Group Actions & Sylow Theory
Group Actions & Sylow Theory
A group action $G\curvearrowright X$ assigns to each $g\in G$ a permutation of $X$ compatibly with the group structure. Orbits partition $X$; stabilizers are subgroups. The orbit-stabilizer theorem $|G|=|\text{Orb}(x)||\text{Stab}(x)|$ is the bridge between group-theoretic and combinatorial counting. Sylow's theorems classify finite groups by their prime-power subgroup structure — the most powerful tool for understanding the structure of finite groups.
Group Actions & Orbit-Stabilizer Theorem
$G$ acts on $X$ if there is $\cdot:G\times X\to X$ with $e\cdot x=x$ and $(gh)\cdot x=g\cdot(h\cdot x)$. Orbit of $x$: $Gx=\{g\cdot x:g\in G\}$. Stabilizer: $G_x=\{g:g\cdot x=x\}\leq G$. Orbit-Stabilizer: $|Gx|=[G:G_x]=|G|/|G_x|$, so $|G|=|Gx|\cdot|G_x|$. Class equation: $|G|=|Z(G)|+\sum_{x\notin Z(G)}[G:C_G(x)]$ where $C_G(x)=\{g:gx=xg\}$ is the centralizer.
Sylow's Theorems
Let $|G|=p^a m$ with $p\nmid m$ (prime $p$). (1) Existence: $G$ has a subgroup of order $p^a$ (Sylow $p$-subgroup). (2) Conjugacy: all Sylow $p$-subgroups are conjugate in $G$. (3) Number: the number $n_p$ of Sylow $p$-subgroups satisfies $n_p\equiv 1\pmod p$ and $n_p\mid m$. Corollary: a group of order $p^a q^b$ (for primes $p,q$) need not be simple — Sylow's theorems often force $n_p=1$ (unique, hence normal Sylow subgroup), making the group solvable.
Example 1
Show no group of order 15 is simple.
Solution: $|G|=15=3\cdot 5$. Sylow: $n_5\equiv 1\pmod 5$ and $n_5|3$, so $n_5\in\{1,3\}$ but $n_5\equiv 1\pmod 5$ forces $n_5=1$. So the Sylow 5-subgroup is unique, hence normal ($G\curvearrowright\{$Sylow 5-subgroups$\}$ by conjugation gives a trivial action). Any group of order 15 has a normal subgroup of order 5, so is not simple. In fact $G\cong\mathbb{Z}_{15}\cong\mathbb{Z}_3\times\mathbb{Z}_5$.
Example 2
Use Burnside's lemma to count the number of distinct colorings of the vertices of a square with 2 colors.
Solution: Burnside: $|X/G|=\frac{1}{|G|}\sum_{g\in G}|X^g|$ where $X^g=\{$colorings fixed by $g\}$. $G=D_4$ (8 symmetries). Rotations: $0°$: all $2^4=16$ colorings fixed; $90°$: need 4 same color: 2; $180°$: opposite vertices same: $2^2=4$; $270°$: 2. Reflections through vertices (2): adjacent pairs same: $2^2=4$; reflections through edges (2): 2 pairs: $2^2=4$. Total: $(16+2+4+2+4+4+4+4)/8=40/8=6$ distinct colorings (up to rotation and reflection).
Practice
- Prove Sylow's first theorem using the action of $G$ on $p$-element subsets.
- Classify all groups of order 12 (there are 5 up to isomorphism).
- Show every group of order $p^2$ (prime $p$) is abelian.
- Use the class equation to prove: if $|G|=p^n$ (prime power $>1$), then $Z(G)\neq\{e\}$.
Show Answer Key
1. Let $|G|=p^a m$ with $p\nmid m$. Consider the action of $G$ on the set $\binom{G}{p^a}$ of $p^a$-element subsets by left multiplication. The number of orbits is $\binom{p^a m}{p^a}/|\text{orbit}|$. Since $p\nmid\binom{p^a m}{p^a}$ (verified by counting $p$-adic valuations), at least one orbit has size not divisible by $p^a$. The stabilizer of that orbit has order $p^a$, giving a Sylow $p$-subgroup.
2. The 5 groups of order 12: $\mathbb{Z}_{12}$, $\mathbb{Z}_2\times\mathbb{Z}_6$, $A_4$, $D_6$ (dihedral of hexagon), and $\text{Dic}_3$ (dicyclic group).
3. Let $|G|=p^2$. The center $Z(G)$ has order $p$ or $p^2$ (by the class equation, $|Z|\neq1$). If $|Z|=p^2$, done. If $|Z|=p$, then $G/Z$ has order $p$ and is cyclic, generated by $gZ$. Every element is $g^a z$ for $z\in Z$. Since $Z$ is central, all elements commute, contradicting $|Z|=p$. So $G$ is abelian.
4. Class equation: $|G|=|Z(G)|+\sum[G:C_G(x_i)]$. Each $[G:C_G(x_i)]$ divides $|G|=p^n$, so each is a power of $p$ (at least $p$). Since $p|\,|G|$ and $p$ divides each conjugacy class size, $p|\,|Z(G)|$. So $|Z(G)|\ge p>1$.