= 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> states
$$
\boxed{|Gx|=\frac{|G|}{|G_x|}.}
$$
To prove it, send a left <coset> $gG_x$ to $g\cdot x$. If $gG_x=hG_x$, 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> $gKg^{-1}$ because it preserves products and inverses. Furthermore $eKe^{-1}=K$ and
$$
(gh)K(gh)^{-1}=g(hKh^{-1})g^{-1}.
$$
Thus 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> is
$$
k=\frac{|G|}{|N_G(H)|}\leq\frac{|G|}{|H|}.
$$
Now suppose $H$ is proper. Put $m=|G|/|H|\geq2$ and list the $k$ distinct <conjugate subgroups> as $H_1,\ldots,H_k$. Each has $|H|$ elements and all contain the same identity. Counting the identity just once, even if other elements overlap, gives
$$
\left|\bigcup_{i=1}^kH_i\right|
\leq1+k(|H|-1)
\leq1+m(|H|-1)
=|G|-m+1<|G|.
$$
Thus \b[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.
Back to article page
