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

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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.