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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?

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What is a presentation matrix? –  Matt N. Dec 8 '12 at 7:49
    
I am referring to this:bookre.org/reader?file=1096064&pg=487 –  alex Dec 8 '12 at 8:05
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1 Answer

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 $G$.

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the first part of your answer is exactly what i did when i said that the corresponding abelian group should be trivial. thanks –  alex Dec 8 '12 at 8:57
    
You are welcome. The second part is interesting (I think) if one knows more about the big picture (how abelian groups fit into the theory of modules). But it is not neccessary when dealing with abelian groups. –  Hans Giebenrath Dec 8 '12 at 9:11
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