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It's about that $A_5$ is the smallest non abelian Group. The task is to determine for what orders < 60 there are only abelian groups. My idea was to go through all those orders and use the main theorem about abelian groups. But.. There must be a more elegant solution right? Thanks in advance :)

wondering1123
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    This isn't exactly easy (beyond the primes). Also, $A_5$ is not the smallest nonabelian group. – Randall Nov 29 '17 at 19:35
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    Also, how would the "main theorem about abelian groups" tell you there are no nonabelian groups of a given order? – Randall Nov 29 '17 at 19:38
  • A hint: there is a nonabelian group of every even order except for 2 and 4. – Randall Nov 29 '17 at 19:40
  • @Randall why is there a non abelian group of every even order except 2 and 4? – wondering1123 Nov 29 '17 at 19:46
  • @wondering1123 do you know of the dihedral groups? If not, this is going to be a killer exercise. – Randall Nov 29 '17 at 19:47
  • Oh yes sorry... O. O dihedral groups are of oder 2n. – wondering1123 Nov 29 '17 at 19:49
  • For orders $\le 4$ and for prime orders, there are only abelian groups. For even $n>4$, see the other comments. Show that there are only $Z_9$ and $Z_3^2$ of order $9$. A standard beginner's exercise is to show that every group of order $15$ is cyclic (hence abelian). Up to $n=143$, check the zero entries in https://en.wikipedia.org/wiki/List_of_small_groups#List_of_small_non-abelian_groups – Hagen von Eitzen Nov 29 '17 at 19:57
  • See also https://oeis.org/A060652 for more info – Hagen von Eitzen Nov 29 '17 at 19:58
  • $S_3$ is a non-abelian group. $A_5$ is the smallest "non-solvable" group. All groups $A_n, n\ge 5$ are non-solvable. Then there is an odd-ball of order 168. – Doug M Nov 29 '17 at 20:04

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