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

*

*$\phi:G\to u_n$ with $\sigma\mapsto \sigma(\lambda)\lambda^{-1}$ is
an injective homomorphism.

*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...
 A: $G$ 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 $\lambda^r\in F$.
A: 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 $\{[a^{0}],[a^{1}],...,[a^{n-1}] \}$.
