Question about orders of alternating and general linear groups. Let $n\geq5$ and consider the alternating group of degree $n$, $A_n$. Now, let $m$ be the smallest positive integer such that $|A_n| \mid |\mathrm{GL}(m,2)|$. I want to find a lower bound for $m$ in terms of $n$. In my preliminary tests ($1<n<200$) it seems that $m \geq n/2$. However, $m$ seems to asymptotically approach $n$ as $n$ grows. For instance, a better bound is given for $100<n<500$, in which case $m \geq 0.9n$. My question is whether this second bound actually holds for all $n>100$ and if so, why? If this is not easy to show, I would be very happy with a proof of just the first bound as well ($m\geq n/2$).
As a bonus question, let $m_3$ be the smallest positive integer such that $|A_n| \mid |\mathrm{GL}(m_3,3)|$. In this case, I only care if the first bound holds. Namely, is it true that for $n\geq5$, $m_3\geq n/2$?
 A: For a prime $p$ write $\nu_p(n)$ for the greatest power of $p$ dividing $n$. Recall that we have Legendre's formula
$$\nu_p(n!) = \sum_{k \ge 1} \left\lfloor \frac{n}{p^k} \right\rfloor = \frac{n - s_p(n)}{p - 1}$$
where $s_p(n)$ is the sum of the digits of $n$ in base $p$. We want to compare this to
$$\nu_p(|GL_m(\mathbb{F}_q)|) = \nu_p \left( q^{ {m \choose 2} } \prod_{i=1}^m (q^i - 1) \right)$$
for $q = 2, 3$. Specifically, if $|A_n|$ divides $|GL_m(\mathbb{F}_q)|$ we'd like to know what lower bound this implies on $m$ in terms of $n$, so we'd like to know when the second expression above is small so that the condition that it is at least as large as the first expression is a strong condition.
We'll ignore the case $p \mid q$ because the bound is not good in this case. If $\gcd(p, q) = 1$, let $d = \text{ord}_p(q)$ be the smallest positive integer such that $q^d \equiv 1 \bmod p$, so that we have $d \mid p - 1$. Then $\nu_p(q^i - 1)$ is only nonzero when $d \mid i$. In this case, if $p$ is odd then lifting the exponent gives
$$\nu_p(q^{dk} - 1) = \nu_p(q^d - 1) + \nu_p(k).$$
This gives
$$\begin{eqnarray*} \nu_p(|GL_m(\mathbb{F}_q)|) &=& \left\lfloor \frac{m}{d} \right\rfloor \nu_p(q^d - 1) + \sum_{k=1}^{ \left\lfloor \frac{m}{d} \right\rfloor} \nu_p(k) \\
 &=& \left\lfloor \frac{m}{d} \right\rfloor \nu_p(q^d - 1) + \nu_p \left( \left\lfloor \frac{m}{d} \right\rfloor ! \right) \\
 &=& \left\lfloor \frac{m}{d} \right\rfloor \nu_p(q^d - 1) + \sum_{k \ge 1} \left\lfloor \frac{m}{dp^k} \right\rfloor \\
 &=& \left\lfloor \frac{m}{d} \right\rfloor \nu_p(q^d - 1) + \frac{\left\lfloor \frac{m}{d} \right\rfloor - s_p \left( \left\lfloor \frac{m}{d} \right\rfloor \right)}{p - 1}. \end{eqnarray*} $$
Now I'm going to specialize to $q = 2, 3$ because I don't know how to analyze this $\nu_p(q^d - 1)$ term in general (it should just equal $1$ most of the time but I don't know when).
Case: $q = 2$. To make the bound for $m$ in terms of $n$ as tight as possible we want $d$ above to be large. It's maximized when $d = p-1$, which means that $2$ is a primitive root $\bmod p$. The sequence of such primes is A001122 and begins $3, 5, 11, 13, \dots$; in particular it contains $p = 101$. We have $2^{100} - 1 \equiv 9292 \bmod 101^2$ which gives $\nu_{101}(2^{100} - 1) = 1$. Write $m' = \left\lfloor \frac{m}{100} \right\rfloor$; this gives
$$\nu_{101}(|GL_m(\mathbb{F}_2)|) = \sum_{k \ge 0} \left\lfloor \frac{m'}{101^k} \right\rfloor = m' + \frac{ m' - s_{101}(m') }{100}$$
so if $|A_n|$ divides $|GL_m(\mathbb{F}_q)|$ then for $n \ge 101$ we have
$$\nu_{101}(n!) = \frac{n - s_{101}(n)}{100} \le m' + \frac{m' - s_{101}(m')}{100}.$$
We have $s_p(n) \le (p-1) \left\lfloor \log_p n \right\rfloor$ so the $s$ terms grow logarithmically and as $n \to \infty$ can be ignored; this tells us that asymptotically we get a bound that looks roughly like $n \le 101 m'$, so $m \ge \frac{100}{101} n$, but with some error. Bounding the $s$ terms more carefully should get us the bound you requested.
Taking larger primes $p$ which both 1) have the property that $2$ is a primitive root and 2) have the property that $\nu_p(2^{p-1} - 1) = 1$ will improve this bound further; we should end up getting $m \ge (1 - \epsilon) n$ for arbitrarily small $\epsilon$ but we need the number of primes satisfying these conditions to be infinite which is an open question.
Case: $q = 3$. Now we want to find primes such that $3$ is a primitive root $\bmod p$. The sequence of such primes is A019334 and begins $2, 5, 7, \dots$; since you only ask for a bound for $n \ge 5$ we'll content ourselves with using $p = 5$. We have $\nu_5(3^4 - 1) = 1$. Write $m' = \left\lfloor \frac{m}{4} \right\rfloor$; this gives
$$\nu_5(|GL_m(\mathbb{F}_3)|) = \sum_{k \ge 0} \left\lfloor \frac{m'}{3^k} \right\rfloor = m' + \frac{m' - s_5(m')}{4}$$
so if $|A_n|$ divides $|GL_m(\mathbb{F}_3)|$ then for $n \ge 5$ we have
$$\nu_5(n!) = \frac{n - s_5(n)}{4} \le m' + \frac{m' - s_5(m')}{4}.$$
As above the $s$ terms only grow logarithmically so as $n \to \infty$ we get something close to $m \ge \frac{4}{5} n$ with some error, and as above bounding the $s$ terms more carefully should give the desired result. Here's a sloppy bound: for $n \ge 15$ we have $s_5(n) \le \frac{7}{19} n$ (the worst case is $n = 19, s_5(n) = 7$), which gives
$$\frac{12}{19} n \le 5m' \le \frac{5m}{4}$$
and hence $m \ge \frac{48}{95} n$. Then you can check $5 \le n \le 14$ individually.
For much more along these lines you can see Serre's Bounds for the orders of the finite subgroups of $G(k)$.
