Multiplicative formula for order of automorphism group I am reading a proof of the following proposition from Dummit and Foote:


Let $E$ be the splitting field over $F$ of the polynomial $f(x) \in F[x]$. Then $$|\textrm{Aut}(E/F)|\leq [E:F]$$
    with equality if $f(x)$ is separable.


Now the proof of this proposition is by induction, for $[E:F] = 1$ this is clear, whereas if $[E:F] > 1$ then $f(x)$ has some irreducible non-linear factor $p(x)$. Then one considers the number of ways in which the identity map from $F$ to itself can be extended to an isomorphism between $F[x]/(p(x)) \cong F(a)$ and $F[x]/(p(x)) \cong F(a')$ where $a$ and $a'$ are two different roots of $p(x)$ in $E$.
Now for the rest of the proof they seem to use the fact that 
$$\left|\textrm{Aut}(E/F(a))\right||\textrm{Aut}(F(a)/F)| = |\textrm{Aut}(E/F)|.$$
Now we know $|\textrm{Aut}(E/F(a))| \leq  [E:F(a)]$ by the induction hypothesis and that $|\textrm{Aut}(F(a)/F)| \leq [F(a):F]$. The reason why I think they use the above fact is because they mention the formula $[E:F] = [E:F(a)][F(a):F]$ and finish the proof of the proposition from there. Where can I find a proof of the fact I stated above, if it is true?
Thanks.
Edit: I just got an idea on how to prove the fact above. Consider the the map
$$\begin{align*} L : \textrm{Aut}(E/F)& \longrightarrow \textrm{Aut}(F(a)/F) \\
\phi& \longmapsto \phi|_{F(a)} \end{align*}.$$
It is easily seen that $L$ is a group homomorphism, and that the kernel of $\phi$ is those elements in $\textrm{Aut}(E/F)$ that are already the identity on $F(a)$, which means that 
$$\ker L = \textrm{Aut}(E/F(a)).$$
As for the image of $L$, I think that $L$ is actually surjective. This is because the automorphisms from $E$ to itself were simply obtained by extending those from $F(a)$ to $F'(a)$, so going in the reverse direction any $\sigma \in \textrm{Aut}(F(a)/F)$ is the result of restricting an automorphism on $E$. Since
$$\begin{eqnarray*} |\textrm{Aut}(E/F)| &=& |\textrm{im} L||\ker L| \\
&=& |\textrm{Aut}(F(a)/F)|\left|\textrm{Aut}(E/F(a))\right| \end{eqnarray*}$$
this finishes my claim. Does this seem right?
 A: The way the argument works is as follows:
Given an automorphism $\phi$ of $E/F$, consider its restriction to $F(\alpha)$. This restriction must maps $F(\alpha)$ to some $F(\alpha')$, with $\alpha'$ another root of $f$ (it could be that $\alpha=\alpha'$, or it could be that $\alpha\neq\alpha'$). Each of the possible restrictions is the restriction of some automorphism $E/F$ (because $E$ is a splitting field), since we can extend any particular isomorphism $F(\alpha)\to F(\alpha')$ to an automorphism of $E$. So in order to count the number of possible automorphism of $E$ over $F$, it suffices to count how many homomorphisms restrict to a particular restriction, and how many possible restrictions there are.
The number of possible restrictions is, by the induction hypothesis, $[E:F(\alpha)]$. And the number of possible restrictions is the number of possible distinct roots of $f(x)$; this is bounded by $\deg(f)=[F(\alpha):F]$.
So we have that there are at most $[F(\alpha):F]$ possible restrictions (equality if all the roots of $f$ are distinct), and each of the restrictions extends in at most $[E:F(\alpha)]$ ways (equality if all the roots of $f$ are distinct). So in conclusion, the number of possible automorphisms is at most $[F(\alpha):F][E:F(\alpha)] = [E:F]$, with equality if and only if we have equality in each factor, if and only if all the roots of $f$ are distinct.
