The maximal free abelian subgroup that can be embedded in $GL(n,\mathbb{Z})$

Question

I am stuck on this problem and cannot seem to find a good reason for drawing the required conclusion. The problem is as follows:

Given $SL(n, \mathbb{Z})$ a subroup in $GL(n, \mathbb{Z})$. How can one prove that the ranks of their maximal free abelian subgroups are the same?

Roughly speaking, This is what I have done. Take $F$ to be the maximal free abelian subgroup of $Sl(n,\mathbb{Z})$. It is a free abelian subgroup of $GL(n,\mathbb{Z})$. Suppose we have an element $x$ in $GL(n,\mathbb{Z})$ that commutes with $F$. now I need to show that $x$ is in $F$ or in just in $SL(n, \mathbb{Z})$ to conclude a contradiction with the fact that $F$ is maximal free abelian in $SL(n, \mathbb{Z})$. Maybe I should use the fact that $Sl(n,\mathbb{Z})$ has index $2$ in $GL(n, \mathbb{Z})$ but I don't know how.

Answer 1

Your own final sentence is a good hint. I'll improve the hint with two questions.

Take $F$ to be a maximal free abelian subgroup of $GL(n,\mathbb Z)$.

• Knowing that $SL(n,\mathbb Z)$ has index 2 in $GL(n,\mathbb Z)$, what can you say about the index of $F' = F \cap SL(n,\mathbb Z)$ in $F$?
• Given a free abelian group $F$ of a certain rank, and given a finite index subgroup $F'$ of $F$, what can you say about the rank of $F'$ in relation to the rank of $F$?