TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck 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

$\bigcup_{g\in G}gHg^{-1}\ne G.$

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 $N_G(H)=\{g\in G:gHg^{-1}=H\}$ (The normalizer $N_G(H)=\{g\in G:gHg^{-1}=H\}$ 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 $x\mapsto gxg^{-1}$ 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 $\lvert A\rvert$ 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 $\lvert B\rvert \le \lvert A\rvert$ , 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 $\lvert A \cup B\rvert = \lvert A\rvert + \lvert B\rvert$ and $\lvert\bigcup_{i \in I} A_i\rvert = \sum_{i \in I}\lvert A_i\rvert$ , 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 $K_1,\ldots,K_m$ 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 $hHh^{-1}=H$ , so [L2] gives $H\le N_G(H)$ .

L1L2L3L4

2.1

Add the sets successively after removing elements already counted. The common identity contributes once and each $K_i\setminus\{e\}$ contributes at most $|H|-1$ , so [L6], [L7], and [L8] give $|\bigcup_iK_i|\le 1+m(|H|-1)$ .

step 1.1L6L7L8

2.2

Put $n=[G:H]$ . Properness gives $n\ge2$ . Since $H\le N_G(H)$ , [L5] gives $m\le n$ , and [L5] also gives $|G|=n|H|$ .

step 1.1L2L3L5L6

3.1

Therefore $|\bigcup_iK_i|\le1+n(|H|-1)=|G|-n+1<|G|$ , so the union is a proper subset of $G$ .

step 2.1step 2.2algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 98 results over 21 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

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