Let $G(n)$ denote number of non isomorphic groups of order $n$. So is $G(2^n)>G(m)$ for $m$ less than $2^n$ always? Ignore $2$.
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3Yes, almost all finite groups are $2$-groups. There are asymptotic formulas, i.e., the number of isomorphism classes of groups of order $p^n$ grows as $p^{\frac{2}{27}n^3 + O(n^{8/3})}$. – Dietrich Burde Apr 07 '17 at 18:06
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1@DietrichBurde is this actually known? It seems that surprisingly little is known about the growth for non prime powers. – Tobias Kildetoft Apr 07 '17 at 18:26
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4@TobiasKildetoft No, this is only a conjecture so far, you are right. There is of course the computation of all groups of order $n\le 2000$ by Bettina Eick et al., where $99%$ of all groups have order $2^{10}$. – Dietrich Burde Apr 07 '17 at 18:44
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1See also this question, and this duplicate of it. – Dietrich Burde Apr 07 '17 at 18:49