Let $M:L:K$ be normal and seperable field extensions, then we know by the fundamental theorem of Galois theory that $$ \frac{\text{Gal}(M:K)}{\text{Gal}(M:L)} \cong \text{Gal}(L:K) $$
I was wondering if the following is true
Let us fix a basis for $M$: $x = \sum_i x_i\alpha_i \quad \forall x\in M$.
Can $\frac{\text{Gal}(M:K)}{\text{Gal(M:L)}}$ be represented by the set
$$ g\in \text{Aut}(M): g(\alpha_i)= \begin{cases} g(\alpha_i) & \forall \alpha_i \in L \\ \alpha_i & \text{otherwise} \end{cases} $$
I suppose the answer is yes, because $g$ permutes the different $\alpha_i$ and we can always negate the permutation on the $\alpha_i$ outside $L$ by an element $h\in \text{Gal(M:L)}$.
Is this reasoning correct?