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Let $\mathbb{Z}_p$ denote the $p$-adic integers.

I know that the groups $GL_n(\mathbb{Z})$ and $GL_n(\mathbb{Z}_p)$ are (topologically for the latter) finitely generated. My question is: what are the minimal number of generators needed to generate each of these groups (topologically for the latter)?

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In 1932/3 B.H. Neumann found a presentation for the automorphism group of the free group of rank $n$ on two generators for $n \ge 4$ and 3 generators for $n=3$. These were based on earlier presentations on 4 generators due to Nielsen. A single additional relation gives a presentation of the quotient ${\rm GL}_n({\mathbb Z})$. You can find these results in the book on Combinatorial group Theory by Magnus, Karass and Solitar.

So, for ${\rm GL}_n({\mathbb Z})$, the answer is 2 for $n \ge 4$ and at most 3 for $n=3$. It is well-known to be 2 for $n=2$. It seems plausible that it is also 2 for $n=3$, and I am sure this must be known. I would expect similar results to hold for ${\rm GL}_n({\mathbb Z_p})$.

You could try asking this on MathOverflow.

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    $\begingroup$ Thanks for the answer! Now that I think of it, since the map $SL_n(\mathbb{Z})\rightarrow SL_n(\mathbb{Z}/p^m\mathbb{Z})$ is surjective, $SL_n(\mathbb{Z})$ is dense in $SL_n(\mathbb{Z}_p)$, so to generate $GL_n(\mathbb{Z}_p)$, it's enough to tack on an element whose determinant topologically generates $\mathbb{Z}_p^\times/\{\pm 1\}$. $\endgroup$ – Ian Malcolm Sep 25 '12 at 21:54
  • $\begingroup$ (I was mainly interested in whether the answer was uniformly bounded in $n$ and $p$) $\endgroup$ – Ian Malcolm Sep 25 '12 at 21:59
  • $\begingroup$ It has been asked on MathOverflow now - mathoverflow.net/questions/181366/…. $\endgroup$ – Dietrich Burde Sep 22 '14 at 8:31
  • $\begingroup$ In particular, the minimal cardinal of a generating subset of $\mathrm{GL}_n(\mathbf{Z})$ is 2 for all $n\ge 2$. $\endgroup$ – YCor Apr 28 '19 at 11:46

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