1,$$ because $|O_{G}(H)||H|$ overcounts the identity element at least once.
Using your notation, we then have
$$\left|\bigcup_{g \in G} gHg^{-1}\right| < x = |O_{G}(H)||H|.$$ Since $|N_{G}(H)| \geq |H|$ (because $H \subseteq N_{G}(H)$) and
$$x|N_{G}(H)| = |G||H|,$$ it follows that
$$\left|\bigcup_{g \in G} gHg^{-1}\right| < x \leq |G|,$$ which proves the desired result.
