2

Let's say we have torsion-free abelian group $G$ of rank $r$, so it has a basis $A = \left\{ g_1,...,g_r \right\}$. Let's also assume there is a subgroup $H$ of $G$ of the same rank $r$, so it has a basis $B = \left\{ h_1,...,h_r \right\}$. I would like to know if there exist integers $m_1,...,m_n$ such that $\left\{ m_1g_1,...,m_ng_n \right\}$ is also a basis of $H$ and am struggling a bit. Any help or guidelines will be greatly appreciated.

Rob Arthan
  • 48,577
James
  • 749
  • 4
  • 12
  • See: smith normal form. Choose any basis for $A,B$ respectively then use the smith normal form algorithm to find invariant factors. – Rubertos Sep 11 '17 at 19:07
  • It's not clear what you are being asked to do. You don't give enough information to determine how the elements of the basis $B$ of $H$ can be expressed as linear combinations of the elements of the basis $A$ of $G$. – Rob Arthan Sep 11 '17 at 19:10
  • @RobArthan I would like to know if there is any way of choosing integers $m_1,....,m_r$ such that $\left{ m_1g_1,...,m_rg_r \right}$ is a basis of $H$. – James Sep 11 '17 at 19:19
  • That's certainly not true in general: in general, a basis of $H$ will comprise linear combinations of the $g_i$ not just simple multiples. You still need to clarify your question: how are the $h_i$ given to you? what does "in terms of the elements in $A$ mean? Please edit your question to make it clear exactly what you are asking. – Rob Arthan Sep 11 '17 at 19:50
  • @RobArthan Yes I edited. I am aware that's certainly not true in general, however I was wondering the fact that $G$ and $H$ have the same rank could make a difference. Thank you. – James Sep 11 '17 at 20:01

1 Answers1

1

Take $r=2$, so $G$ is the free abelian group on $g_1$ and $g_2$. If $H$ is generated by $h_1 = 2g_1 + 2g_2$ and $h_2 = 6g_2$, then $H$ also has rank $2$. Now look at the intersections of $H$ with the subgroups $\langle g_i \rangle$ generated by the $g_i$. $H \cap \langle g_1 \rangle = \langle 6g_1 \rangle$ and $H \cap \langle g_2\rangle = \langle 6g_2\rangle$. But $H \neq \langle 6g_1, 6g_2\rangle$ (since the latter group does not contain $2g_1 + 2g_2$). So even though the ranks are equal, you can't expect $H$ to be generated by multiples (rather than linear combinations) of the $g_i$.

Rob Arthan
  • 48,577