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For a group $G$ acting on a set $X$ and a point $x\in X$, the orbit map is $G\to X$, $g\mapsto gx$. Its image is the orbit of a group action of $x$, and its fibres are cosets of the stabilizer subgroup of $x$.

We first prove the needed instance of Cauchy theorem for groups, rather than invoke it. Let a prime number $q$ divide $|G|$ and consider the setThere are $|G{|}^{q-1}$ such tuples: choose the first $q-1$ entries freely, and the last is forced. Cyclic rotation preserves this set, because moving the first entry to the end conjugates the product, which remains $1$. Each rotation orbit of a group action has either one or $q$ elements. Indeed a nontrivial shift fixing a tuple generates all shifts, since its step has an inverse modulo the prime $q$. The fixed tuples are precisely $(g,\dots ,g)$ with $g^q=1$.

Counting nonfixed orbits in multiples of $q$, and using $q\mid |G{|}^{q-1}$, shows that the number of solutions of $g^q=1$ is divisible by $q$. The identity supplies one solution, so there is a nonidentity solution. If its order of a group element is $d$, divide $q$ by $d$: the remainder would give a smaller positive exponent producing the identity, so $d\mid q$. Hence $d=q$. This proves the required prime-order existence result by the cyclic-tuple proof of Cauchy theorem.

Our hypothesis says that this nonidentity element has order $p$, so $q=p$. Every prime divisor of $|G|$ is therefore $p$, which givesThus a finite group of prime exponent has prime-power order. The case $n=0$ includes the trivial group. No group-theoretic counting theorem was assumed in the tuple argument.

For a group action of a finite group $G$ on a set and a point $x$, the orbit-stabilizer theorem iswhere $G_x=\{g:gx=x\}$ is the stabilizer subgroup. The map $g{G}_x\mapsto gx$ is well defined because elements of the same coset act identically on $x$. If $gx=hx$, then $h^{-1}g\in G_x$, so their cosets agree; every orbit point has the form $gx$, giving surjectivity. This is a bijection. Lagrange's theorem, $|G|=[G:H]|H|$ for any subgroup $H$, yields the counting formula.

Apply the conjugation action of the finite p-group $G$ to its nontrivial normal subgroup $N$. Normality ensures the action stays inside $N$. The singleton orbits of a group action are exactly the elements of $N\cap Z(G)$. Every larger orbit of a group action has size a positive power of $p$, by the orbit-stabilizer theorem and Lagrange's theorem. ThereforeSince $N$ is nontrivial, its order is divisible by $p$. The intersection contains the identity and has cardinality a positive multiple of $p$, so it contains a nonidentity element. This proves the central intersection property of normal subgroups of finite p-groups.

For the proper subgroup $H$, let $H$ act by left multiplication on the left cosets $G/H$. A coset $gH$ is fixed by every $h\in H$ exactly when $g^{-1}Hg\subseteq H$. Both groups have the same finite size, so this inclusion is equality; the fixed cosets are precisely those with $g\in N_G(H)$. Hence their number is $[N_G(H):H]$. All non-singleton orbits of a group action have size divisible by $p$, soSince $H$ is proper, $[G:H]$ is a positive power of $p$ greater than one. The fixed-coset count is nonzero, because $H$ itself is fixed, and is divisible by $p$. Therefore the normalizer strictly contains $H$:This is the normalizer condition for finite p-groups. The proof also covers $H=\{1\}$, whose action fixes all cosets.

For a group action, the orbit of a group action of $x$ is $Gx=\{gx:g\in G\}$, and its stabilizer subgroup is $G_x=\{g:gx=x\}$. The orbit-stabilizer theorem gives $|Gx||G_x|=|G|$: the map $g{G}_x\mapsto gx$ is a well-defined bijection from stabilizer cosets to the orbit.

Within one orbit, the stabilizer subgroups are conjugate, since $G_{hx}=h{G}_x{h}^{-1}$. Their sizes are therefore equal. Summing stabilizer sizes over a single orbit gives $|Gx||G_x|=|G|$, and summing over all $n$ orbits yieldsNow count the pairs $(g,x)$ fixed by the action in two ways. Holding $x$ fixed gives $|G_x|$ choices of $g$, whereas holding $g$ fixed gives $|\mathrm{Fix}(g)|$ choices of $x$. ThusCombining the two identities proves Burnside lemma:The finiteness hypotheses justify these cardinality sums. If $X$ is empty, every sum and the orbit count are zero, so the formula still applies.

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 and stabilizer of $x$ areFor a finite group, the mapis 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

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, letThe 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 becauseEvery orbit has size one or $p$. The fixed points are exactly the tuples $(g,\dots ,g)$ 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 $11$, soDistinct members of $\mathcal{X}$ share only the identity and each contributes ten nonidentity elements. Hence $1+10|\mathcal{X}|\le 33$, forcing $|\mathcal{X}|=1$. Thus $H$ is a normal subgroup.

Conjugation now defines a group homomorphismBecause $H\cong C_{11}$, 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 $G$. Cauchy's theorem also supplies $x$ of order $3$. If $h$ generates $H$, then $h$ and $x$ commute and $hx$ has order $\mathrm{lcm}(11,3)=33$. Therefore

A Sylow subgroup of a finite group $G$ is a subgroup whose order is the largest power of a prime $p$ dividing $|G|$. The Sylow theorems say that such subgroups exist, that every $p$-subgroup lies in one, that all Sylow $p$-subgroups are conjugate, and that their number $n_p$ divides $|G|/p^a$ and obeys $n_p\equiv 1(\mathrm{mod}p)$.

For the last congruence, let one Sylow $p$-subgroup $P$ act by conjugation on the set of all Sylow $p$-subgroups. Every orbit other than a fixed point has size divisible by $p$. If $Q$ is fixed, then $P,Q\le N_G(Q)$. They are Sylow subgroups of this normalizer, so they are conjugate within $N_G(Q)$; because every element of $N_G(Q)$ fixes $Q$ under conjugation, this forces $P=Q$. Thus there is exactly one fixed point and $n_p\equiv 1(\mathrm{mod}p)$.

Now let $H<A_n$ have index $m$ with $1<m<n$. The coset action gives a homomorphismIts kernel is a normal subgroup of the simple group $A_n$. It cannot be all of $A_n$, since the action is transitive and nontrivial, so it is trivial. This would embed $A_n$ into $S_m$, contrary to $|A_n|=n!/2>m!=|S_m|$. Hence no such subgroup exists.

Suppose finally that a group $G$ of order $90$ were simple. The Sylow count satisfiesand simplicity excludes $n_5=1$, so $n_5=6$. Conjugation on these six subgroups gives a nontrivial homomorphism $G\to S_6$, which simplicity makes injective. Its image lies in $A_6$, because the composite with the sign homomorphism $S_6\to \{\pm 1\}$ must be trivial. It would therefore be a subgroup of $A_6$ of index $360/90=4$, contradicting the result just proved. Thus

A group action on a set $X$ is transitive when every two points lie in the same orbit of a group action, equivalently when for all $x,y\in X$ there is a group element $g$ with $gx=y$.