Questions: The Class Equation

Question 1

A group G has order p⁴ where p is prime. What does the class equation guarantee about Z(G)?

• A. Z(G) is trivial — only the identity commutes with all elements in most p-groups
• B. Z(G) has order divisible by p, since every term in the class equation divides p⁴ and each non-central term is divisible by p
• C. Z(G) must equal G, making G abelian
• D. Z(G) could have any size — the class equation gives no information about prime-power groups

Answer: (B)

The class equation states |G| = |Z(G)| + Σ|Cᵢ| where each non-central class size equals |G|/|C_G(x)| and divides p⁴. Since each non-central class has size greater than 1 and divides p⁴, each such size is divisible by p. The total |G| = p⁴ is also divisible by p. So |Z(G)| = p⁴ − (sum of terms each divisible by p) must itself be divisible by p, giving |Z(G)| ≥ p > 1. The center is non-trivial. This argument works for any p-group and is the heart of the result.

Question 2

An element x commutes with every element of G. What is the size of the conjugacy class of x?

• A. Equal to |G| — central elements appear across every conjugacy class
• B. Equal to |G| / |Z(G)| — central elements spread proportionally through the group
• C. 1 — the conjugacy class of x contains only x itself, since gxg⁻¹ = x for all g
• D. Equal to the index of the centralizer of x in G

Answer: (C)

If x ∈ Z(G), then gxg⁻¹ = x for every g ∈ G — conjugation moves nothing. The conjugacy class of x is its orbit under conjugation, which contains only x. Singleton orbits correspond exactly to central elements; non-central elements have conjugacy classes of size greater than 1. This is why the class equation separates into |Z(G)| (the total contribution of singleton classes) plus the sum of non-singleton class sizes.

Question 3

The class equation is a new, independent theorem about finite groups that requires different techniques from the orbit-stabilizer theorem.

• T. True
• F. False

Answer: False

The class equation IS the orbit-stabilizer theorem applied to the conjugation action of G on itself: g · x = gxg⁻¹. The orbits under this action are the conjugacy classes, and the stabilizer of x is its centralizer C_G(x). The orbit-stabilizer theorem gives |conjugacy class of x| = |G|/|C_G(x)|. Summing over all orbits and separating singletons (central elements) yields the class equation directly. No new technique is required — it is an application of existing machinery.

Question 4

In any finite group G, the size of every conjugacy class divides |G|.

• T. True
• F. False

Answer: True

By the orbit-stabilizer theorem applied to conjugation, the size of the conjugacy class of x equals |G| / |C_G(x)|, where C_G(x) is the centralizer of x — a subgroup of G. By Lagrange's theorem, |C_G(x)| divides |G|, so |G| / |C_G(x)| also divides |G|. This divisibility is what makes the class equation a powerful constraint: when |G| = pⁿ, every class size is a power of p, forcing the center to be non-trivial.

Question 5

Explain why a p-group (a group of prime power order pⁿ) must have a non-trivial center. Walk through the class equation argument.

Model answer: Write the class equation: |G| = |Z(G)| + Σ|Cᵢ| where the sum is over non-central conjugacy classes. Since |G| = pⁿ, every conjugacy class size divides pⁿ (by orbit-stabilizer), so each non-central class has size pᵏ for some k ≥ 1 — each is divisible by p. The total |G| = pⁿ is divisible by p. So |Z(G)| = pⁿ − (sum of multiples of p) must itself be divisible by p. Since p ≥ 2 divides |Z(G)|, we have |Z(G)| ≥ p > 1, so the center is non-trivial.

This is modular arithmetic forcing group structure. The class equation expresses |G| as a sum; every term except |Z(G)| is divisible by p for a p-group. Since the whole sum is divisible by p, the remaining term must also be divisible by p — an elementary but powerful congruence argument. The conclusion seeds further results: groups of order p² are abelian, and the class equation argument recurs throughout the proof of the Sylow theorems.