Is there such a number N such that any group of order N is simple? Is there such a number N, such that any group of order N is simple?
 A: There are abelian groups of any order, and an abelian group which is not of prime order is not simple.
A: If $N$ is prime, then any group of order $N$ is unique up to isomorphism and is simple and cyclic! You can also take $N=1$, the trivial group is also simple.
A: A more subtle question would be: are there any natural numbers $N$, such that all the non-abelian groups of order $N$ are simple?
Note that for infinite many $N$, there do not exist any non-abelian groups of order $N$, for example if $N$ is the square of a prime or gcd$(N,\varphi(N))=1$. But suppose the set $\{G: G \text{ is a non-abelian group with } |G|=N\}$ is non-empty. Can it be the case? The answer is no! We may assume $N \gt 2$. Let $G$ be a non-abelian simple group of order $N$. Then $|N|$ must be even (this is due to the deep Odd-Order Theorem of Feit and Thompson), say $N=2M$ with $M \gt 1$. If $M \geq 3$, then the dihedral group of order $N$, $D_{M}=\langle a,b: a^{M}=1=b^2, b^{-1}ab=a^{-1}\rangle$, is a non-abelian group of order $N$ and is certainly not simple (for example $\langle a \rangle \lhd D_{M}$). Hence $M=2$, so $N=4$, but all groups of order $4$ are abelian, a contradiction.
Fun fact to know: there are non-isomorphic simple groups of the same order: $A_8 \cong PSL(4,2)$ and $PSL(3,4)$ of order $20160$.
