Are there finite groups $G$ and $H$ such that:
- $n:=|G|=|H|$.
- $G$ is simple.
- $H$ is not simple.
- for every $d\mid n$, $G$ and $H$ have the same number of elements of order $d$. ?
2026-03-28 04:42:50.1774672970
Is there a simple and a non-simple group with same numbers of elements of each order.
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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.