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

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 $\mathrm{P\Gamma L}(V)$.


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?

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The definition of ${\rm P \Gamma L}(V)$ that I have always used is ${\rm \Gamma L}(V)/Z({\rm GL}(V))$ (i.e. factor out scalars), so it seems possible that the authors have made a mistake with the definition. –  Derek Holt Mar 2 '13 at 13:44
Thanks @Derek, with this definition the question makes much more sense. I'll look in other books to compare definitions. –  Dan Shved Mar 2 '13 at 13:48
In general, the center of $GL$ is isomorphic to $F^\times$ and that of $\Gamma L$ is isomorphic to $K^\times$ where $K$ is the prime field of $F$. Then indeed there is no natural way to obtain $PGL\toP\Gamma L$ as one needs $F^\times\to K^\times$ for that. For finite extensions, this might be the norm, but even that does not extend nicely to $GL$. So Derek's hint might be it. –  Hagen von Eitzen Mar 2 '13 at 14:44
@DerekHolt Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. –  Julian Kuelshammer Jun 23 '13 at 7:57