Irreducible polynomial of degree $8$ Would you please help me to prove this problem.
Let $G=\mathrm{Hol}(C_8)$. And let $f(x)\in \mathbb{Q}[x]$ be an irreducible polynomial of degree $8$, $L$ a splitting field of $f(x)$ over $\mathbb{Q}$ and suppose that $\mathrm{Gal}(L/\mathbb{Q})\cong G$.

Show that there is a subgroup of $S_8$ isomorphic to $G$.
  I am reviewing for my test and i find this good exercice 

Thanks.
 A: This statement of the result 
$S_8$ contains a subgroup isomorphic to $G=\mathrm{Hol}(C_8)$ 
does not have anything to do with Galois theory. And, indeed, you can see that it is true without mentioning polynomials at all. Let $C \cong C_8$ be the subgroup of $S_8$ generated by the $8$-cycle $(12345678)$. The group of automorphisms of $C_8$ is the group of units modulo $8$ (also known as the Klein $4$-group). These are achieved via the conjugation action within $S_8$ by the subgroup $V$ generated by 
$$(24)(37)(68), \ (28)(37)(46), \ \text{and} \ (26)(84).$$ 
The subgroup generated by $C$ and $V$ is isomorphic to $G$ (it is enough to check that the conjugates of our $8$-cycle by the elements of $V$ produce precisely its first, third, fifth, and seventh powers).
It is also true that this occurrence of $G$ as a subgroup of $S_8$ is unique up to conjugacy: start with the simple observation that there is a unique up to conjugacy element of $S_8$ of order $8$. Let $H$ be a subgroup of $S_8$ isomorphic to $G$. It contains an $8$-cycle, which we may assume is $(12345678)$. Since $H \cong G$, there must be some $v \in H$ with
$$v (12345678) v^{-1}=(12345678)^3=(14725836).$$ Since $C$ is self-centralizing, any such $v$ has the form $$v=(24)(37)(68) c$$ for some $c \in C$. It follows that $(24)(37)(68) \in H$, and similarly the other elements of order two written down above belong to $H$. This implies that $H$ is the group we wrote down before, by order considerations.
A: Let us give it a try (haven't taught Galois Theory in the last years...:)) :
Suppose we have a field $\;k\;$ and an irreducible $\;f(x)\in k[x]\;$ with $\;\deg f=n\;$, say . We suppose everything is separable here, and let $\;L\;$ be the splitting field (the one and unique up to $\;k\,-$ isomorphism) of $\;f\;$ over $\;k\;$, so $\;L/k\;$ is a Galois extension.
Since $\;G:=Gal(L/k)\;$ acts transitivily over the roots of $\;f\;$ in $\;L\;$ (in fact, this is an iff condition: a polynomial in $\;k[x]\;$ with at least one root in $\;L\;$ is irreducible iff $\;G\;$ acts transitively over its roots. This makes sense since $\;L/k\;$ is normal and thus an irredubile polynomial having a root there automatically has all its roots in $\;L\;$) , we can embed $\;G\;$ into $\;S_n\;$ in the obvious way (observe all the roots of $\;f\;$ are distinct since we assume separability!) ...and that's all.
Of course, the above relies heavily on some basic facts from Galois Theory.
