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Let $a,b\in\mathbb N$ be coprime. By manipulating with the matrix $\operatorname{diag}(a,b)$, prove that the cyclic group $Z_{ab}$ is isomorphic to the direct sum $Z_a\oplus Z_b$.

I guess I should obtain the matrix $\operatorname{diag}(1,ab)$ from the matrix $\operatorname{diag}(a,b)$ by elementary (integer) row and column operations (which can be done by a theorem on Smith's normal form), but I don't know how to do that since $a,b$ needn't be invertible, so I cannot multiply rows/columns by them.

janmarqz
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user557
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2 Answers2

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You can certainly multiply rows and columns by $a$ and $b$, if all you do is add that multiple to another row or column. Less cryptically put: $$ \det\begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix} = 1 $$ so it is safe to multiply by it. First however, let $s,t\in\Bbb Z$ be so that $sa+tb=1$, which you can do because they are coprime. So, consider $$ \begin{pmatrix} s & -t \\ b & a \end{pmatrix} \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix} \begin{pmatrix} 1 & 0 \\ -1 & 1\end{pmatrix} \begin{pmatrix} 1 & tb \\0 & 1 \end{pmatrix} =\begin{pmatrix} sa & -tb \\ ba & ab \end{pmatrix} \begin{pmatrix} 1 & 0 \\ -1 & 1\end{pmatrix} \begin{pmatrix} 1 & tb \\0 & 1 \end{pmatrix} =\begin{pmatrix} 1 & -tb \\ 0 & ab \end{pmatrix} \begin{pmatrix} 1 & tb \\0 & 1 \end{pmatrix} =\begin{pmatrix} 1 & 0 \\ 0 & ab \end{pmatrix} $$ where we have multiplied from the left and from the right only by invertible matrices.

Jesko Hüttenhain
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  • Thanks. Your proof is clear, but I don't have intuition on how could one come up with left multiplication by the matrix with rows $(s,-t), (b,a)$. That is, I don't see which elementary row operations left multiplication by this matrix corresponds to. – user557 Mar 31 '18 at 20:34
  • Well, it is a matrix formulation of an argument which always boils down to expressing $1$ as a $\Bbb Z$-linear combination of $a$ and $b$. – Jesko Hüttenhain Mar 31 '18 at 20:43
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Hint: To obtain that $1$ in the diagonal, use Bezout's identity: there are integers $u, v$ such that $ua+vb=1$.

Berci
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