Quotient group maps to outer automorphism group of normal subgroup

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Statement

Suppose $G$ is a group and $N$ is a normal subgroup of $G$. There is a natural choice of homomorphism of groups from the quotient group $G/N$ to the outer automorphism group $\operatorname{Out}(N)$:

$G/N \to \operatorname{Out}(N)$

defined as follows: for any element of $G/N$, pick an element $g \in G$ in that coset of $N$. Conjugation by $g$ induces an automorphism of $N$, i.e., an element of the automorphism group $\operatorname{Aut}(N)$. Although the automorphism depends on the choice of $g$ in the coset of $N$, the coset of $\operatorname{Inn}(N)$ in $\operatorname{Aut}(N)$ for that element is independent of $g$.

More explicitly, there is a composite map:

$G \to \operatorname{Inn}(G) \to \operatorname{Aut}(N)$

Under this map, the image of the subgroup $N$ of $G$ lies inside $\operatorname{Inn}(N)$. Thus, we get an induced map from the quotient group $G/N$ to the quotient group $\operatorname{Aut}(N)/\operatorname{Inn}(N) = \operatorname{Out}(N)$.