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Let $G$ and $H$ be two groups such that

  1. $|G|=|H|$;
  2. for every natural number $n$ the number of elements of order $n$ in $G$ and $H$ are equal;
  3. $H$ is a solvable group.

Is $G$ solvable?

or

Is there any counterexample?

Condition $2$ requires two groups are the same order type. If $G$ is a finite group and $n\in \mathbb{N}$, then $G(n):=\{x\in G|x^{n}=1\}$. We say that the groups $G$ and $H$ are of the same order type if $|G(n)|=|H(n)|$, for all $n\in \mathbb{N}$.

J. G. Thompson has a problem on the same order type group:

Let $G$ and $H$ be two groups of the same order type such that $H$ is solvable. Is $G$ solvable?

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    $\begingroup$ You have changed the question, but it is still missing any context or motivation. $\endgroup$ – Derek Holt Apr 5 at 5:27
  • $\begingroup$ @Derek Holt: Motivation of the question added. $\endgroup$ – M. T Apr 5 at 11:03
  • $\begingroup$ So you are just re-asking Thompson's question? I guess I should vote to reopen then! $\endgroup$ – Derek Holt Apr 5 at 11:30

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