Inner group automorphism

The inner group automorphism $I n n (𝐺) ⊴ A u t (𝐺)$ is a normal subgroup given by conjugation by an element, #m/def/group i.e. if $ˆ 𝑔 \in I n n (𝐺)$ then $ˆ 𝑔 (ℎ) = 𝑔 ℎ 𝑔 - 1$ for some $𝑔 \in 𝐺$ . It is hence the image of conjugation as a group action $̂ \cdot : 𝐺 \to A u t (𝐺)$ .

Proof of normal subgroup

That $I n n (𝐺)$ is a subgroup follows from the fact that it is the image of the homomorphism $̂ \cdot : 𝐺 \to A u t (𝐺)$ . Let $𝑔 \in 𝐺$ and $𝜑 \in A u t (𝐺)$ . Then for any $ℎ \in 𝐺$

(𝜑 ̂ 𝑔 𝜑 - 1) (ℎ) = (𝜑 ̂ 𝑔) (𝜑 - 1 (ℎ)) = 𝜑 (𝑔 𝜑 - 1 (ℎ) 𝑔 - 1) = 𝜑 (𝑔) ℎ 𝜑 (𝑔) - 1 = ̂ 𝜑 (𝑔) (ℎ)

thus $𝜑 ̂ 𝑔 𝜑 - 1 = ̂ 𝜑 (𝑔) \in I n n (𝐺)$ .

By the First isomorphism theorem, this is isomorphic to $𝐺 / 𝑍 (𝐺)$ , where the divisor is the centre.

#state/tidy | #lang/en | #SemBr