Group theory commutator and solvable groups

Question

let G be a group such that it contains 2 members $a, b \in G$ that statisfy:

1. $a = p^{-1} b p$ where $p \in G$
2. $a = q^{-1} [a,b]q $ where $q \in G$
3. $a,b,[a,b]\neq e$

where $[a,b]$ is the commutator of $a,b$ and $e$ the identity element of the group

can G be solvable (finite/not finite)?

Answer 1

Assume that $G$ is solvable and let $G=G_1 > G_2 > \cdots G_n > G_{n+1}=1$ be the derived series of $G$.

There is a unique $k$ with $1 \le k \le n$ such that $a \in G_k \setminus G_{k+1}$. If $b$ is conjugate to $a$ in $G$, then we must also have $b \in G_k \setminus G_{k+1}$. But then $[a,b] \in G_{k+1}$, so $[a,b]$ cannot be conjugate to $a$.

So the answer to the question is no.