Let $k>0$ and $l \geq 0$ be integers and denote by $\varphi$ the Euler function. Prove that there exists an algebraic number field $K$ for which $\mathcal{O}_K^*$ is isomorphic to $\mathbb{Z}/k\mathbb{Z} \oplus \mathbb{Z}^l$ if and only if $k$ is even and $\varphi(k)$ divides $2(l+1)$.
By Dirichlet unit theorem we have $$\mathcal{O}_K^* \cong \mathbb{Z}^{r_1+r_2-1} \times \mu_{\mathcal{O}_K}$$ So we need $l=r_1+r_2-1$. Also, we have $\mu_{\mathcal{O}_K} \subset \left<\xi_m\right>$ with $\xi_m$ the largest integer for which the $m$-th cyclotomic field $\mathbb{Q}(\xi_m)$ can be embedded in $K$, $[\mathbb{Q}(\xi_m):\mathbb{Q}]= \varphi(m)$ which divides $[K:\mathbb{Q}]$.
If $r_1$ the number of real embedding $>0$ then $\mu_{\mathcal{O}_K}=\{\pm 1\}$ and $k=2$ and of course $\varphi(2) \vert 2(l+1)$. If $r_1=0$ then $l=r_2-1$ and obtain $ \varphi(m) |2(l+1)=[K:\mathbb{Q}]$. So in this case can we choose $K$ equal to $\mathbb{Q}(\xi_m)$?
It relates to Unit group isomorphic to $\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}$, in this case we have $K$ is any quadratic extension of $\mathbb{Q_{\xi_4}}$ but not equal $\mathbf{Q}(\zeta_8)$ or $\mathbf{Q}(\zeta_{12})$. However i don't know how to deal with general case.
If $k$ is odd and a number field containes the $k$-th roots of unity, then it also contains the $2k$-th roots of unity. Thus $k$ is even.
If $k = 2$, take a totally real field with degree $n = \ell + 1$.
If $k \ge 4$, $K$ contains $F = {\mathbb Q}(\zeta_k)$; this field has degree $\phi(k)$ and is totally complex. In this case, the degree of the (totally complex) number field $K$ must be equal to $2\ell+2$ for the unit rank to be $\ell$. Since $F$ is a subfield, $\phi(k) \mid 2(\ell+1)$.
Finding fields satisfying these conditions is easy if a little tedious.