Restriction of automorphism to subgroup invariant under it and its inverse is automorphism

From Groupprops

Statement

Suppose $G$ is a group, $H$ is a subgroup, and $σ$ is an automorphism of $G$ such that both $σ$ and $σ - 1$ leave $H$ invariant. Then, $σ$ restricts to an automorphism of $H$ .

Related facts

Restriction of automorphism to subgroup not implies automorphism: If $σ$ is an automorphism of a group $G$ and $H$ is a subgroup of $G$ such that $\sigma(H) \subseteq H$ . The restriction of $σ$ to $H$ need not be an automorphism of $H$ .
For any group-closed automorphism property (i.e., any property of automorphisms such that for any given group the automorphisms satisfying the property form a group), any subgroup invariant under all automorphisms with the property satisfies the additional condition that the restriction of each such automorphism to the subgroup is an automorphism of the subgroup. A subgroup property that can be expressed this way is termed an auto-invariance property. Examples of this are:
- The property of being a characteristic subgroup is the invariance property with respect to all automorphisms. The restriction of any automorphism of the whole group to a characteristic subgroup is an automorphism of the subgroup.
- The property of being a normal subgroup is the invariance property with respect to all inner automorphisms. The restriction of any inner automorphism of the whole group to a normal subgroup is an automorphism of the subgroup.

Proof

Given: A group $G$ , a subgroup $H$ , an automorphism $σ$ of $G$ such that $\sigma(H) \subseteq H$ and $\sigma^{-1}(H) \subseteq H$ .

To prove: $σ(H) = H$ and the restriction of $σ$ to $H$ is an automorphism of $H$ .

Proof:

Since $σ$ is an automorphism of $G$ , so is $σ - 1$ , and their composite (both ways) is the identity map on $G$ . In other words, $σ(σ - 1 (g)) = g$ and $σ - 1 (σ(g)) = g$ for all $g \in G$ .
By our assumption, the restrictions $σ | H$ and $σ - 1 | H$ are both functions from $H$ to itself. Further, we have that $σ(σ - 1 (h)) = h$ and $σ - 1 (σ(h)) = h$ for all $h \in H$ . Thus, $σ | H$ and $σ - 1 | H$ are two-sided inverses of each other, and are thus both bijections. In particular, $σ(H) = H$ .
Finally, since $σ$ is a homomorphism, so is $σ | H$ . Thus, $σ | H$ is a bijective homomorphism from $H$ to itself, and is hence an automorphism of $H$ .