Are $\mathbb{Z}^*_{28}$, $\mathbb{Z}^*_{4}×\mathbb{Z}^*_{7}$, $\mathbb{Z}^*_{2}×\mathbb{Z}^*_{14}$ cyclic groups?

Question

Are $\mathbb{Z}^*_{28}$, $\mathbb{Z}^*_{4}×\mathbb{Z}^*_{7}$, $\mathbb{Z}^*_{2}×\mathbb{Z}^*_{14}$ cyclic groups? I know how to determine when they are in the normal $\mathbb{Z}_{n}$ form. And I know for the last two, it might have some thing to do with relatively prime numbers. But I still have problem figuring it out.

Answer 1

• $(\Bbb Z/28)^\times$ is not cyclic, in fact the ring $(\Bbb Z/28,+,\cdot)$ is isomorphic to the cartesian product of the rings $\Bbb Z/4$ and $\Bbb Z/7$, chinese remainder theorem (and compatibility of the operations (at least in one direction, but then also the other one)).

• So let us analyze the units subgroups of the rings $\Bbb Z/4$ and $\Bbb Z/7$ closer. We have $(\Bbb Z/4)^\times= (\pm1,\cdot)\cong(\Bbb Z/2,+)$, generated by an element of order two. And the ring $\Bbb Z/7$ is even a field, the field with seven elements, its units form a cyclic group or order $6$, $3$ modulo seven is a generator. So $(\Bbb Z/7)^\times\cong(\Bbb Z/6,+)$. And in the theorem of invariant factors we have a group $\Bbb Z/2\times \Bbb Z/6$, with invariant factors $2|6$. Not cyclic, two invariant factors.

• Now in the last example, we can ignore the trivial group $(\Bbb Z/2)^\times$, so we have to consider the units of the ring $\Bbb Z/14$. It is isomorphic as a ring with $\Bbb Z/2\times \Bbb Z/7$. We pass to the units, so we pass to the units in each cartesian factor, the first factor has the trivial group as a unit group, the second one, as mentioned, cyclic.

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Note: There are computer algebra systems that help to see clear the structure, and also experiment to understand in detail (by getting the hands "dirty"). For instance in sage:

sage: R = Zmod(28)
sage: G = R.unit_group()
sage: G.order()
12
sage: euler_phi(28)
12
sage: G.is_cyclic()
False
sage: print "G has %s generators" % len(G.gens() )
G has 2 generators
sage: 
sage: H = Zmod(14).unit_group()
sage: H.is_cyclic()
True