$\DeclareMathOperator\Aut{Aut}\DeclareMathOperator\End{End}\DeclareMathOperator\GL{GL}\DeclareMathOperator\Z{Z}\DeclareMathOperator\Sp{Sp}\DeclareMathOperator\Span{Span}\DeclareMathOperator\ker{ker}\DeclareMathOperator\coker{coker}\DeclareMathOperator\im{im}\newcommand{\Id}{\mathrm{Id}}$
Notation
For a group $L$, denote its center by $\Z(L)$ and its abelianisation by $L_1 = L/L'$.
An automorphism $A \in \Aut(L)$
restricts to an automorphism $A_{\Z} \in \Aut(\Z(L))$ of the center and
quotients to an automorphism $A_1 \in \Aut(L_1)$ of the abelianisation.
We thus have $\mathbb{Z}$-linear maps $(A_1 - \Id) \in \End(L_1)$ and $(A_{\Z} - \Id) \in \End(\Z(L))$.
Main questions
I am very interested in the following questions :
QA: Suppose that $L_1 / \im(A_1 - \Id)$ is torsion, what can we say about $\Z(L) / \im(A_{\Z} - \Id)$ ?
QA1: For which groups $L$ can we conclude that $\Z(L)/\im(A_{\Z} - \Id)$ is torsion ?
QA2: Can we construct examples where $L_1/\im(A_1 - \Id)$ is any torsion abelian group
and $\Z(L)/\im(A_{\Z} - \Id)$ is any abelian group ?
Can we do so even with $L$ a finitely generated nilpotent group of class 2 ?
Thoughts and related questions
Of course if $L$ is abelian the we have $Z(L)/\im(A_z-\Id)=L/im(A-Id)$ so we can realize all torsion abelian groups.
I tried constructing examples of QA2 where L is a nilpotent group of class 2.
The extension $1 \to L' \to L \to L_1 \to 1$ is central, namely $L' \subset \Z(L)$.
Define the skew-symmetric bilinear map $w : L_1 \times L_1 \to L'$
by $w(x_1, y_1) = [x,y]$ for any lifts $x,y \in L$ of $x_1, y_1 \in L_1$.
It satisfies $\ker(w^\flat) = \Z(L)/L'$ and $\im(w) = L'$.
We must have $A'(w(x_1,y_1)) = w(A_1(x_1),A_1(y_1))$ so $A_1$ uniquely determines $A'$ and thus $A$.
- QB: Is this enough to show that $L_1/\im(A_1 - \Id)$ is torsion $\implies$ $L'/\im(A'-\Id)$ is torsion ?
(It would then follow that $\Z(L)/\im(A_{\Z}-\Id)$ is torsion.)
This also leads to ask :
- QC: Which $A_1 \in \GL(L_1)$ extend to $A \in \Aut(A)$, in other terms
when does the formula $A'(w(a,b)) = w(A_1(a), A_1(b))$ yield a well defined element $A'\in \GL(L')$.
A necessary condition is that $w(a,b)=0 \implies w(A_1(a), A_1(b))=0$, namely that $A_1$ preserves $w$-orthogonality.
Understanding QC would yield conditions which may help answering QB.
Examples and a solution to QA2
The Heisenber group is the central extension of $L_1 = \mathbb{Z}^2$ by $L' = \mathbb{Z}$ with generators $x,y$ for $L_1$ and $z = [x,y]$ for $L'$.
Now any $A_1 \in \Sp_{2}(\mathbb{Z})$ will yield $A'(z)=z$, so $\coker(A'-\Id)=\mathbb{Z}$, and we may choose the $A_1$ so that $\coker(A_1-\Id)$ is torsion.
This example generalises with the center $\mathbb{Z}$ replaced by $\mathbb{Z}/n$. Then by taking direct sums we may realize any abelian group, giving a partial positive answer to QA2. Is that right ?
