 Solution

A group action is a map $(g,x)\mapsto g\cdot x$ from $G\times X$ to $X$ satisfying $e\cdot x=x$ and $(gh)\cdot x=g\cdot (h\cdot x)$. Define the orbit of a group action $Gx=\{g\cdot x:g\in G\}$ and the stabilizer subgroup $G_x=\{g:g\cdot x=x\}$. The latter is a subgroup: the identity fixes $x$, products of elements fixing $x$ fix $x$, and inverses do too.

For finite $G$, the orbit-stabilizer theorem statesTo prove it, send a left coset $g{G}_x$ to $g\cdot x$. If $g{G}_x=h{G}_x$, then $h^{-1}g\in G_x$ and $g\cdot x=h\cdot x$. Conversely, equality of the two images implies $h^{-1}g\in G_x$, hence equality of the cosets. Every point of the orbit of a group action has such a preimage. This is therefore a bijection, and each coset has $|G_x|$ elements, proving the formula.

For the action on subgroups, conjugation maps a subgroup $K$ to a subgroup $gK{g}^{-1}$ because it preserves products and inverses. Furthermore $eK{e}^{-1}=K$ andThus this is the conjugation action. The stabilizer subgroup of $H$ is its normalizer $N_G(H)$, and $H\subseteq N_G(H)$ since each $h\in H$ conjugates $H$ onto itself. Hence the number $k$ of distinct conjugate subgroups isNow suppose $H$ is proper. Put $m=|G|/|H|\ge 2$ and list the $k$ distinct conjugate subgroups as $H_1,\dots ,H_k$. Each has $|H|$ elements and all contain the same identity. Counting the identity just once, even if other elements overlap, givesThus some element of $G$ lies in no conjugate of $H$. This proves conjugates of a proper subgroup do not cover a finite group by an explicit count; it does not require the conjugates to intersect only at the identity.