How is $\mathrm{PGL}(V)$ a subgroup of $\mathrm{P\Gamma L}(V)$?

Question

I've stumbled upon a strange exercise while reading "Notes on Infinite Permutation Groups" by Bhattacharjee, Möller, Macpherson and Neumann. If you have the book, the exercise is 7(ix) on page 66.

Let me start with the definitions. Let $F$ be an arbitrary field, and $V$ a finite-dimensional vector space over $F$.

Definition 1. A linear transformation on $V$ is a mapping $\varphi :\ V \to V$ such that $$ (\lambda x +\mu y)\varphi = \lambda (x \varphi) + \mu (y \varphi) $$ for all $\lambda, \mu \in F$ and all $x, y \in V$. Note that we use notation suitable for right actions here, i.e. we write $x \varphi$ instead of the probably more usual $\varphi(x)$. Also, the composition of transformations is defined accordingly, i.e. $x(\varphi \circ \psi) = (x \varphi)\psi$.

The general linear group $\mathrm{GL}(V)$ consists of all the invertible linear transformations on $V$. Taking the quotient by the centre, we obtain the projective linear group $\mathrm{PGL}(V) = \mathrm{GL}(V) / Z(\mathrm{GL}(V))$.

Definition 2. A semilinear transformation on $V$ is a mapping $\varphi :\ V \to V$ such that there exists an automorphism $\sigma$ of $F$, for which $$ (\lambda x +\mu y)\varphi = (\lambda^\sigma) (x \varphi) + (\mu^\sigma) (y \varphi) $$ for all $\lambda, \mu \in F$ and all $x, y \in V$.

Not surprisingly, the semilinear group $\mathrm{\Gamma L}(V)$ consists of all the invertible semilinear transformations on $V$, and its quotient by the centre is the projective semilinear group $\mathrm{P\Gamma L}(V) = \mathrm{\Gamma L}(V) / Z(\mathrm{\Gamma L}(V))$.

Here is the exercise that gives me trouble:

Show that the group $\mathrm{GL}(V)$ is normal in $\mathrm{\Gamma L}(V)$, and that $\mathrm{PGL}(V)$ is normal in $\mathrm{P\Gamma L}(V)$.

Work done. I don't have any trouble proving that $\mathrm{GL}(V) \lhd \mathrm{\Gamma L}(V)$, it follows straight from the definitions above. But I'm confused by the second part of the question, the one about projective groups.

This is what I don't understand: how can $\mathrm{PGL}(V)$ be a normal subgroup of $\mathrm{P\Gamma L}(V)$ if it is not its subgroup in the first place? I don't see a natural way to build an injection from $\mathrm{PGL}(V)$ to $\mathrm{P\Gamma L}(V)$.

It would be easy if this inclusion were true: $Z(\mathrm{GL}(V)) \leq Z(\mathrm{\Gamma L}(V))$. Then I would easily build a map $\alpha: \mathrm{PGL}(V) \to \mathrm{P\Gamma L}(V)$ that would make the diagram commute: $$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} \mathrm{GL}(V) & \ra{i} & \mathrm{\Gamma L}(V) \\ \da{ } & & \da{ } \\ \mathrm{PGL}(V) & \ra{\alpha} & \mathrm{P \Gamma L}(V) \end{array} $$ where $i$ is inclusion and vertical arrows are factorization.

The problem is, $Z(\mathrm{GL}(V)) \not\leq Z(\mathrm{\Gamma L}(V))$ in the general case. To see this, let $V=F$ be a one-dimensional space. Let $\sigma$ be a non-trivial automorphism of the field $F$. Note that $\sigma$ is automatically a semilinear transformation of $V$.

Now take $\varphi:\ V \to V$ to be multiplication by $\lambda \in F$, where $\lambda^\sigma \neq \lambda$. Then $\varphi$ commutes with every linear transformation on $V$, but it does not commute with the semilinear $\sigma$. So $\varphi$ belongs to $Z(\mathrm{GL}(V))$, but not to $Z(\mathrm{\Gamma L}(V))$.

Is my approach naive? Where should the map $\mathrm{PGL}(V) \to \mathrm{P\Gamma L}(V)$ come from?

Answer 1

The given definition of $\operatorname{P\Gamma L}(V)$ is not the usual one: Typically $\operatorname{P\Gamma L}(V)$ is not defined as $\operatorname{\Gamma L}(V)/Z(\operatorname{\Gamma L}(V))$, but as the quotient group $\operatorname{\Gamma L}(V)/Z(\operatorname{GL}(V))$. Now it is clear that $\operatorname{PGL}(V)$ is a subgroup of $\operatorname{P\Gamma L}(V)$.

The slightly subtle point is that the center $Z(\operatorname{\Gamma L}(V))$ is not the same as the center $Z(\operatorname{GL}(V))$. The latter one is given by all diagonal maps $v\mapsto \lambda v$ with $\lambda\in F^\times$. However, for the center of $\operatorname{\Gamma L}(V)$ we get the additional condition that $\sigma(\lambda) = \lambda$ for all field automorphisms $\sigma\in\operatorname{Aut}(F)$. So $Z(\operatorname{\Gamma L}(V))$ only consists of the scalar matrices whose diagonal entry lies in the unit group of the prime field of $F$.

Given the way how $\operatorname{PGL}(V)$ is defined ("take the group and mod out its center"), the definition of $\operatorname{P\Gamma L}(V)$ might come as a surprise. The motivation for this slightly unexpected definition the is the following:

The natural action of $\operatorname{\Gamma L}(V)$ on $V$ induces an action on the subspace lattice of $V$. However, this action is not faithful. The kernel of the action is given by $Z(\operatorname{GL}(V))$. Modding out the kernel, we arrive at the group $\operatorname{P\Gamma L}(V)$ which acts faithfully on the subspace lattice of $V$. Now by the fundamental theorem of projective geometry, the so-defined group $\operatorname{P\Gamma L}(V)$ is precisely the automorphism group of the subspace lattice of $V$, as long as $\dim V \geq 3$.

Conclusion

The right way to look at the transition from the groups $\operatorname{GL}(V)$ and $\operatorname{\Gamma L}(V)$ to the projective variants $\operatorname{PGL}(V)$ and $\operatorname{P\Gamma L}(V)$ is: Look at the natural action on the projective geometry (aka the subspace lattice of $V$) an mod out the kernel. This is also reflected by the letter "P" for projective in the nomenclature of those groups.