Polynomials over the rationals with alternating group as Galois group.

Suppose $f(x) \in \mathbb{Q} [x]$ is irreducible, and of degree $n$. Let $K$ denote its splitting field over $\mathbb{Q}$, and let $G = \text{gal} (K : \mathbb{Q})$ be its Galois group.

Now, if $G$ as a subgroup of $S_n$ (acting on the roots of $f$) is the alternating group $A_n$, then the irreducibility of $f$ over $\mathbb{Q}$ ensures that $G$ acts transitively on the roots of $f$. In other words, $G$ is $n$-transitive. But it is a well known fact that $A_n$ is at most $(n-2)$-transitive. Contradiction??

My question is this: Why does the above argument not prove that the Galois group of $f$ cannot be $A_n$? Are any of my facts/assumptions wrong? I am aware that the family of alternating groups can be realized as Galois groups over $\mathbb{Q}$, so I'm quite stumped.

One observation I had is that if $\alpha$ is a primitive element for $K$, and $g$ is its minimal polynomial, then the degree of $g$ would be $|G| = n!/2$, so that $G$ can be identified as a subgroup of $S_{n!/2}$ isomorphic to $A_n$, but I don't see the impact it has (if at all) on the earlier argument.

• no, transitive means $1$-transitive. – mercio Oct 29 '15 at 14:02
• aha, I completely misunderstood what $k$-transitive meant. thanks for clearing that up. – adiddy15 Oct 29 '15 at 14:09