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.

  • $\begingroup$ For $n$ having prime factors $p_1, p_2, \cdots, p_n$, we have $\phi(n) = n \left(1 - \dfrac1{p_1}\right) \left(1 - \dfrac1{p_2}\right) \cdots \left(1 - \dfrac1{p_n}\right)$. $\endgroup$ – DHMO Apr 18 '17 at 7:01
  • 2
    $\begingroup$ I think you've miscalculated something: $U(21)$ has $12$ elements. $\endgroup$ – Alex Wertheim Apr 18 '17 at 7:01

$|U(49)| = \phi(49) = 49 \times \dfrac67 = 42$ 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 $14$:

$$6 \mapsto 36 \mapsto 20 \mapsto 22 \mapsto 34 \mapsto 8 \mapsto 48 \mapsto 43 \mapsto 13 \mapsto 29 \mapsto 27 \mapsto 15 \mapsto 41 \mapsto 1$$


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