Timeline for Abstract Commensurator Group of $\mathbb{Z}^n$ $Comm(\mathbb{Z}^n)\cong GL(n,\mathbb{Q})$?

6 events

when toggle format	what		by	license	comment
May 23, 2012 at 18:32	comment	added	Peter		Oh. hehe. Sorry. I was a little bit confused. I thought that I have to multiply $\alpha$ with $m$. So I didn't know how to do that. But know I'm fine. Thanks.
May 23, 2012 at 17:46	comment	added	Igor Rivin		I am puzzled by your question. The map $\alpha$ is only defined on $H.$ The map $\alpha \circ m$ is given by $\alpha \circ m (x) = \alpha(m x).$ Since $m x \in H$ you are good.
May 23, 2012 at 17:37	comment	added	Peter		Sorry. But I don't see, why this map is an injectic endomorphism. I know that $m\mathbb{Z}^n\leq H$ because $[G:H]=m$ and so the we can restrict $\alpha$ to this subgroup. But how is this map defined on the other elements $\mathbb{Z}^n\backslash m\mathbb{Z}^n$. And if we multiply with $m$, we imagine $\alpha$ as matrix with integers in $\mathbb{Q}$, right?
May 23, 2012 at 15:18	comment	added	Igor Rivin		$m$ is multiplication by $m.$
May 23, 2012 at 14:30	comment	added	Peter		Thanks. In the proof, what is meant with the $\alpha m=\alpha\mid_{m\mathbb{Z}^n}\circ m$ What does the $\circ$ mean? Is it composition? If so, which map is meant by $m$?
May 23, 2012 at 13:49	history	answered	Igor Rivin	CC BY-SA 3.0