= Solution
A <group action> of $G$ on $X$ is a map $(g,x)\mapsto gx$ such that $ex=x$ and $(gh)x=g(hx)$. The <orbit of a group action>[orbit] and <stabilizer subgroup>[stabilizer] of $x$ are
$$
Gx=\{gx:g\in G\},
\qquad
G_x=\{g\in G:gx=x\}.
$$
For a <finite group>, the map
$$
gG_x\longmapsto gx
$$
is a well-defined <bijection> from the left cosets of $G_x$ to $Gx$. Each coset has $|G_x|$ elements, so the <orbit-stabilizer theorem> is
$$
\boxed{|G|=|Gx|\,|G_x|}.
$$
The <Cauchy theorem for groups> says that if a <prime number> $p$ divides $|G|$, then $G$ has an element of order $p$. To prove it, let
$$
X=\{(g_1,\ldots,g_p)\in G^p:g_1\cdots g_p=e\}.
$$
The first $p-1$ entries determine the last, so $|X|=|G|^{p-1}$, which is divisible by $p$. The <cyclic group> $C_p$ acts on $X$ by cyclically rotating the entries; rotation preserves the product condition because
$$
g_2\cdots g_pg_1=g_1^{-1}(g_1\cdots g_p)g_1=e.
$$
Every orbit has size one or $p$. The fixed points are exactly the tuples $(g,\ldots,g)$ with $g^p=e$. Their number is therefore divisible by $p$. Since the identity gives one fixed point, there is another, and its entry has order $p$.
Now let $|G|=33$. Cauchy's theorem gives a subgroup $H$ of order $11$. Let $H$ act by the <conjugation action> on the set $\mathcal X$ of subgroups of order $11$. It fixes $H$. If it also fixed $K\ne H$, then $H$ would normalize $K$; since $H\cap K=\{e\}$, the product $HK$ would be a subgroup of order $121$, which is impossible. Every other $H$-orbit in $\mathcal X$ therefore has size $11$, so
$$
|\mathcal X|\equiv1\pmod{11}.
$$
Distinct members of $\mathcal X$ share only the identity and each contributes ten nonidentity elements. Hence $1+10|\mathcal X|\leq33$, forcing $|\mathcal X|=1$. Thus $H$ is a <normal subgroup>.
Conjugation now defines a <group homomorphism>
$$
G\longrightarrow\operatorname{Aut}(H).
$$
Because $H\cong C_{11}$, its <automorphism group> has order $10$. By the <Lagrange theorem>, the image has order dividing both $33$ and $10$, so the image is trivial and $H$ lies in the <center of a group>[center] of $G$. Cauchy's theorem also supplies $x$ of order $3$. If $h$ generates $H$, then $h$ and $x$ commute and $hx$ has order $\operatorname{lcm}(11,3)=33$. Therefore
$$
\boxed{G=\langle hx\rangle\cong C_{33}}.
$$
Back to article page
