IA-automorphism group of finite nilpotent group has precisely the same prime factors of order as the derived subgroup

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Statement

Suppose G is a finite nilpotent group. Then, the IA-automorphism group of G (i.e., the subgroup of the automorphism group of G comprising the IA-automorphisms -- those automorphisms that preserve each of the cosets of the derived subgroup) is a finite group and the prime factors of its order are precisely the same as the prime factors of the order of the derived subgroup of G.

Facts used

  1. Equivalence of definitions of finite nilpotent group
  2. IA-automorphism group of finite p-group is p-group

Proof

Given: A finite nilpotent group G.

To prove: The prime factors of the order of the IA-automorphism group of G are precisely the same as the prime factors of the order of the derived subgroup of G.

Proof: By Fact (1) and some more work, we can see that an automorphism of G is an IA-automorphism if and only if it induces IA-automorphisms on each of the Sylow subgroups of G. We can work out from this that the IA-automorphism group of G is isomorphic to the external direct product of the IA-automorphism groups of each of the Sylow subgroups of G. By Fact (2), for each prime factor p of G, the corresponding IA-automorphism group is a p-group, so the IA-automorphism group of G is isomorphic to a direct product of groups whose orders are powers of primes appearing in the prime factorization of G.

It remains to show that a prime appears with a positive power if and only if it also appears as a factor in the order of the derived subgroup. This follows from the observation that, for a prime number p, the IA-automorphism group for the p-Sylow subgroup is trivial if and only if the corresponding p-Sylow subgroup in G is abelian, which in turn is equivalent to saying that the derived subgroup of G does not have p dividing its order. This completes the proof.