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

## Statement

Suppose $G$ is a group, $H$ is a subgroup, and $\sigma$ is an Automorphism (?) of $G$ such that both $\sigma$ and $\sigma^{-1}$ leave $H$ invariant. Then, $\sigma$ restricts to an automorphism of $H$.

## Related facts

• Restriction of automorphism to subgroup not implies automorphism: If $\sigma$ is an automorphism of a group $G$ and $H$ is a subgroup of $G$ such that $\sigma(H) \subseteq H$. The restriction of $\sigma$ 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 $\sigma$ of $G$ such that $\sigma(H) \subseteq H$ and $\sigma^{-1}(H) \subseteq H$.

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

Proof:

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