$G$ finite group with $H \leq G$ with $G = \bigcup_{g \in G} gHg^{-1} \implies H=G$ 
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*Let $G$ be a finite group and let $H$ be a subgroup of $G$. If $G = \bigcup_{g \in G} gHg^{-1}$, then I have to show $H=G$.

*Let $N(H) =\left\{ g \in G \ \bigl| \ gHg^{-1}=H\right\}$.
What I have done is the following :


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*If $g \in H$ then I know that $gHg^{-1} =H$.

*And if $g_{1}Hg_{1}^{-1} = g_{2}Hg_{2}^{-1}$ then I have $g_{2}^{-1}g_{1}Hg_{1}^{-1}g_{2}=H$ which says $g_{1}^{-1}g_{2} \in N(H)$ or in other words $g_{2} \in g_{1}N(H)$. 

*From the above $2$ things I am able to infer that if $g_{k} \in G$ is in some coset of $N(H)$ the union is same. Means say if $g_{k} \in g_{1}N(H)$ then $g_{1}Hg_{1}^{-1} = g_{k}Hg_{k}^{-1}$ while looking at the union we don't have to count $g_{k}$ and $g_{1}$ seprately as they are both the same. 
I am not able to proceed further. The point is I don't really know what I have to contradict in order to conclude $H=G$. 
 A: There are $|G/N_G(H)|$ distinct subgroups of $G$ of form $gHg^{-1}$, where we denote by $N_G(H)$ the subgroup formed by the elements $g$ such that $gHg^{-1} = H$. 
Note that $H \subset N_G(H)$ and so $|G/N_G(H)| \le |G|/|H|$. From here for the cardinality for the cardinality of the union $\cup_{g\in G} gHg^{-1}$ we get 
$$|\cup_{g\in G} gHg^{-1}| \le |G/N_G(H)| \cdot |H| \le |G/H| \cdot |H| = |G|$$
If $|G/N_G(H) | = 1$ we get $|\cup_{g\in G} gHg^{-1}| = |H| \le |G|$ and we have 
$|\cup_{g\in G} gHg^{-1}| = |G|$ if and only if $H= G$, done. 
If $|G/N_G(H) | > 1$  then we have more than one subgroup of form $gHg^{-1}$.  But then we have for the cardinality of the union a strict inequality because all these subgroups have in common at least the element $e$. Therefore we have $|\cup_{g\in G} gHg^{-1}|< |G|$.
Note that for infinite groups $G$ it is entirely possible to have subgroups $H$ whose conjugates cover the whole $G$, for instance $G = U(n)$ the unitary matrices of size $n$ and $H = D$ the diagonal matrices with elements in $S^1$. 
A: You need to use finiteness (since the result in not true without this hypothesis). You've got a union over conjugates of $H$ indexed by (essentially) representatives of $G/N(H)$. Now try to assemble the facts you know about the orders of (conjugates of) $H$, of $N(H)$ and of $G$ to conclude that (unless $H=N(H)=G$) the union has strictly less elements than $G$ has. The strictness comes from the fact that two conjugates cannot be disjoint, as they have at least $e$ in common, so there is some loss in taking the union.
