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

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Left multiplication on $G / H$ is transitive, has stabiliser $H$ at $H$ , and has kernel ${Core}_{G} (H)$

Statement

Let $H \leq G$ . Left multiplication defines a transitive action of $G$ on the coset set $G / H$ by

$g \cdot (a H) : = (g a) H .$

The stabilizer of the point $H$ is $H$ . The corresponding homomorphism $ρ : G \to Sym (G / H)$ has

$\ker ρ = {Core}_{G} (H) .$

Facts & Assumptions

Given: A group $G$ and a subgroup $H \leq G$ .

[L1]

A left action satisfies the identity and multiplication laws and is transitive when some group element carries any chosen point to any other (Left group actions, transitive actions, and faithful actions).

[L2]

Every action yields a homomorphism into the symmetric group of the acted-on set (Actions of $G$ on $X$ correspond exactly to homomorphisms $G \to Sym (X)$ ).

[L3]

The elements of $G / H$ are the left cosets $a H$ (Left and right cosets $g H$ and $H g$ of a subgroup, The coset set $G / H$ and the index $[G : H]$ of a subgroup).

[L4]

One has $a H = b H$ exactly when $a^{- 1} b \in H$ ( $x \in a H$ iff $a^{- 1} x \in H$ , and $a H = b H$ iff $a^{- 1} b \in H$ ).

[L5]

The kernel of a homomorphism consists of the elements mapped to the identity (The kernel and image of a group homomorphism).

[L6]

The core is ${Core}_{G} (H) = ⋂_{a \in G} a H a^{- 1}$ (The core ${Core}_{G} (H) = ⋂_{g \in G} g H g^{- 1}$ of a subgroup).

[L7]

The core is a normal subgroup contained in $H$ ( ${Core}_{G} (H)$ is the largest normal subgroup of $G$ contained in $H$ ).

Proof

technique · direct

1.1

If $a H = b H$ , then $a^{- 1} b \in H$ by [L4], and $(g a)^{- 1} (g b) = a^{- 1} b \in H$ , so $(g a) H = (g b) H$ and the rule is well-defined. It satisfies $e \cdot (a H) = a H$ and $g \cdot (k \cdot (a H)) = (g k) a H = (g k) \cdot (a H)$ ; moreover $a \cdot H = a H$ , so the action is transitive, and $g \cdot H = H$ exactly when $g \in H$ .

L1L3L4

2.1

By [L2], the action defines $ρ : G \to Sym (G / H)$ . By [L5], an element $k$ lies in $\ker ρ$ exactly when $k \cdot (a H) = a H$ for every $a \in G$ , that is, when $(k a) H = a H$ for every $a$ .

step 1.1L2L5

3.1

By [L4], $(k a) H = a H$ is equivalent to $a^{- 1} k a \in H$ , or to $k \in a H a^{- 1}$ . Requiring this for every $a$ gives $k \in ⋂_{a} a H a^{- 1} = {Core}_{G} (H)$ , so $\ker ρ = {Core}_{G} (H)$ , which is normal by [L7].

step 2.1L4L6L7∎

Depends on

Used by

Dependency tree · two levels

24 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

P. Brosnan, Undergraduate Algebra Notes, 3.14: G-Sets, Proposition 3.102 and Corollary 3.104 (standard reference, not scraped)
K. Conrad, Group Actions, Theorem 6.8 (standard reference, not scraped)

Left multiplication on G/HG/H is transitive, has stabiliser HH at HH, and has kernel Core⁡G(H)CoreG​(H)