Suppose $H$ is a finite 2-group and $H$ is a subgroup of $GL(k,\Bbb Z)$. What can be the maximum order of $H$? We can embed $(\mathbb Z _2)^k$ in $GL(k;\Bbb Z)$ as diagonal elements. So $(\mathbb Z _2)^k$ is a subgroup of $GL(k;\Bbb Z)$ of order $2^k$. So maximum order of $H\ge 2^k$. I can not proceed further.
A detailed proof will be very helpful.
Thank you so much in advance.
I think Derek Holt's example from the comment should be the largest. Let's show that the largest 2-subgroup must be contained in (a conjugate of) the subgroup of monomial matrices.
Any finite subgroup will also be a finite subgroup in $GL_n(\mathbb{F}_p)$ for all sufficiently large $p$. So we just need to find a large $p$ such that the subgroup of monomial matrices $N(T) = \mathbb{F}_p^\times \wr S_n$ contains a $2$-sylow subgroup of $GL_n(\mathbb{F}_p)$, which of course just means that it has odd index.
The order of $\mathbb{F}_p^\times \wr S_n$ is $ (p-1)^n\times n!$, and the order of $GL_n(\mathbb{F}_p)$ is $ [n]!_p\times (p-1)^n\times p^N$ where $[n]!_p = [n]_p[n-1]_p\dots[2]_p$ with $[k]_p = p^{k-1}+p^{k-2}+\dots+1$.
In particular we just need to pick a prime $p$ such that the $2$-adic valuation of $n!$ and $[n]!_p$ are the same. But $[n]!_p$ is equal a $f(p)$ for $f \in \mathbb{Z}[x]$ so in particular if we take $p \equiv 1 \text{ (mod } 2^M)$ with $M$ larger than the $2$-adic valuation of $n!$ (i.e. $M=2n$ works) then $f(p) \equiv f(1) \equiv n! \text{ (mod } 2^M)$ so they have the same $2$-adic valuation.