Consecutive Galois extensions

$Proposition$: An extension $K/F$ is normal if and only if $H \triangleleft Gal(E/F)$

Now let $E/F$ be a finite galois extension and let $F=K_0 \leqslant K_1 \leqslant K_2 \leqslant .....\leqslant K_n=E$ consecutive extensions.

Let $H_i$ (for $i=0,1,2,,,n$) subgroup of $Gal(E/F)$ which corresponds to $K_i$ from the Galois correspondence (i.e $K_i \longleftarrow \longrightarrow H_i$)

Prove that $\forall i \in \{1,2,,,n\}$, we have that $K_i / K_{i-1}$ is a galois extension if and only if $H_i \triangleleft H_{i-1}$

I was going to proceed by induction to $n$.

For $n=1$ its valid because of the proposition .

Thus i assumed that the statement is valid $\forall 1<i<n$

But i find a difficulty to proceed because i cannot find the inductive step.

Can someone help me to solve this and complete the proof?

It has nothing to do with induction. Assume $F\subseteq L\subseteq K\subseteq E$ with $E/F$ Galois. Let $H$ be the subgroup fixing $L$ and $N$ the subgroup fixing $K$. The $K/L$ is Galois iff $N \triangleleft H$. Assume first that $K/L$ is Galois let $\tau \in N$ and $\sigma \in H$. Then $\sigma$ maps $K$ into $K$. So if $x\in K$ then $\sigma(x)\in K$ and thus $\sigma^{-1}\tau\sigma(x)=x$, since $\tau$ fixes $K$ and thus $\sigma^{-1}\tau\sigma\in N$.
Assume conversely that $N\triangleleft H$ then with the same notation as above, $\sigma^{-1}\tau\sigma\in N$ and so $\sigma^{-1}\tau\sigma(x)=x$. So $\tau\sigma(x)=\sigma(x)$ and so $\sigma(x)$ is fixed by $H$ and is thus in $K$.
• @capo Because if an automorphism fixes $K$, then it fixes $L$. Fixes means pointwise fixes. – Rene Schipperus Nov 18 '16 at 13:40
• Every auto of $K$ extends to an auto of $E$. Thus autos of $E$ and $K$ are the same on $K$. – Rene Schipperus Nov 18 '16 at 15:33