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It is not very hard to see that for each prime power $q$ and natural numbers $n,h$, we have an embedding $$\iota \colon \mathrm{GL}(n,q^h) \hookrightarrow \mathrm{GL}(nh, q),$$ obtained by choosing a basis for the finite field $\mathbb{F}_{q^h}$ over the base field $\mathbb{F}_q$. It is not very hard either to see that this does not induce an embedding of the corresponding projective groups $\mathrm{PGL}(n,q^h)$ into $\mathrm{PGL}(nh, q)$. Of course, this does not exclude the possibility that $\mathrm{PGL}(n,q^h)$ is nevertheless isomorphic to a subgroup of $\mathrm{PGL}(nh, q)$, and that is precisely my question:

For which values of $q,n,h$ with $h > 1$ is $\mathrm{PGL}(n,q^h)$ isomorphic to a subgroup of $\mathrm{PGL}(nh, q)$?

Notice that comparing the order of both groups does not rule out anything at all.

I believe I read somewhere that $\mathrm{PGL}(n,q^h)$ is isomorphic to a subgroup of $\mathrm{PGL}(n^h, q)$ (but I forgot the precise construction), so this would at least tell that the answer is positive for $n=h=2$. (Edit: This fact is now confirmed by Derek Holt's answer.)

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    $\begingroup$ According to Magma, ${\rm PGL}(n,q^h)$ is isomorphic to a subgroup of ${\rm PGL}(nh,q)$ for $n = 2$, $h = 3$ and $2 \leq q \leq 5$, but ${\rm PGL}(3,4)$ is not isomorphic to a subgroup of ${\rm PGL}(6,2)$. $\endgroup$ Apr 28, 2015 at 17:06
  • $\begingroup$ Should "I believe I read somewhere that $\operatorname{PGL}(n, q^h)$ is isomorphic to a subgroup of $\operatorname{PGL}(n h, q)$" be "… that $\operatorname{PGL}(2, q^2)$ is isomorphic to a subgroup of $\operatorname{PGL}(4, q)$" (as in @DerekHolt's answer mathoverflow.net/a/204173/2383)? $\endgroup$
    – LSpice
    Apr 28, 2015 at 17:38
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    $\begingroup$ Originally it read "${\rm PGL}(n,q^h)$ is isomorphic to a subgroup of ${\rm PGL}(n^h,q)$", which I believe is correct, but that has seems to have been changed! $\endgroup$
    – Derek Holt
    Apr 28, 2015 at 17:46
  • $\begingroup$ @DerekHolt, I'm sorry. I was the one who changed it, because I thought that it was a typo; but I should have asked. I'll un-change it now. EDIT: Darn, it's too few characters to make a new edit. Is there any way that I or someone else could request that it be reverted? $\endgroup$
    – LSpice
    Apr 28, 2015 at 18:04
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    $\begingroup$ I have changed it back to $n^h$. I just retyped some of the words to apparently increase the amount of editing. $\endgroup$
    – Derek Holt
    Apr 28, 2015 at 18:15

2 Answers 2

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Yes, ${\rm PGL}(2,q^2)$ is a subgroup of ${\rm PGL}(4,q)$, but I would guess that that is an exception, and in general there is no such embedding.

${\rm GL}(4,q)$ contains the subgroup that I denote by ${\rm CO}^-(4,q)$, which is the conformal orthogonal group of minus-type (and equal to the normalizer in ${\rm GL}(4,q)$ of ${\rm GO}^-(4,q)$).

The projective image ${\rm PCO}^-(4,q)$, which is of course a subgroup of ${\rm PGL}(4,q)$, happens to be isomorphic to ${\rm P \Gamma L}(2,q^2)$, which contains ${\rm PGL}(2,q^2)$ as a subgroup of index $2$.

So this is a separate embedding, and is not related to the semiliear embedding ${\rm GL}(2,q^2) \to {\rm GL}(4,q)$.

The general embedding ${\rm PGL}(n,q^h) \to {\rm PGL}(n^h,q)$ can be defined as follows. Let $M$ be the natural module for $G={\rm GL}(n,q^h)$ and let $\phi$ be the field automorphism of $G$ induced by the automorphism $x \mapsto x^q$ of the field. Then $M \otimes M^\phi \otimes M^{\phi^2} \otimes \cdots \otimes M^{\phi^{h-1}}$ is a module of dimension $n^h$ that is stabilized by $\phi$ and hence can be realized over ${\mathbb F}_q$. So we get a homomorphism ${\rm GL}(n,q^h) \to {\rm GL}(n^h,q)$, which is not always injective, but scalars map to scalars, so it induces the required embedding ${\rm PGL}(n,q^h) \to {\rm PGL}(n^h,q)$.

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  • $\begingroup$ Thank you, that is very helpful and confirms my guess. I'm still wondering whether there is an (easy?) argument showing that the answer is negative for all other values of $n$ and $h$. Do you have any ideas? $\endgroup$ Apr 29, 2015 at 7:26
  • $\begingroup$ But there are some cases where the semilinear embedding ${\rm GL}(n,q^h) \to {\rm GL}(nh,q)$ does induce ${\rm PGL}(n,q^h) \to {\rm PGL}(nh,q)$. For example ${\rm PGL}(3,27) \to {\rm PGL}(9,3)$. This depends on $\gcd(q^h-1,n)$ and $\gcd(q-1,nh)$. It shouldn't be hard to write down the precise conditions for that. I might do that later. $\endgroup$
    – Derek Holt
    Apr 29, 2015 at 7:57
  • $\begingroup$ I'm confused. If we multiply a matrix in $\mathrm{GL}(3,27)$ by a scalar in $\mathbb{F}_{27} \setminus \mathbb{F}_3$, then the corresponding matrix in $\mathrm{GL}(9,3)$ is not obtained by multiplying by a scalar, so I don't see how it can induce a well-defined map $\mathrm{PGL}(3,27) \to \mathrm{PGL}(9,3)$. What am I overlooking? $\endgroup$ Apr 29, 2015 at 8:07
  • $\begingroup$ It might be that $\mathrm{SL}_n$ is very closely related to $\mathrm{PGL}_n$ in those cases and the embedding is obtained via $\mathrm{SL}(n, q^h)\rightarrow \mathrm{GL}(nh, q)$. $\endgroup$
    – peliukas
    Apr 29, 2015 at 8:09
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    $\begingroup$ Yes, maybe it is incorrect to say that the embedding is induced. It turns out that the image of ${\rm GL}(3,27)$ in ${\rm PGL}(9,3)$ splits as a direct product $C_{13} \times {\rm PGL}(3,27)$, so it does occur as a subgroup, but only because of the splitting. $\endgroup$
    – Derek Holt
    Apr 29, 2015 at 9:02
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Assume that $n$ is prime to $q^h-1$. Then $\mathrm{SL}(n, q^h)\cap \mathbb F_{q^h}^{\times}$ is trivial and the map $\mathrm{SL}(n, q^h)\rightarrow \mathrm {PGL}(n, q^h)$ is an isomorphism, as $x\mapsto x^n$ induces an iso on $\mathbb{F}_{q^h}^{\times}$. The composition

$$ \mathrm{SL}(n, q^h)\rightarrow \mathrm{GL}(n, q^h)\rightarrow \mathrm{GL}(nh, q)\rightarrow \mathrm{PGL}(nh, q),$$ where the middle arrow is the one described in the question, is injective, because the kernel is equal to $\mathrm{SL}(n, q^h)\cap \mathbb F^{\times}_{q}$.

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    $\begingroup$ In fact ${\rm SL}(n,q^h)$ and ${\rm PGL}(n,q^h)$ are btoh simple groups in this case. I think it might be enough to have $n$ coprime to $(q^h-1)/(q-1)$. $\endgroup$
    – Derek Holt
    Apr 29, 2015 at 10:27

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