On the solvability group

Question

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?