Group actions, class equation, p-groups

In one sentence

Orbit-stabiliser converts group theory into counting, the class equation is that count applied to conjugation, and Sylow's theorems are what let you decide the structure of a group from its order alone.

Why the exam asks it

Because "show no group of order $n$ is simple" is a self-contained puzzle with a standard method, and because the two Sylow conditions — $n_p \equiv 1 \pmod p$ and $n_p \mid m$ — are easy to state and easy to apply in the wrong order.

The idea

A group $G$ acts on a set $X$ by $g\cdot x$, compatibly with the operation. The orbit of $x$ is $\{g\cdot x\}$ and the stabiliser $G_x = \{g : g\cdot x = x\}$ is a subgroup.

Orbit-stabiliser: $|{\rm Orb}(x)| = [G : G_x]$.

Every orbit size therefore divides $|G|$, which is the fact that does the work. Orbits partition $X$, so $|X| = \sum |{\rm Orb}|$ over representatives.

The class equation

Take $G$ acting on itself by conjugation. Orbits are conjugacy classes, stabilisers are centralisers, and fixed points are exactly the centre:

$$|G| = |Z(G)| + \sum_{i}[G : C(x_i)],$$

summed over representatives of the classes with more than one element. Each term in the sum is $> 1$ and divides $|G|$.

Two standard consequences:

• A $p$-group has a non-trivial centre. Every term in the sum is divisible by $p$, and so is $|G|$, so $p \mid |Z(G)|$.
• Every group of order $p^2$ is abelian. The centre is non-trivial, and $G/Z(G)$ cyclic forces $G$ abelian, so $|Z(G)| = p$ is impossible.

That second argument uses a lemma worth memorising in its own right: if $G/Z(G)$ is cyclic then $G$ is abelian. It is the engine of most small-order classification.

Sylow

Let $|G| = p^k m$ with $p \nmid m$.

1. Existence: Sylow $p$-subgroups of order $p^k$ exist.
2. Conjugacy: any two are conjugate, and every $p$-subgroup lies in one.
3. Counting: $n_p \equiv 1 \pmod p$ and $n_p \mid m$.

The immediate corollary used everywhere: $n_p = 1$ $\iff$ the Sylow $p$-subgroup is normal, because conjugacy of Sylow subgroups means a unique one is fixed by conjugation.

Order 15. $n_5 \mid 3$ and $n_5 \equiv 1 \pmod 5$, so $n_5 \in \{1,3\}$ and $3 \not\equiv 1$, forcing $n_5 = 1$. Similarly $n_3 = 1$. Both Sylow subgroups are normal, intersect trivially, and $G \cong C_3 \times C_5 \cong C_{15}$. Every group of order $15$ is cyclic.

Order 12. $n_3 \mid 4$ and $n_3 \equiv 1 \pmod 3$, so $n_3 \in \{1,4\}$. Both occur — $C_{12}$ has $n_3 = 1$, and $A_4$ has $n_3 = 4$ — so counting alone does not finish, and this is why $A_4$ keeps appearing. But no group of order $12$ is simple: if $n_3 = 4$ then the four Sylow $3$-subgroups contribute $8$ elements of order $3$, leaving exactly $4$, which must form the unique Sylow $2$-subgroup, hence normal.

That counting move — count elements of prime order, see what is left — is the standard technique for non-simplicity.

Burnside

$$\#\text{orbits} = \frac{1}{|G|}\sum_{g\in G}|{\rm Fix}(g)|,$$

the tool for counting colourings up to symmetry.

Where intuition breaks

"$n_p \equiv 1 \pmod p$, so $n_p = 1$." It is one congruence and one divisibility, and often both $1$ and a larger value satisfy them — order $12$ with $n_3 \in \{1,4\}$. Options quote one condition and drop the other.

"$n_p = 1$ means the group is abelian." It means that Sylow subgroup is normal. $A_4$ has a normal Sylow $2$-subgroup and is not abelian.

"A $p$-group is abelian." It has non-trivial centre. $D_4$ has order $8$, centre of order $2$, and is not abelian. Order $p^2$ forces abelian; $p^3$ does not.

"$G/Z(G)$ cyclic, so $Z(G) = G$." It gives $G$ abelian, which then makes $Z(G) = G$ — but the useful direction is the implication, and its usual role is to prove $|Z(G)| = p$ impossible for $|G| = p^2$.

"Orbits all have the same size." Only for transitive actions. In general orbit sizes vary and each divides $|G|$.

"The class equation says $|G| = \sum$ class sizes." It does, and the point is that the centre contributes $|Z(G)|$ classes of size $1$. Writing $|Z(G)|$ separately is what makes the divisibility argument work.

"Every group of order $p^k$ has a normal subgroup of every order $p^j$." True for $p$-groups, and worth recognising — it is one of the few places the converse of Lagrange does hold.

The exam's angle

1. Factor the order and list Sylow possibilities. Apply both conditions; most questions are decided by the pair.
2. If some $n_p = 1$, you have a normal subgroup and the group is not simple. That is usually the whole question.
3. If no $n_p = 1$, count elements. Distinct Sylow $p$-subgroups intersect trivially for prime order, so $n_p(p-1)$ elements have order $p$; see whether the total exceeds $|G|$.
4. For $p$-groups reach for the class equation — non-trivial centre is almost always the first step.
5. $G/Z(G)$ cyclic $\Rightarrow$ abelian closes a surprising number of arguments.

The night before

• Orbit-stabiliser: $|{\rm Orb}(x)| = [G:G_x]$; orbit sizes divide $|G|$.
• Class equation: $|G| = |Z(G)| + \sum [G:C(x_i)]$, each term $>1$ and dividing $|G|$.
• $p$-groups have non-trivial centre. $|G| = p^2 \Rightarrow$ abelian. $p^3$ does not — $D_4$.
• $G/Z(G)$ cyclic $\Rightarrow$ $G$ abelian.
• Sylow: exist; all conjugate; $n_p \equiv 1 \pmod p$ and $n_p \mid m$. $n_p = 1 \iff$ normal.
• Order $15$: forced cyclic. Order $12$: $n_3 \in \{1,4\}$, both occur, but never simple.
• Counting elements of order $p$ is the standard non-simplicity move.
• Burnside: orbits $= \frac1{|G|}\sum|{\rm Fix}(g)|$.