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

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 $G$.

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.