IA-automorphism group of nilpotent group equals stability group of lower central series

For an individual automorphism
Suppose $$G$$ is a nilpotent group. Then, the following are equivalent for an automorphism $$\sigma$$ of $$G$$:


 * 1) $$\sigma$$ is an IA-automorphism of $$G$$, i.e., it induces the identity map on the abelianization of $$G$$.
 * 2) $$\sigma$$ is a stability automorphism for the lower central series of $$G$$.

Note that $$G$$ being nilpotent is important only in so far as it guarantees that the lower central series reaches the trivial subgroup. A slight variant of the statement would be true for non-nilpotent groups, but we wouldn't use the jargon of stability automorphism.

For subgroups of the automorphism group
Suppose $$G$$ is a nilpotent group. The following subgroups of the automorphism group of $$G$$ are equal:


 * 1) The subgroup of IA-automorphisms of $$G$$, i.e., automorphisms that induce the identity map on the abelianization of $$G$$.
 * 2) The stability group (i.e., the group of stability automorphisms) for the lower central series of $$G$$.

Background fact on iterated commutator mapping
The proof basically follows from the fact that for any positive integer $$n$$, the left-normed iterated commutator gives a surjective $$n$$-linear map from the abelianization of $$G$$ to the quotient group $$\gamma_n(G)/\gamma_{n+1}(G)$$ between successive members of the lower central series. Here, $$\gamma_n(G)$$ is the $$n^{th}$$ member of the lower central series of $$G$$. Explicitly, it is a $$n$$-linear map of abelian groups:

$$G/G' \times G/G' \times \dots G/G' \to \gamma_n(G)/\gamma_{n+1}(G)$$

This mapping is canonical, hence covariant with automorphisms.

(1) implies (2)
Given: An IA-automorphism $$\sigma$$ of a nilpotent group $$G$$.

 To prove: $$\sigma$$ induces the identity map on each of the successive quotients of the lower central series, i.e., for every $$n$$, $$\sigma$$ induces the identity map on each of the groups of the form $$\gamma_n(G)/\gamma_{n+1}(G)$$ where $$\gamma_n(G)$$ is the $$n^{th}$$ member of the lower central series of $$G$$.

Proof: The covariance of the iterated commutator mapping with respect to $$\sigma$$, along with its surjectivity to $$\gamma_n(G)/\gamma_{n+1}(G)$$, guarantees that since $$\sigma$$ fixes $$G/G'$$ pointwise, it also fixes the image $$\gamma_n(G)/\gamma_{n+1}(G)$$ pointwise.

(2) implies (1)
Given: A nilpotent group $$G$$. An automorphism $$\sigma$$ induces the identity map on each of the successive quotients of the lower central series, i.e., for every $$n$$, $$\sigma$$ induces the identity map on each of the groups of the form $$\gamma_n(G)/\gamma_{n+1}(G)$$ where $$\gamma_n(G)$$ is the $$n^{th}$$ member of the lower central series of $$G$$.

To prove: $$\sigma$$ is an IA-automorphism of $$G$$.

Proof: This follows directly from setting $$n = 1$$.