Isoclinic groups have same derived length

Statement

Suppose $G_{1}$ and $G_{2}$ are isoclinic groups. Then, the following are true:

$G_{1}$ is a solvable group if and only if $G_{2}$ is a solvable group.
If $G_{1}$ and $G_{2}$ are both solvable and nontrivial, then they have the same derived length. If either of them is trivial, the other may be nontrivial but must still be abelian (in which case we have derived lengths of zero and one).

Related facts

Proof

Given: Isoclinic groups $G_{1}$ and $G_{2}$ .

To prove: $G_{1}$ is solvable if and only if $G_{2}$ is, and if so, they have the same derived length if both are nontrivial. If either is trivial, the other may be nontrivial but must be abelian.

Step no.	Assertion/construction	Facts used	Previous steps used
1	$G_{1}$ is solvable if and only if its derived subgroup is solvable, and if so, the derived length of $G_{1}$ is one more than the derived length of its derived subgroup (unless $G_{1}$ is trivial).	definition of solvable group, via the derived series.	--
2	$G_{2}$ is solvable if and only if its derived subgroup is solvable, and if so, the derived length of $G_{2}$ is one more than the derived length of its derived subgroup (unless $G_{2}$ is trivial).	definition of solvable group, via the upper central series.	--
3	The derived subgroup of $G_{1}$ is isomorphic to the derived subgroup of $G_{2}$ .	definition of isoclinism	$G_{1}$ is isoclinic to $G_{2}$ .
4	$G_{1}$ is solvable if and only if $G_{2}$ is solvable, and they have the same derived length unless one of them is trivial.		Steps (1)-(3)
5	If either group is trivial, the derived subgroup of both must be trivial, so both must be abelian.		Step (3).