Show that $\displaystyle \min(F,a)=\prod_{1}^{r}(x-\tau_{i}(a))$ 
Let $K$ be a Galois extension of $F$, and let $a \in K$. Let $n = [K:F]$, $r = [F(a):F]$, and $H = \mathrm{Gal}(K/F(a))$. Let $\tau_{1},...,\tau_{r}$ be the left coset representative of $H$ in $G$. Show that $\displaystyle \min(F,a)=\prod_{1}^{r}(x-\tau_{i}(a))$.

If $\sigma \in \mathrm{Gal}(K/F)$ so, $\displaystyle \sigma\left(\prod_{1}^{r}(x-\tau_{i}(a))\right) = \prod_{1}^{r}\sigma(x-\tau_{i}(a)) = \prod_{1}^{r}(x-\sigma(\tau_{i}(a))) = ... = \prod_{1}^{r}(x-\tau_{i}(a))$. Then, $\displaystyle \prod_{1}^{r}(x-\tau_{i}(a)) \in F[x]$ and has degree $r$ and, since $\deg(\min(F,a)) = r$,
$$\displaystyle \min(F,a)=\prod_{1}^{r}(x-\tau_{i}(a))$$
up to a constant.

If I'm right, my question is: how to show that $``\displaystyle \prod_{1}^{r}(x-\sigma(\tau_{i}(a))) = ... = \prod_{1}^{r}(x-\tau_{i}(a))"$?
I found this suggestion here, but I don't know how to prove it. Thanks for the advance!
 A: Fix $\sigma\in G$. Use the fact that for all $1\leq i\leq r$, there exists a unique $1\leq j(i)\leq r$ and a unique $\mu_{i}\in H$ so that $$\sigma\tau_i = \tau_{j(i)} \mu_{i}.$$
This comes from the fact that $\sigma\tau_i$ is just another element of $G$, and hence belongs to some coset $\tau_{j(i)}H$. Then 
$$ \sigma\left(\prod_{i=1}^{r}(x-\tau_{i}(a))\right) 
= \prod_{i=1}^{r}(x-\sigma(\tau_{i}(a))) 
= \prod_{i=1}^{r}(x- \tau_{j(i)} \mu_{i}(a))
= \prod_{i=1}^{r}(x-\tau_{j(i)}(a))
=\prod_{i=1}^{r}(x-\tau_{i}(a)).$$
The second to last equality follows since the elements of $H$ fix $a$. The last equality follows since, as sets, $\{\tau_i\}=\{\tau_{j(i)}\}$. 
For completeness, we conclude by saying that $G$ fixes the polynomial $\prod_{i=1}^{r}(x-\tau_{i}(a))$, and hence $G$ fixes it coefficients. Factoring this polynomial into irreducibles over $F$ we see that $\prod_{i=1}^{r}(x-\tau_{i}(a))$ must itself be irreducible, otherwise,  looking at the factor containing $(x-a)$, we would have $\min(F,a)<r$. So we have a monic irreducible polynomial with coefficients in $F$ of degree equal to $[F(a):F] = \deg \min(F,a)$ with $a$ as a root. Therefore, $\min(F,a) = \prod_{i=1}^{r}(x-\tau_{i}(a))$.
