f irreducible polynomial with $p-2$ real roots $\Rightarrow$ $Gal(\mathbb{Q}_{f}/\mathbb{Q}) \cong S_{p}$ I have no idea what to do to show the following
Let $p\ge 5$ be a prime number
1) Let $H\subset S_{p}$ be a subgroup of the symmetric group. Assume that $p$ divides the order of $H$ and that $H$ contains a transposition. Show that $H=S_{p}$.
2) Let $f(X)\in \mathbb{Q}[X]$ be an irreducible polynomial of degree $p$ with exactly $p-2$ real roots and $\mathbb{Q}_{f}$ is the splitting field of $f$ in an algebraic closure $\overline{\mathbb{Q}}$ of $\mathbb{Q}$. Show that $Gal(\mathbb{Q}_{f}/\mathbb{Q})$ is isomorphic to $S_{p}$. 
Thanks in advance for any help.
 A: 1) Let $h$ be the order of the given subgroup $H$ of $S_p$. By hypothesis $h=p.n$, so that any $\sigma \in H$ verifies ${\sigma}^h=1={({\sigma}^n)}^p=1$. It follows that $H$ contains a $p$-cycle, which is $\gamma={\sigma}^n$.  Suppose moreover that $H$ contains a transposition $\tau$. By rearranging the notation if necessary (it suffices to replace $\gamma$ by a power), we may write $\gamma=(1, 2, ...,p)$ and $\tau=(1, 2)$. Considering then the repeated conjugates ${\gamma}_1=\tau \gamma \tau=(2,1,...,p),{\tau}_1={{\gamma}_1}^{-1}\tau{\gamma}_1=(1,3),{\gamma}_2={\tau}_1 {\gamma}_1 {\tau}_1=(2,3,1,...,p), {\tau}_2={{\gamma}_2}^{-1}{\tau}_1{\gamma}_2=(1,4)$, etc. we obtain all the transpositions $(1,2), (1,3),..., (1,p)$. These generate $S_p$, so $H=S_p$.
2) Let $f(X)\in \mathbf Q[X]$, irreducible of degree $p$ and admitting exactly $p-2$ real roots, and let $H$ be the Galois group of the splitting field ${\mathbf Q}_f$ of $f$. Since ${\mathbf Q}_f$ contains $\mathbf Q (a)$, where $a$ is a root of $f$, the order of  $H$ is divisible by $p$. Besides, the complex conjugation permutes the two non real roots of $f$, so $H$ contains a transposition. By 1), $H\cong S_p$  ./.
