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Can we embedd $GL(2n,q)$ into $GL(n,q^2)$ for $n\in \mathbb{N}$ and $q=p^m$, $p$ a prime? If yes, how?

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Can you embed $GL_2(p)$ into $GL_1(p^2)$? Hint: What are their respective orders? Always think about the simplest non-trivial example before asking a general question. – Alex B. Mar 29 '11 at 6:51
If you had instead asked whether we could embed ${\rm GL}(n,q^2)$ into ${\rm GL}(2n,q)$, then the answer would have been different. – Derek Holt Mar 29 '11 at 11:46
did you not just completely change the question? Please do not do that. – Mariano Suárez-Alvarez Mar 30 '11 at 4:21

As others have mentioned, this is clearly not true in special cases: GL(2,2) ≅ Sym(3) clearly does not embed in GL(1,4) ≅ Alt(3), and indeed the embedding should go the other way (in general even, not just for n=1, q=2).

If q is a prime power, and n is a positive integer, then GL(2n,q) never embeds in GL(n,qq). One way to see this is the size of the Sylow p-subgroup. GL(k,pi) has a Sylow p-subgroup of order p(ik(k−1)/2). So in our cases, we have GL(2n,q) has a Sylow p-subgroup of order qn(2n−1) and GL(n,qq) has a Sylow p-subgroup of order qn(n−1)/2. Since n(n−1)/2 < n(2n−1) whenever n ≥ 1, we never have an embedding.

In fact, a little more is true: homomorphisms from GL(n,q) into GL(k,qj) are actually conjugate to homomorphisms from GL(n,q) into GL(k,q). You don't need to make the field bigger in order to map a GL(n) into a GL(k), at least not for finite fields.

Of course there are inclusions of GL(n,qq) into GL(2n,q), since an n-dimensional vector space over GF(qq) is also a 2n-dimensional vector space over GF(q). You can even have inclusions of GL(n,q) into GL(k,q) for k ≥ n that are not conjugate to the obvious embedding. In all cases though, the dimension cannot go down, and there is no sense making the field bigger, only occasionally making it smaller.

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