# The Galois Group of $x^n-1$ is isomorphic to $(Z/nZ)^x$

I think that the problem statement on this practice question is incorrect, but I don't want to assume that.

The statement is: Let F be a field whose characteristic does not divide n. Prove that f(x) = $x^n − 1$ is separable over F and that the Galois group of f is isomorphic to a subgroup of $(Z/nZ)^×$.

Proving that f is separable: f'(x) = $nx^n-1$ is relatively prime to f(x) since the factors of f' are just n-1 copies of x. If a function is relatively prime to its derivative in a field, then it is separable over that field.

While proving that the Galois group is isomorphic to a subgroup of $(Z/nz)^x$ I ran into some trouble. I think it is just isomorphic to $(Z/nz)^x$ not a subgroup.

My proof is: Since f is separable in F, we must have that L the splitting field for f is a subfield of F. Since all splitting fields are unique up to isomorphism we know L $\cong$ $\mathbb Q$[w, $w^2$, ..., $w^n$,] where w is the primitive nth root of unity.

These roots form a cyclic group of order n, and Aut(Z/nZ) $\cong$ $(Z/nZ)^×$ So the Galois Group (which is the group of permutations of the roots) is isomorphic to $(Z/nZ)^×$.

My issue is that it is isomorphic to the multiplicative group itself, not a subgroup and in a followup problem I am asked to construct a field for $x^15-1$ where the Galois group is a proper subgroup of (Z/15Z)$^x$.

Any insight on this would be awesome! Thanks.

• A group is always a subgroup of itself. – Lord Shark the Unknown Mar 26 '18 at 18:22

It is true that the Galois group $G_n$ of the splitting field of $x^n-1$ is isomorphic to $(\Bbb Z/n\Bbb Z)^*$. That's not so easy (see for instance Washington's book on cyclotomic fields), but is is much more elementary to prove that $G$ embeds in $(\Bbb Z/n\Bbb Z)^*$. Given an $n$-th root of unity $\zeta$, $\sigma\in G_n$ maps $\zeta$ to $\zeta^a$ for some $a$, and $\sigma\mapsto a$ is the embedding you seek. It's more difficult to show it's surjective.
To get a smaller Galois group than say $(\Bbb Z/15\Bbb Z)^*$ take the fixed field of a subgroup of the Galois group of $\Bbb Q(\zeta_{15})$ but I suppose that would give an extension whose Galois group was a quotient rather than a subgroup of $(\Bbb Z/15\Bbb Z)^*$.