Let $G$ be a group acting on a set $\Omega$ faithfully. Then 2closure of $G$ denoted by $G^{(2)}$ is the largest subgroup of the symmetric group of $\Omega$ with the same orbits as $G$ on $\Omega\times\Omega$. Clearly $G\leq G^{(2)}$. If $G=G^{(2)}$ then $G$ is called 2closed. I am collecting some grouptheoretic properties of $G$ which inherited to $G^{(2)}$. For example, I know that if $G$ is abelian (pgroup,nilpotent, or has odd order), then so is $G^{(2)}$. Also I am looking for (grouptheoretic) properties of $G$ which makes it to be 2closed. For example, I know that every semiregular permutation group is 2closed. Also I guess that every cyclic permutation group is 2closed. I know that the basic properties of closures of permutation groups are given in Wielandt's book, "Permutation groups through invariant relations and invariant functions, lectures given at The Ohio State University, Columbus, Ohio, 1969", but unfortunately I dont have access it. Now my questions are requests are: 1) Which grouptheoretic properties of $G$ inherited to $G^{(2)}$?(other than the above properties) For example, for which solvable groups $G$, $G^{(2)}$ is solvable? 2) Which group theoretic properties of $G$ make it 2closed?(other than when $G$ is cyclic) 3) may it possible that one send me the file of Wielandt's book?

1$\begingroup$ I can't answer your question, but there are certainly examples in which $G$ is solvable but $G^{(2)}$ is not. There is one of degree $16$ for example. $\endgroup$ – Derek Holt Apr 2 '16 at 15:40

1$\begingroup$ @Derek: You don't need to go as far as degree 16.  Already the 2closure of $G := \langle (1,2,3,4,5), (1,2,4,3) \rangle \cong {\rm C}_5 \rtimes {\rm C}_4$ is ${\rm S}_5$, and hence not solvable. $\endgroup$ – Stefan Kohl Apr 2 '16 at 15:46

$\begingroup$ Dear Derek and Stefan, every 2  transitive Solvable proper subgroup of a symmetric group is not 2  closed. $\endgroup$ – majid arezoomand Apr 2 '16 at 15:55

$\begingroup$ For some reason I was looking for an example with at least three orbits! $\endgroup$ – Derek Holt Apr 2 '16 at 19:09

2$\begingroup$ This seems a little broad to me. $\endgroup$ – verret Apr 3 '16 at 0:12
Let's call an abstract group $2$closed if all of its faithful permutation representations are $2$closed.
You observed that cyclic groups are $2$closed.
Also a group $G$ for which the only faithful permutation representation is the regular representation must be $2$closed, because in any faithful action there must be a regular orbit, so if $x$ is in this orbit then the stabilizer of $x$ in $G^{(2)}$ must also be trivial, so $G = G^{(2)}$.
The only finite groups with that property are cyclic groups of prime power order and (generalized) quaternion groups. (I am not sure about infinite examples, but the group ${\mathbb Z}_{p^\infty}$ is one such for each prime $p$.)
So the (generalized) quaternion groups $Q_{2^n}$ for $n \ge 3$ are $2$closed and I think the direct product of $Q_{2^n}$ with any cyclic group of odd order is also $2$closed.
I have not been able to think of any other finite examples. $Q_{2^n} \times C_2$ is not $2$closed for example. But I would not care to make a conjecture about it, even for finite $p$groups.

$\begingroup$ Dear Derek, Thank you for your answer. May I have more details about your claims? $\endgroup$ – majid arezoomand Apr 4 '16 at 5:47

$\begingroup$ Does the property you mention in the 4th paragraph of your answer must be as follows: Every (faithful) permutation representation has a trivial pointstabilizer? $\endgroup$ – majid arezoomand Apr 6 '16 at 11:27

2$\begingroup$ Yes that's equivalent to saying that at least one of its orbits must be regular. For finite groups, the claim in this paragraph follows from the standard result that the only finite $p$groups with a unique subgroup of order $p$ are the cyclic and (generalized) quaternion groups. $\endgroup$ – Derek Holt Apr 6 '16 at 12:02