TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck 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 $\operatorname{Core}_G(H)$

Statement

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

$g\cdot(aH):=(ga)H.$

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

$\ker\rho=\operatorname{Core}_G(H).$

Facts & Assumptions

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

[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 $G\to\operatorname{Sym}(X)$ ).

[L3]

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

[L4]

One has $aH=bH$ exactly when $a^{-1}b\in H$ ( $x\in aH$ iff $a^{-1}x\in H$ , and $aH=bH$ 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 $\operatorname{Core}_G(H)=\bigcap_{a\in G}aHa^{-1}$ (The core $\operatorname{Core}_G(H)=\bigcap_{g\in G}gHg^{-1}$ of a subgroup).

[L7]

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

Proof

technique · direct

1.1

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

L1L3L4

2.1

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

step 1.1L2L5

3.1

By [L4], $(ka)H=aH$ is equivalent to $a^{-1}ka\in H$ , or to $k\in aHa^{-1}$ . Requiring this for every $a$ gives $k\in\bigcap_a aHa^{-1}=\operatorname{Core}_G(H)$ , so $\ker\rho=\operatorname{Core}_G(H)$ , which is normal by [L7].

step 2.1L4L6L7∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 49 results over 12 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

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/HG/H is transitive, has stabiliser HHH at HHH, and has kernel Core⁡G(H)\operatorname{Core}_G(H)CoreG​(H)