# Prove that $|\operatorname{Gal}(F(\lambda)/F)|=\left|\frac{\langle a \rangle}{\langle a \rangle \cap H}\right|.$

Question: Let $$F$$ be a field contains primitive nth root of unity $$\zeta$$. Define the multiplicative group with order $$n$$ to be $$u_n=\{z\in F : z^n=1\}$$. Let $$a\in F\setminus\{0\}$$ and $$\lambda \in \bar F$$ such that $$\lambda$$ is a root of $$x^n-a=0$$.

Suppose $$G=\operatorname{Gal}(F(\lambda)/F)$$. I want to prove

1. $$\phi:G\to u_n$$ with $$\sigma\mapsto \sigma(\lambda)\lambda^{-1}$$ is an injective homomorphism.
2. Suppose $$H=\{y^n: y^n\in F^*\}$$. Prove that $$|\operatorname{Im}(\phi)|=\left|\frac{\langle a \rangle}{\langle a \rangle \cap H}\right|.$$

Here is my work on 1. : Let $$\sigma,\tau \in G.$$ $$\phi(\sigma\circ\tau)=\sigma(\tau(\lambda))\lambda^{-1}$$ $$\phi(\sigma)\phi(\tau)=\sigma(\lambda)\lambda^{-1}\cdot\tau(\lambda)\lambda^{-1}.$$

They seems to be not equal... Is the multiplication $$\sigma(\lambda)\tau(\lambda)$$ just the notation of permutation decomposition? For checking the injectivity, if $$\sigma \in ker(\phi)$$, then $$\sigma(\lambda)\lambda^{-1}=1$$. So $$\sigma(\lambda)=\lambda$$. Can I say $$\sigma$$ is just the identity permutation?

For 2., By the last part and fundamental homomorphism theorem 1, $$\operatorname{Im}(\phi) \cong G$$. Then I divide the problem into 2 cases

• $$a \neq 1$$
• $$a=1$$.

The case $$a=1$$ is trivial because it makes $$F(\lambda)=F$$, then $$|G|=[F(\lambda):F]=1$$ and also $$\left|\frac{\langle a \rangle}{\langle a \rangle \cap H}\right|=1$$. But when $$a\neq 1, |\operatorname{Im}(\phi)|=[F(\lambda)/F]=n$$ but $$\left|\frac{\langle a \rangle}{\langle a \rangle \cap H}\right|=\frac{|\langle a \rangle|}{\operatorname{lcm}(|\langle a \rangle|,|H|)}.$$ They look very different...

I mean if $$\lambda$$ is a root of the given polynomial, then you can view it as $$a^{(1/n)}\zeta$$, and here we are taking the principal $$n^{th}$$ root for $$a$$. Any $$\sigma \in G$$ will raise $$\zeta$$ to some power, so $$\sigma(\lambda)= a^{(1/n)}\zeta^{r}$$, and say $$\tau(\lambda)=a^{1/n}\zeta^{s}$$. So $$\phi(\sigma \circ \tau)(\lambda)=\sigma\circ\tau(\lambda)\lambda^{-1}=\sigma(a^{1/n}\zeta^{s})\lambda^{-1}=\zeta^{s-1}\sigma(a^{1/n}\zeta)\lambda^{-1}=\zeta^{s-1}(a^{1/n}\zeta^{r})\lambda^{-1}=\zeta^{s+r-2}.$$ On the other hand we get $$\sigma(\lambda)\lambda^{-1}\tau(\lambda)\lambda^{-1}=\frac{a^{1/n}\zeta^{r} a^{1/n}\zeta^{s}}{a^{1/n}\zeta a^{1/n}\zeta}=\zeta^{r+s-2}$$ as needed.
Yes, what you have done is enough to show the kernel is trivial, since if sigma fixes $$\lambda$$, and your field extension is generated by $$\lambda$$ then it fixes everything.
As for the final bit you have left, I believe you can see that $$\left|\frac{\langle a \rangle}{\langle a \rangle \cap H}\right|=n$$ by using the second group isomorphism theorem to deduce that it has the same size as $$\frac{\langle a\rangle H}{H}$$. $$\langle a \rangle=\{a^{i} \ | \enspace i \in \mathbb{Z} \}$$, and since you are taking modulo all $$n^{th}$$ powers, then your cosets are determined by the power of $$i$$ modulo $$n$$, i.e. your cosets are of the form $$\{[a^{0}],[a^{1}],...,[a^{n-1}] \}$$.
$$G$$ is cyclic generated by $$g(\lambda)=\zeta^m \lambda$$. Take $$\sigma=g^a,\tau=g^b$$.
For a primitive $$n$$-th root of unity to exist $$F(\lambda)/F$$ is separable (thus Galois)
$$\lambda$$'s $$F$$ minimal polynomial is $$\prod_{d=1}^{|G|} (x-\zeta^{md} \lambda)=x^{|G|} -\lambda^{|G|}$$ where $$|G|=n/\gcd(n,m)$$ must be the least integer such that $$\lambda^r\in F$$.