most recent 30 from http://mathoverflow.net 2013-06-19T05:09:23Z http://mathoverflow.net/feeds/question/91346 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/91346/how-to-inject-pgl-n-k-in-pgl-n-1-k Rajkarov 2012-03-16T02:03:27Z 2012-03-16T10:07:33Z

<p>How to construct an injection of $PGL(n,k)$ in $PGL(n+1,k)$ if $GL(n,k)$ injects in $GL(n+1,k)$. I think it depends on the field k. For example, if we put $\varphi:GL(n,k)\longrightarrow GL(n+1,k)$ defined by : </p> <p>$$\varphi(g)=\left( \begin{array}{cc} g &amp; 0 \cr 0 &amp; \chi(g) \cr \end{array} \right)$$</p> <p>where $\chi$ is a caracter of $GL(n,k)$, that is a morphism of groups of $GL(n,k)$ to $k^{\times}$, then $\chi$ has the form $\chi=\phi\circ det$, where $\phi$ is an endomorphism of $k^{\times}$.</p> <p>Then, $\varphi$ induces a morphisme of groups of $PGL(n,k)$ in $PGL(n+1,k)$ if and only if the morphism $\phi$ satisfies : For all $x\in k^{\times}$, $\phi(x)^{n}=x$. </p> <p>For $k=\mathbb{R}$ that is imposible.</p>

http://mathoverflow.net/questions/91346/how-to-inject-pgl-n-k-in-pgl-n-1-k/91362#91362 Derek Holt 2012-03-16T10:07:33Z 2012-03-16T10:07:33Z

<p>You can't do it in general. A quick computer calculation (I used Magma) shows that ${\rm PGL}(4,4)$ has no subgroup isomorphic to ${\rm PGL}(3,4)$. (It does have one isomorphic to ${\rm SL}(3,4)$.)</p> <p>I suspect that there is an embedding ${\rm PGL}(2,K) \to {\rm PGL}(3,K)$, but that is coming from the irreducible orthogonal action of ${\rm GL}(2,K)$ on a 3-dimensional module.</p>