# On the solvability group

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?

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