About abelian groups If the determinant of the presentation matrix for an abelian group is one, does it mean that the abelian group is simply the trivial group?
 A: Yes.
Using the Smith normal form (which the book uses for the proof of the structure theorem), we can assume that the presentation matrix $P$ is of the form
$$ P = \begin{pmatrix} d_1 & 0 & \dotsc & 0 \\ 
                     0 & d_2 & \ddots & 0 \\
                     \vdots & \ddots & \ddots & 0 \\
                     0 & 0 & 0 & d_n \end{pmatrix} \in \mathbb Z^{n\times n}
$$
with integers $d_i$ such that $d_i\mid d_{i+1}$ for $i \leq 1 \leq n-1$. Then we know that $G$ is isomorphic to the product
$$ (\mathbb Z / d_1 \mathbb Z) \oplus \dotsb \oplus (\mathbb Z / d_n \mathbb Z) = 
\bigoplus_{i=1}^n \mathbb Z/ d_i \mathbb Z.$$
If the determinant of $P$ is $1$, all $d_i$ have to be $1$. But this implies that all $\mathbb Z / d_i \mathbb Z$ are trivial. Thus $G$ is trivial.
If you are familiar with $\mathbb Z$-modules:
Assume your finitely generated abelian group $G$ gives rise to a surjective morphism $\mathbb Z^n \to G$ with kernel $K$. Thus $\mathbb Z^n/K \cong G$. So we only have to consider $\mathbb Z^n / K$. The base change matrix between a basis of $\mathbb Z ^n$ and $K$ is exactly a presentation matrix of $G$.
Now comes the real deal. Whenever you have finitely generated $\mathbb Z$-modules $N \subseteq M$ such that the quotient $M/N$ is finite, its order $\left|M/N\right|$ is given by $\left|\det(T)\right|$, where $T$ is the transformation matrix between a basis of $M$ and a basis of $N$. This can be shown using the Smith normal form, which you have hopefully encountered.
Hence $\left|G\right| = \left|\mathbb Z^n/K\right| = \left|\det(P)\right| = 1$, where $P$ denotes a presentation matrix of $G$.
