We have this theorem:
** Let $F$ be a free abelian group of rank $n$ and let $H$ be a subgroup
of $F.$ There exists a basis $\{x_1,…,x_n\}$ of $F$ and integers $d_1,…,d_r > 0 $ such
that
• $di\vert d_{i+1}$ for $i = 1,…,r$
• $\{d_1x_1,\ldots,d_rx_r\}$ is a basis of H.**
I am confused with this. $\mathbb{Z}*\mathbb{Z}$ is a free abelian group with basis $(1,0)$ and $(0,1)$. Let $H=2\mathbb{Z}*3\mathbb{Z}$ be a subgroup of $\mathbb{Z}*\mathbb{Z}$ with basis $2(1,0)$ and $3(0,1)$. $\mathbf {But}$ $2$ does not divide $3$. What is the problem here?
Best Answer
In your example, take $\;B:=\{\,(2,3)\,,\,(1,2)\,\}\;$ , then show:
1) $\;B\;$ is a $\,\Bbb Z\,-$ linearly independent set and is thus a basis of $\;\Bbb Z\times\Bbb Z\;$ ;
2) $\;C:=\left\{\,1\cdot(2,3)\,,\,6\cdot(1,2)\,\}=\{\,(2,3)\,,\,\,(6,12)\,\right\}\;$ are $\;\Bbb Z\,-$ linearly independent and a basis of $\;2\Bbb Z\times3\Bbb Z\;$ as required.