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A finite groups of $\mathrm{GL}_n(\mathbb C)$ of exponent $m$ necessarily have order $C$ verifying $C\leqslant m^n$ and $n! m^n$ divides $C$, but this condition is not sufficient, for instance $\mathrm{GL}_3(\mathbb C)$ has no subgroup of exponent $7$ and order $42$.

So can we determine fully the possible orders of the subgroups of $\mathrm{GL}_n(\mathbb C)$ of exponent $m$?

~~~~~~~~~~~~OLD QUESTION - Which was obvious, sorry and thanks Pete ~~~~~~~~~~~~~~~~

After some computation and search it seems that finite subgroups of $\mathrm{GL}_3(\mathbb C)$ of exponent $5$ have order $5$, $5^2$ or $5^3$. Similarly, finite subgroups of $\mathrm{GL}_3(\mathbb C)$ of exponent $7$ have order $7$, $7^2$ or $7^3$.

So, for $n >2$ and a prime $p >2$, is it true that finite subgroups of $\mathrm{GL}_n(\mathbb C)$ of exponent $p$ necessarily have order of the form $p^k$? I apologize in advance if this is obvious and I missed it.

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  • $\begingroup$ you intended to say for $0\leq k \leq n$ I suppose. $\endgroup$
    – Xandi Tuni
    Commented May 15, 2010 at 15:11
  • $\begingroup$ Any finite group $G$ of prime exponent $p$ has $p$-power order: if not, there exists a prime $\ell \neq p$ dividing $\# G$ , and then by Cauchy's theorem, $G$ has an element of order $\ell$, contradiction. $\endgroup$
    – Pete L. Clark
    Commented May 15, 2010 at 15:19
  • $\begingroup$ @Pete: Sure, but why is $(\mathbb Z/7)^{1'000'000}$ not a subgroup of $\mathrm{GL}_{999'999}(\mathbb C)$? I think that's the question. $\endgroup$
    – Xandi Tuni
    Commented May 15, 2010 at 15:24
  • $\begingroup$ @Xandi: That's not the question. But I agree that maybe it should be the question...let's wait and see what transpires. $\endgroup$
    – Pete L. Clark
    Commented May 15, 2010 at 15:24
  • $\begingroup$ Indeed, first question was obvious, I just edited and posted the question that was of interest. $\endgroup$
    – Portland
    Commented May 15, 2010 at 15:35

1 Answer 1

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The paper:

Herzog, Marcel; Praeger, Cheryl E. "On the order of linear groups of fixed finite exponent." J. Algebra 43 (1976), no. 1, 216–220. MR424960 DOI:10.1016/0021-8693(76)90156-3

contains the important bound, if G ≤ GL(n,F) where F has characteristic coprime to |G|, then |G| ≤ exp(G)n. Obviously these bounds can be obtained over large enough F (containing exp(G) roots of unity), as the diagonal subgroup generated by eth roots of unity has exponent e and order en.

The exponent of a finite group divides the order of the group: the exponent of a group is the product of the exponents of its Sylow subgroups, and a p-group always contains an element whose order is equal to the exponent of the group, so by Lagrange the exponent divides the order. Also, every prime dividing the order of the group divides the exponent of the group, by Cauchy's theorem.

In particular, the possible orders of finite groups of exponent p, p a prime, that are contained in GL(n,C) are exactly p1, p2, …, pn. The elementary abelian subgroups generated by diagonal matrices whose entries are pth roots of unity shows the existence, and Herzog and Praeger eliminate all other orders. Note that when p is large, these are all possible anyways, so that the theorem is probably only interesting for small p.

For instance, GL(3,C) contains no non-abelian group of exponent 5.

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  • $\begingroup$ Thanks Jack, I knew of that paper, although I had a hard time with Blichfelt lemma (p. 218). I wonder if there is a way around? $\endgroup$
    – Portland
    Commented May 15, 2010 at 16:09
  • $\begingroup$ @ Jack, How do show that "For instance, GL(3,C) contains no non-abelian group of exponent 5." Because GL(3,C) contains for instance a non-abelian group of exponent 3 (isomorphic to $(\mathbb{Z} /3 \mathbb{Z} \times \mathbb{Z} /3 \mathbb{Z} ) \rtimes \mathbb{Z} /3 \mathbb{Z}$). $\endgroup$
    – Portland
    Commented May 15, 2010 at 16:33
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    $\begingroup$ It'll take a bit to write up a simpler proof of Blichfeldt, but the idea of that section of Herzog-Praeger is very standard and effective: (1) if G is a subgroup of GL(M), M a vector space, then M is a G-module, (2) handle the case where M is reducible first (lemma 2.1), (3) handle the case where M is imprimitive second (Blichfeldt), and (4) handle the primitive cases. 1,2,3 are standard and usually "easy", and the groups that survive to 4 usually have a very special structure. Blichfeldt's lemma basically says no primitive group is a nonabelian p-group, which might be familiar for perm reps. $\endgroup$
    – Jack Schmidt
    Commented May 15, 2010 at 17:07
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    $\begingroup$ The specific case of exponent 5 in GL(3,C) follows from two facts: (1) if a group has a faithful representation that is a direct sum of 1-dimensional representations, then it is abelian, and (2) if a group has an irreducible representation of dimension d, then d divides the order of the group. A group of exponent 5 has order a power of 5, so the dimension of its representations are powers of 5. The only power of 5 less than or equal to 3 is 5<sup>0</sup> = 1, and so the only 5-subgroups of GL(3,C) are abelian. The general case is Blichfeldt's theorem 14.1 in Isaacs's Character Theory book. $\endgroup$
    – Jack Schmidt
    Commented May 15, 2010 at 17:12
  • $\begingroup$ Perhaps it is interesting to note that both theorems of Blichfeldt mentioned here can be found in his group theory book with Dickson and Miller. The proofs there don't require much character theory IIRC, but are much longer than the ones you can find in a Robinson or Isaacs. The book is available freely (and legally) from Google Books. $\endgroup$
    – Steve D
    Commented May 16, 2010 at 22:20

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