Let $G$ be a finite group and $n$ be a natural number. Set $T(G)=\{g\in G|$ $% g^{n}=1\}$ and $L_{n}(G)=|T(G)|$. Two finite groups $G_{1}$ and $G_{2}$ are called of the same order type if and only if $L_{n}(G_{1})=L_{n}(G_{2})$ for all $n$.
If $G_{1}$ and $G_{2}$ are the same order type, then we have 1) $% |G_{1}|=|G_{2}|$ 2) The set of all element orders of $G_{1}$ and $G_{2}$ are equal.
It is well-known when $G_{1}$ is a non-abelian simple group $G_{2}$ is too.
In 1987, J. G. Thompson put forward the following problem.
Thompson's Problem. Suppose that $G_{1}$ and $G_{2}$ are two finite groups of the same order type. If $G_{1}$ is solvable, is it true that $G_{2} $ is also necessarily solvable?
Rulin Shen in the paper: On same order type groups, SA"{U}. Fen Bilimleri Dergisi, 15. Cilt, 2. Say\i, s.156-158, 2011, proved that Thompson's problem is true for nilpotent and supersolvable groups. Also, he proved that problem is true for solvable groups with more than one connected prime graph component (please see Shen R, Shi W. On Thompson problem (in Chinese). Sci Sin Math, 2010, 40(6): 533-537).
Now this problem remains only for the case where the prime graph is connected. Has there ever been proof of this residual state? If not, I would appreciate it if you have an idea to prove it, and share it.