TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11

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The conjugates of a proper subgroup do not cover a finite group

Statement

If $H$ is a proper subgroup of a finite group $G$ , then

$⋃_{g \in G} g H g^{- 1} \neq G .$

Thus some element of $G$ lies in no conjugate of $H$ .

Facts & Assumptions

Given: A finite group $G$ and a proper subgroup $H < G$ .

[L1]

There are $m = [G : N_{G} (H)]$ distinct conjugates of $H$ (The conjugates of $H$ are in bijection with $G / N_{G} (H)$ and, for finite $G$ , number $[G : N_{G} (H)]$ ).

[L2]

The normalizer is $N_{G} (H) = {g \in G : g H g^{- 1} = H}$ (The normalizer $N_{G} (H) = {g \in G : g H g^{- 1} = H}$ of a subgroup).

[L3]

The normalizer is a subgroup of $G$ ( $C_{G} (x)$ and $N_{G} (H)$ are subgroups of $G$ ).

[L4]

Conjugation is an automorphism, so every conjugate of $H$ has cardinality $∣ H ∣$ (Conjugation $x \mapsto g x g^{- 1}$ is an automorphism).

[L5]

For a finite group and subgroup, $∣ G ∣ = [G : H] ∣ H ∣$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ , The coset set $G / H$ and the index $[G : H]$ of a subgroup, The cardinality $∣ A ∣$ of a finite set).

[L6]

A subset of a finite set is finite, has no larger cardinality, and has equal cardinality only when it is the whole set (A subset of a finite set is finite, with $∣ B ∣ \leq ∣ A ∣$ , and equality holds if and only if $B = A$ ).

[L7]

The cardinality of a finite disjoint union is the sum of the cardinalities of its parts (The sum rule: a finite disjoint union is finite with $∣ A \cup B ∣ = ∣ A ∣ + ∣ B ∣$ and $∣ ⋃_{i \in I} A_{i} ∣ = \sum_{i \in I} ∣ A_{i} ∣$ , and a sum over a finite index set splits along a partition).

[L8]

Finite sums over finite index sets are well-defined (The sum $\sum_{i \in S} a_{i}$ over a finite index set, and its product form).

Proof

technique · direct

1.1

Let $K_{1}, \dots, K_{m}$ be the distinct conjugates of $H$ , where $m = [G : N_{G} (H)]$ by [L1]. Each $K_{i}$ has $∣ H ∣$ elements by [L4] and contains $e$ . For $h \in H$ , subgroup closure gives $h H h^{- 1} = H$ , so [L2] gives $H \leq N_{G} (H)$ .

L1L2L3L4

2.1

Add the sets successively after removing elements already counted. The common identity contributes once and each $K_{i} ∖ {e}$ contributes at most $∣ H ∣ - 1$ , so [L6], [L7], and [L8] give $∣ ⋃_{i} K_{i} ∣ \leq 1 + m (∣ H ∣ - 1)$ .

step 1.1L6L7L8

2.2

Put $n = [G : H]$ . Properness gives $n \geq 2$ . Since $H \leq N_{G} (H)$ , [L5] gives $m \leq n$ , and [L5] also gives $∣ G ∣ = n ∣ H ∣$ .

step 1.1L2L3L5L6

3.1

Therefore $∣ ⋃_{i} K_{i} ∣ \leq 1 + n (∣ H ∣ - 1) = ∣ G ∣ - n + 1 < ∣ G ∣$ , so the union is a proper subset of $G$ .

step 2.1step 2.2algebra∎

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Sources

K. Conrad, Group Actions, Theorem 6.10 (standard reference, not scraped)