IA-automorphism group of finite nilpotent group has precisely the same prime factors of order as the derived subgroup
Suppose is a finite nilpotent group. Then, the IA-automorphism group of (i.e., the subgroup of the automorphism group of 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 .
- Equivalence of definitions of finite nilpotent group
- IA-automorphism group of finite p-group is p-group
Given: A finite nilpotent group .
To prove: The prime factors of the order of the IA-automorphism group of are precisely the same as the prime factors of the order of the derived subgroup of .
Proof: By Fact (1) and some more work, we can see that an automorphism of is an IA-automorphism if and only if it induces IA-automorphisms on each of the Sylow subgroups of . We can work out from this that the IA-automorphism group of is isomorphic to the external direct product of the IA-automorphism groups of each of the Sylow subgroups of . By Fact (2), for each prime factor of , the corresponding IA-automorphism group is a -group, so the IA-automorphism group of is isomorphic to a direct product of groups whose orders are powers of primes appearing in the prime factorization of .
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 , the IA-automorphism group for the -Sylow subgroup is trivial if and only if the corresponding -Sylow subgroup in is abelian, which in turn is equivalent to saying that the derived subgroup of does not have dividing its order. This completes the proof.