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Are there finite groups $G$ and $H$ such that:

  1. $n:=|G|=|H|$.
  2. $G$ is simple.
  3. $H$ is not simple.
  4. for every $d\mid n$, $G$ and $H$ have the same number of elements of order $d$. ?
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up vote 5 down vote accepted

No, there is no such example. It has been shown that if 1., 2. and 4. hold, then $G$ and $H$ are isomorphic. This is not an easy result, as the proof requires classification of finite simple groups. See the slides here for references and more information.

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great. So orders of elements determine whether a group is simple but do not determine whether it is abelian. Do you know any formal book about it!? – user59671 Feb 11 '13 at 8:55
I don't know if there is a book that deals with this stuff, but I doubt it. But there are many articles about the subject, and questions on this site and MO. There are some properties you can deduce from the number of elements of each order. Suppose $G$ and $H$ have the same number of elements of each order. You can show that if $G$ is nilpotent, then $H$ must be nilpotent as well. Also, if $G$ is supersolvable, then $H$ must be solvable. I think it is still an open question whether $G$ solvable implies $H$ solvable when $G$ and $H$ have the same number of elements of each order. – Mikko Korhonen Feb 11 '13 at 18:48

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