2.3 (1966)

No. The group $G = \langle x, y, z, t \mid x^4 = y^4 = z^2 = t^3 = 1, [x, y] = z, [x, z] = [y, z] = 1, x^t = y, y^t = x^{-1}y^{-1} \rangle$ is $\Gamma$-quasi-nilpotent, but not quasi-nilpotent. Since $G/\Phi(G) \cong A_4$, the intersection of any two maximal subgroups $A$ and $B$ of $G$ equals $\Phi(G)$, whence $N_A(A \cap B) \neq A \cap B \neq N_B(A \cap B)$; thus $G$ is $\Gamma$-quasi-nilpotent. On the other hand, if $A_1 = \langle zx^2, zy^2, t \rangle$ and $B_1 = \langle z, t \rangle$, then $N_{A_1}(A_1 \cap B_1) = A_1 \cap B_1 = \langle t \rangle$; hence $G$ is not quasi-nilpotent. (V. D. Mazurov, 1973.)