TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02

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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.
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$\operatorname{Inn}(G)$ is a normal subgroup of $\operatorname{Aut}(G)$

Statement

$\operatorname{Inn}(G)$ is a normal subgroup of $\operatorname{Aut}(G)$ .

Facts & Assumptions

Given: A group $G$ .

[L1]

Inner automorphisms are the maps $c_g$ (Inner automorphisms and $\operatorname{Inn}(G)$ ).

[L2]

$\operatorname{Aut}(G)$ is a group under composition (The automorphisms of a group form a group under composition).

[L3]

A conjugation-stable subgroup is normal (Equivalent characterisations of a normal subgroup by conjugates and left and right cosets).

[L4]

The inverse of a bijective homomorphism is a homomorphism (The inverse of a bijective group homomorphism is a group homomorphism).

Proof

technique · direct

1.1

For $\varphi\in\operatorname{Aut}(G)$ and $c_g\in\operatorname{Inn}(G)$ , direct evaluation gives $(\varphi c_g\varphi^{-1})(x)=\varphi(g)x\varphi(g)^{-1}=c_{\varphi(g)}(x)$ .

L1L2L3L4givenalgebra

2.1

Thus conjugation by every element of $\operatorname{Aut}(G)$ carries $\operatorname{Inn}(G)$ into itself; applying the same statement to $\varphi^{-1}$ gives equality.

step 1.1L1L2L3L4givenalgebra

3.1

The conjugation closure in step 2.1 proves normality.

step 2.1∎

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: 33 results over 13 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

Milne, Group Theory, Automorphisms of Groups (standard reference, not scraped)

Inn⁡(G)\operatorname{Inn}(G)Inn(G) is a normal subgroup of Aut⁡(G)\operatorname{Aut}(G)Aut(G)