3.15 (1969)

No, in both cases (A. I. Budkin, V. A. Gorbunov, Algebra and Logic, 14 (1975), 73–84). Another example. The group $G = \langle a, b \mid (b^2)^a = b^3 \rangle$ is non-Hopfian as proved in (G. Baumslag, D. Solitar, Bull. Amer. Math. Soc., 68, no. 3 (1962), 199–201) and therefore is not metabelian. We choose elements $a, b, x_1, x_2, x_3, x_4 \in G$ such that $w = [[x_1, x_2], [x_3, x_4]] \neq 1$, add to them 1, their inverses, and all the initial segments of the word $w$ in the alphabet $\{x_i\}$, and all the initial segments of the word $a^{-1} b^2 a b^{-3}$ and of each of the words $x_i$ in the alphabet of $\{a, b\}$. Let $M$ be the resulting model. If $M$ was embeddable into a finite group $G_0$, then $G_0$ would have to be metacyclic, and also would have to have elements satisfying $[[x_1, x_2], [x_3, x_4]] \neq 1$, which is impossible. Hence $G$ is not $U$-embeddable into finite groups. Since the group $G/G^{(k)}$ is also non-Hopfian for a suitable $k$ (ibid.), it is not $U$-embeddable into finite groups for similar reasons. (Yu. I. Merzlyakov, 1969.)