Galois group of $x^{28}-1$ over $\mathbb{Q}$

Question

The Galois group of $x^{28}-1$ over $\mathbb{Q}$ is $\mathbb{Z}/28\mathbb{Z}^\times\cong C_2\times C_6$. But without knowing a priori, how can I tell that $\mathbb{Z}/28\mathbb{Z}^\times$ is isomorphic to $C_2\times C_6$? I know the question of determining the group structure of $\mathbb{Z}/n\mathbb{Z}^\times$ is hard in general, but is there some clever trick I can do, on an exam for example, to easily verify that $C_2\times C_6$ is actually correct, and that it isn't something like $C_{12}$? I know that being abelian narrows it down a lot, but is that the best we have? Is there some other property I can exploit?

Answer 1

By the Chinese Remainder Theorem, $\mathbb{Z}_{28} \cong \mathbb{Z}_7\times\mathbb{Z}_4$. The units will be elements of the form $(u_1,u_2)$ with $u_1$ a unit in $\mathbb{Z}_7$, and $u_2$ a unit in $\mathbb{Z}_4$.

The structure of the unit group of $\mathbb{Z}_{p^n}$ with $p$ a prime is well understood: for odd primes, it is cyclic of order $(p-1)p^{n-1}$. For $p=2$, if $n=1$ the group is trivial, and if $n\geq 2$, then the unit group is isomorphic to $C_{2^{n-2}}\times C_2$.

So in this case, you immediately get that $(\mathbb{Z}_{28})^*$ is isomorphic to $C_{6}\times C_2$, the $C_6$ factor coming from the $\mathbb{Z}_7$ part, and the $C_2$ factor from the $\mathbb{Z}_4$ factor.

Determining the group structure of $(\mathbb{Z}_n)^*$ is basically as hard as factoring $n$. Unless you want the resulting group expressed in terms of its invariant factors or primary divisors, in which case you basically need to know how to factor $n$ and how to factor $p-1$ for every prime $p$ that divides $n$.