Embeddings of finite groups into $\mathrm{GL}_n(\mathbb{Z})$

Solution 1:

For n ≤ 100, k is exactly equal to oeis:A152455, and I think it would have been reasonable to conjecture it always holds, but it would have been quite wrong.

The group of order 144 with id 114 (an "almost double cover" of AGL(1,9), the unique non-split downward extension of AGL(1,9) by a normal subgroup of order 2) has minimum rational degree 16, but the cyclic group has minimum rational degree 14. Similarly, the almost double cover of AGL(1,25) has minimum rational degree 32 compared to the cyclic group of the same order with minimum rational degree 30; 2.AGL(1,49) has 64 instead of 60; 2.AGL(1,81) has 96 instead of 74; 2.AGL(1,121) has 128 instead of 124; 2.AGL(1,169) has 176 instead of 172. ^{_{2.AGL(1,9) has been checked carefully, the others need Schur indices checked, but I suspect the infinite family should follow from a single proof.}}

Certainly as in Derek Holt's answer, one can compute the minimum dimension of a faithful integral representation of a cyclic group as an additive version of the Euler phi function. That is, Derek's answer holds for prime powers as k( p^e ) ≥ (p−1)p^e−1, and using rational canonical form, shows that this lower bound for k is additive on relatively-prime prime powers.

Since this grows fairly quickly, it seems that it is also often an upper bound for the minimum dimension of a faithful integral representation of any group of order n. I just checked the character tables (with Schur indices) to verify the bound for n ≤ 143, but there is a group of order 144 that is a counterexample. There are no other counterexamples up to order n ≤ 200 (but another counterexample of order 1200).

There are some integers n such that every group of order n is cyclic, and so where we have exactly computed k, not just a lower bound. As Derek mentioned, taking n prime shows 1 is an accumulation point of u_n. Taking a product of i distinct primes such that no prime divides the Euler phi function of the rest, one gets an n with u_n ≈ (ip)/(pⁱ) → 0, as i increases. In particular, 0 is an accumulation point of u_n as n varies over:

3,
3⋅5,
3⋅5⋅17,
3⋅5⋅17⋅23,
3⋅5⋅17⋅23⋅29,
3⋅5⋅17⋅23⋅29⋅53,
3⋅5⋅17⋅23⋅29⋅53⋅83,
3⋅5⋅17⋅23⋅29⋅53⋅83⋅89, …

with the last displayed value of n having 0.000000004 as its value of u_n.

Solution 2:

If $n=p$ is prime, then $k(n)=n-1$. This follows from the fact that the minimal polynomial of a non-identity matrix of order $p$ must divide the cyclotomic polynomial $\Phi(p) = (x^p-1)/(x-1)$, which is irreducible over the integers.