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Is there such a number N, such that any group of order N is simple?

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    $\begingroup$ Take for example $N=2$. $\endgroup$ – André Nicolas Aug 7 '15 at 5:55
  • $\begingroup$ @AndréNicolas Do you have an example s.t N is not prime? $\endgroup$ – 6666 Aug 7 '15 at 6:56
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    $\begingroup$ Well, there is the even duller $N=1$. There is no composite $N$, since there is always an abelian group of order $N$, and if $N$ is composite any abelian group of order $N$ is not simple. $\endgroup$ – André Nicolas Aug 7 '15 at 7:20
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    $\begingroup$ But, according to standard terminology, the group of order $1$ is not simple, for the same sorts of reasons that $1$ is not a prime number. $\endgroup$ – Derek Holt Aug 7 '15 at 8:06
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    $\begingroup$ For that matter there is the even duller $N = 0$. There are no such groups, so a statement about "any group of order 0" is vacuously true ;-) $\endgroup$ – Steve Jessop Aug 7 '15 at 12:37
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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 $20160$.

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  • $\begingroup$ Thank you very much, this answer is very intriguing $\endgroup$ – 6666 Aug 8 '15 at 11:52
  • $\begingroup$ You are welcome Joseph! You asked a good question (+1 from me!) and actually the relation between the order of a group and its intrinsic group theoretical structure has been studied a lot - see for example citeseerx.ist.psu.edu/viewdoc/… $\endgroup$ – Nicky Hekster Aug 8 '15 at 12:42
  • $\begingroup$ Thank you, you are really helpful! I will see that! $\endgroup$ – 6666 Aug 8 '15 at 13:00
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There are abelian groups of any order, and an abelian group which is not of prime order is not simple.

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  • $\begingroup$ Wow, this answer eliminates lots of groups! $\endgroup$ – coldnumber Aug 7 '15 at 8:59
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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.

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  • $\begingroup$ Do you have an example s.t N is not prime? $\endgroup$ – 6666 Aug 7 '15 at 6:53
  • $\begingroup$ The trivial group is not simple. $\endgroup$ – Derek Holt Aug 7 '15 at 8:06
  • $\begingroup$ @DerekHolt Ok, thank you $\endgroup$ – 6666 Aug 7 '15 at 8:35

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