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How to construct an injection of $PGL(n,k)$ in $PGL(n+1,k)$ if $PGL(n,k)$ injects in $PGL(n+1,k)$. I think it depends on the field k. For example, if we put $\varphi:GL(n,k)\longrightarrow GL(n+1,k)$ defined by :

$$\varphi(g)=\left( \begin{array}{cc} g & 0 \cr 0 & \chi(g) \cr \end{array} \right)$$

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 $k^{\times}$.

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 $x\in k^{\times}$, $\phi(x)^{n}=x$.

For $k=\mathbb{R}$ that is imposible.

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 :

$$\varphi(g)=\left( \begin{array}{cc} g & 0 \cr 0 & \chi(g) \cr \end{array} \right)$$

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 $k^{\times}$.

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 $x\in k^{\times}$, $\phi(x)^{n}=x$.

For $k=\mathbb{R}$ that is imposible.

How to construct an injection of $PGL(n,k)$ in $PGL(n+1,k)$ if $PGL(n,k)$ injects in $PGL(n+1,k)$. I think it depends on the field k. For example, if we put $\varphi:GL(n,k)\longrightarrow GL(n+1,k)$ defined by :

$$\varphi(g)=\left( \begin{array}{cc} g & 0 \\ 0 & \chi(g) \\ \end{array} \right)$$

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 $k^{\times}$.

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 $x\in k^{\times}$, $\phi(x)^{n}=x$.

For $k=\mathbb{R}$ that is imposible.

How to construct an injection of $PGL(n,k)$ in $PGL(n+1,k)$ if $PGL(n,k)$ injects in $PGL(n+1,k)$. I think it depends on the field k. For example, if we put $\varphi:GL(n,k)\longrightarrow GL(n+1,k)$ defined by :

$$\varphi(g)=\left( \begin{array}{cc} g & 0 \cr 0 & \chi(g) \cr \end{array} \right)$$

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 $k^{\times}$.

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 $x\in k^{\times}$, $\phi(x)^{n}=x$.

For $k=\mathbb{R}$ that is imposible.

How to construct an injection of $PGL(n,k)$ in $PGL(n+1,k)$ if $PGL(n,k)$ injects in $PGL(n+1,k)$. I think it depends on the field k. For example, if we put $\varphi:GL(n,k)\longrightarrow GL(n+1,k)$ defined by :

$$\varphi(g)=\left( \begin{array}{cc} g & 0 \\ 0 & \chi(g) \\ \end{array} \right)$$

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 $k^{\times}$.

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 $x\in k^{\times}$, $\phi(x)^{n}=x$.

For $k=\mathbb{R}$ that is imposible.

How to construct an injection of $PGL(n,k)$ in $PGL(n+1,k)$ if $PGL(n,k)$ injects in $PGL(n+1,k)$. I think it depends on the field k. For example, if we put $\varphi:GL(n,k)\longrightarrow GL(n+1,k)$ defined by :

$$\varphi(g)=\left( \begin{array}{cc} g & 0 \\ 0 & \chi(g) \\ \end{array} \right)$$

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 $k^{\times}$.

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 $x\in k^{\times}$, $\phi(x)^{n}=x$.

For $k=\mathbb{R}$ that is imposible.