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How does we obtain a small (not necessarily smallest) value of $n$ such that a $p$-group of order $p^m$ can be embedded in $\operatorname{GL}(n,\mathbb{F}_p)$?

[For embedding of a finite group $G$ in $S_n$, we can compute a small value of $n$ by looking at the minimum index of a subgroup $H$ containing no proper normal subgroup of $G$. Here, this value of $n$ need be necessarily smallest.]

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4 Answers 4

An easy to embed $p$-group is the Sylow $p$-subgroup of $\operatorname{GL}(n,p)$. It has $m = n(n−1)/2$, the largest possible $m$. A hard to embed $p$-group is the cyclic group of order $p^m$. It embeds in $\operatorname{GL}(n,p)$ only if $n ≥ p^{m−1}+1$.

Easy groups have minimal $n = Θ(\sqrt{m})$ and hard groups have minimal $n = Θ(p^m)$, so there is a very wide range of bounds based on the structure of the group.

In my experience, the single worst feature of a $p$-group for a small embedding is its exponent. Cyclic groups are nearly regular anyways, so there is very little room to be "worse" asymptotically.

The proof of the exponential embedding lower bound is just a calculation with Jordan canonical forms, and applies to all fields of characteristic $p$.

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I guess nilpotence class (and derived length) of the $p$-group have some relevance, since the upper unitriangular subgroup of ${\rm GL}(n,p)$ has nilpotence class $n-1,$ and similar (known) bounds exist for its derived length. –  Geoff Robinson Dec 16 '11 at 0:11

Embed your group in a permutation group $S_n$ using Cayley's theorem, and then embed $S_n$ into a $GL_n$ using permutation matrices.

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2  
You mean $GL_n$, right? –  Qiaochu Yuan May 15 '11 at 15:30
    
@Qiaochu, Yup! Fixed. ${}$ –  Mariano Suárez-Alvarez May 15 '11 at 18:48

For p > 2, check http://ysharifi.wordpress.com/2011/02/17/gln-z-has-only-finitely-many-finite-subgroups/, it is not hard to show that a group of order n can be embedded in $GL(n,\mathbb{F}_q)$ with q an arbitrary odd prime power.

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Here I think "small" should be interpreted as "growing not too fast in $m$." The Sylow $p$-subgroup of $\text{GL}_n(\mathbb{F}_p)$ has order $p^{{n \choose 2}}$ so the theoretical lower bound here is $n = \Omega(\sqrt{m})$, and the embedding given by Cayley's theorem is substantially worse (exponential in $m$). –  Qiaochu Yuan May 15 '11 at 15:35

Here's a theorem due to Minkowski-Schur-Blichtfeld which might be of some interest for you:

Theorem. Let $G$ be a subgroup of $GL_n(\mathbb Z)$ for some $n$. We suppose that $G$ is a $p$-group of order $p^m$. Then $m\leqslant N(p,n)$, with $$N(p,n)= \left[\frac{n}{p-1}\right] + \left[\frac{n}{p(p-1)}+ \right]\cdots $$ Conversely, for any prime $p$ and any $n>0$, there exists a subgroup $G$ of $GL_n(\mathbb Z)$ of order $p^{N(p,n)}$.

Another remark, due to Minkowski, is that if $G$ is a finite subgroup of $GL_n(\mathbb Z)$, then the natural restriction arrow $G \to GL_n(\mathbb F_p)$ is injective whenever $p\geqslant 3$. This enables to relate more precisely the last theorem to your question.

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Yes, but it is not true that every finite subgroup of $\text{GL}_n(\mathbb{F}_p)$ pulls back to a finite subgroup of $\text{GL}_n(\mathbb{Z})$, so I don't see how this is relevant. –  Qiaochu Yuan May 16 '11 at 4:25

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