Automorphism group for a finitely generagted abelian group

Question

$\newcommand{\Aut}{\mathrm{Aut}}$ Let $G$ be a finitely generated abelian group, we know that $G\cong \Bbb Z^n\oplus\bigoplus \Bbb Z_{p_i^{e_i}}$.

I already know that $\Aut\Bbb Z_n\cong (\Bbb Z_n)^\times$. For the torsion-free part I know if I regard $\Bbb Z$ as a ring I will have $\Aut\Bbb Z^n\cong\mathrm{GL}_n(\Bbb Z)$ like in this post

My questions are:

1. Does the latter result ($\Aut\Bbb Z^n\cong\mathrm{GL}_n(\Bbb Z)$) hold when we are only interested in its group structure? In proving it I used that the surjective ring homomorphism preserves multiplicative identity, which seems not working if I just deal with group homomorphism.

2. How can I composite these two result to compute a general $\Aut G$? I know $\Aut G\times H\cong\Aut G\times\Aut H$ seldom holds, except some special cases like $\gcd(|G|,|H|)=1$ or something else...

Specificly, I'd like to show that there are only 4 finitely generated abelian groups satisfying $|\Aut G|=2$. But I'm also interested in a general solution to the above 2 questions.

Thanks in advance!

Answer 1

1. Recall that being an abelian group and a $\mathbb{Z}$-module are equivalent notions, so $\mathrm{Aut}(\mathbb{Z}^n)$ is the group of $\mathbb{Z}$-module automorphisms of $\mathbb{Z}^n$. The claim is that in terms of matrices with respect to the standard basis, this is precisely the group $\mathrm{GL}_n(\mathbb{Z})$ of $n \times n$ matrices with integer coefficients and determinant $\pm 1$. Suppose $\phi \in \mathrm{Aut}(\mathbb{Z}^n)$ has matrix $M$. Clearly $M$ has integer entries, so $\det(M) \in \mathbb{Z}$. Since $\phi$ is an automorphism, the same argument applied to $\phi^{-1}$ gives $\det(M^{-1}) \in \mathbb{Z}$. Thus $\det(M) = \pm 1$. For the converse, the key observation is that $M \mathrm{adj}(M) = \det(M) I$ where $\mathrm{adj}(M)$ is the adjugate matrix. Thus when $\det(M) = \pm 1$ and $M$ has integer entries, the same is true of $M^{-1}$, so such $M$ represents an automorphism.

2. Well certainly $\mathrm{Aut}(G) \times \mathrm{Aut}(H) \leq \mathrm{Aut}(G \times H)$, which is enough to rule almost everything out for your desired application.

Regarding (2), I got curious about this situation and worked the following out. I wrote it up before noticing the link in the comments, which gives a more general result; I'll leave it since it's explicit.

Lemma: Let $R$ be an integral domain and let $U, V$ be $R$-modules. Suppose $U$ is torsion-free and $V$ is torsion. Then $$\mathrm{Aut}_R(U \times V) \cong \begin{pmatrix}\mathrm{Aut}_R(U) & 0 \\ \mathrm{Hom}_R(V, U) & \mathrm{Aut}_R(V)\end{pmatrix}.$$

Proof: ($\supseteq$) Let $\alpha \in \mathrm{Aut}(V)$, $\beta \in \mathrm{Hom}(U, V)$, $\gamma \in \mathrm{Aut}(V)$. Define $\phi \colon U \times V \to U \times V$ by $\phi(u, v) = (\alpha(u), \beta(u) + \gamma(v))$. It is easy to check this is in $\mathrm{End}(U \times V)$. Its inverse is $\psi(u, v) = (\alpha^{-1}(u), -\gamma^{-1}\beta\alpha^{-1}(u) + \gamma^{-1}(v))$. (This argument uses almost none of our assumptions.)

($\subseteq$) Let $\phi \in \mathrm{Aut}(U \times V)$. Suppose $\phi(0, v) = (a, b)$. Since $v, b$ are torsion, we have some $0 \neq r$ such that $r \cdot v = 0 = r \cdot b$. Acting by $r$ now gives $(0, 0) = (r \cdot a, 0)$, so $a = 0$ since $U$ is torsion-free. That is, $\phi(0, v) = (0, b)$.

Now write $\phi(u, v) = (\pi_U(u, v), \pi_V(u, v))$ where $\pi_U, \pi_V$ are the projection morphisms. We just showed $\pi_U(0, v) = 0$. Hence for any $v_1, v_2 \in V$, we have $$\pi_U(u, v_1) - \pi_U(u, v_2) = \pi_U(0, v_1-v_2) = 0,$$ so $\pi_U(u, v) = \alpha(u)$ is independent of $v$. Since $\phi$ is bijective, now $\alpha$ must be as well, so $\alpha \in \mathrm{Aut}(U)$.

Let $\beta(u) = \pi_V(u, 0)$ and let $\gamma(v) = \pi_V(0, v)$, so $\pi_V(u, v) = \beta(u) + \gamma(v)$. Setting $u=0$ or $v=0$ shows that $\beta, \gamma$ are $R$-linear, so $\beta \in \mathrm{Hom}(V, U)$. Since $\phi(0, v) = (0, \gamma(v))$ and $\phi$ is injective, so is $\gamma$. Moreover, if $v_0 \in V$, then $\phi(u, v) = (0, v_0)$ for unique $u, v$. We must have $u=0$, so $v_0 = \gamma(v)$ and $\gamma$ is surjective. Thus $\gamma \in \mathrm{Aut}(V)$. $\Box$

Now, you want to apply the lemma when $R=\mathbb{Z}$, $U = \mathbb{Z}^n$, $V = \oplus_i \mathbb{Z}_{p_i^{e_i}}$. The hypotheses are clearly satisfied. The result is $$\mathrm{Aut}(\mathbb{Z}^n \times \oplus_i \mathbb{Z}_{p_i^{e_i}}) = \begin{pmatrix}\mathrm{GL}_n(\mathbb{Z}) & 0 \\ \oplus_i \mathbb{Z}_{p_i^{e_i}} & \mathrm{Aut}(\oplus \mathbb{Z}_{p_i^{e_i}})\end{pmatrix}.$$

The upper-left and lower-left terms are what they are. For the lower-right term, using the usual $\gcd$ property, it breaks up as a product of automorphism groups over distinct primes. For each fixed prime, I think it's messy.

Answer 2

Note that we can view $\mathbb{R}^n$ as tiled by congruent fundamental parallelepipeds corresponding to the integer lattice, where each parallelepiped has corners at points that differ by elements of $\mathbb{Z}^n$. Any automorphism $\varphi: \mathbb{Z}^n \to \mathbb{Z}^n$ shifts lattice points to lattice points bijectively.

We extend $\varphi$ to a piecewise-linear map on $\mathbb{R}^n$ by sending each lattice point $x$ to $\varphi(x)$ and then defining the image of each fundamental parallelepiped in the simplest affine way consistent with those vertex images. Because there is no overlap in images of distinct parallelepipeds (they must remain disjoint except at boundary points, or else injectivity on the lattice itself would fail), this extension is forced to be a globally linear bijection on $\mathbb{R}^n$. That linear bijection preserves the set of integer points and carries the original integer-lattice tiling to another integer-lattice tiling, which forces all images of basis vectors to remain integral and forces the linear map to have determinant $\pm 1$.

Indeed, if the determinant were not $\pm 1$, then either the images of entire parallelepipeds would overlap improperly (if $|\det| < 1$) or would fail to tile $\mathbb{R}^n$ without gaps (if $|\det| > 1$), contradicting bijectivity on the lattice. Thus any automorphism $\varphi$ of $\mathbb{Z}^n$ extends to a linear map represented by an integer matrix of determinant $\pm 1$, and conversely any integer matrix of determinant $\pm 1$ obviously gives a lattice-preserving bijection $\mathbb{Z}^n \to \mathbb{Z}^n$.

Hence $\operatorname{Aut}(\mathbb{Z}^n) \cong \mathrm{GL}_n(\mathbb{Z})$.