Question:
show that there is a U-group containing a subgroup isomorphic to $\mathbb{Z}_{14}$.
The order of $\mathbb{Z}_{14}$ is 14.
By the fundamental theorem of cyclic group:
$\mathbb{Z}_{14}$ has subgroups of order $k=\left \{ 1,2,7,14 \right \}$.
Using the phi-Euler totient:
$\mathbb{Z}_{14}$ has 1 element of order 1, 1 element of order 2. 6 element of order 7 and 6 elements of order 6.
Isomorphism preserves order of group elements so it suffices to determine a subgroup of order 14 corresponding a $U\left ( n \right )$-group for some $n > 1 \in \mathbb{Z}^{+}$.
Laboriously, $U\left ( 21 \right )$ has order 14 and hence it has a unique subgroup of order 1,2,7, 14.
The order of a cyclic subgroup is the order of any group element in $\mathbb{Z}_{14}$.
Using the Phi-Euler totient function: one can determine again the number of elements of certain order in $U\left ( 21 \right )$.
But is there a more efficient way to determine a U-group of a certain order corresponding to some $n>1 \in \mathbb{Z}$?
Thank in advance.
$|U(49)| = \phi(49) = 49 \times \dfrac67 = 42$ which is a multiple of $14$.
Now, to find a subgroup of order $14$, we need to solve $x^{14}=1$, which gives $x=7k\pm1$.
We can pick $6$ to generate $\langle 6\rangle$, which is a subgroup of order $14$:
$$6 \mapsto 36 \mapsto 20 \mapsto 22 \mapsto 34 \mapsto 8 \mapsto 48 \mapsto 43 \mapsto 13 \mapsto 29 \mapsto 27 \mapsto 15 \mapsto 41 \mapsto 1$$