Skip to main content

Timeline for Conjugacy classes of PGL(3,Z)

Current License: CC BY-SA 3.0

5 events
when toggle format what by license comment
Oct 9, 2013 at 19:44 comment added YCor yes because the obvious homomorphism $SL_3(\mathbb{Z})\to PGL_3(\mathbb{Z})$ is an isomorphism
Oct 9, 2013 at 19:11 comment added tj_ @Yves: Thanks for pointing out that $GL_3(\mathbb{Z}) \not\cong PGL_3(\mathbb{Z})$. I understand $GL_3(\mathbb{Z}) \cong SL_3(\mathbb{Z})\times \lbrace \pm 1\rbrace$. But do we really have $GL_3(\mathbb{Z}) \cong PGL_3(\mathbb{Z}) \times \lbrace \pm 1\rbrace$ ?
Oct 9, 2013 at 19:03 history edited tj_ CC BY-SA 3.0
Added the case PGL_3(Z)
Oct 9, 2013 at 17:44 comment added YCor @tj: no, $GL_3(\mathbb{Z})$ is not isomorphic to $PGL_3(\mathbb{Z})$, but to $PGL_3(\mathbb{Z})\times\{\pm 1\}$.
Oct 9, 2013 at 16:03 history answered tj_ CC BY-SA 3.0