Use Galois theory to explain the following. Let $G$ be a group, $H\unlhd G$, $K\unlhd H$. Then, we cannot always say $K\unlhd G$.

Question

Let $G$ be a group, $H$ be normal subgroup of $G$, $K$ be normal subgroup of $H$. Then, we cannot always say $K$ is normal in $G$.

There are counterexamples, for example, here:

Are normal subgroups transitive?

But I want to relate this phenomenon to Galois theory.

For example, with $L/F/K$, both being Galois extensions, does not mean that $L/K$ is Galois.

This might not have to do with this case, but I guess the titled statement can be explained by Galois correspondence.

Could you explain the titled statement in terms of Galois theory?

Answer 1

It seems you just want an explicit example that one can see in both worlds?

You use some letters for both fields and groups so my notation is a little different.

Let $G = \langle s,r | s^2, r^4, sr=rs^{-1} \rangle$ be the dihedral group of order $8$.

Let $M = \mathbf{Q}(\sqrt[4]{2},i)$. Then $G = \mathrm{Gal}(M/\mathbf{Q})$, where, for the natural embedding of $M$ into $\mathbf{C}$, the element $s$ acts by complex conjugation and $r$ fixes $i$ by sends $\sqrt[4]{2}$ to $i \sqrt[4]{2}$.

Consider the fields

$$L = \mathbf{Q}(\sqrt[4]{2}), K = \mathbf{Q}(\sqrt{2}), \mathbf{Q} = \mathbf{Q}.$$

Consider the groups

$$A = \langle s \rangle, B = \langle s,r^2 \rangle, G = G.$$

Then $M^{A} = \mathbf{Q}(\sqrt[4]{2}) = L$, $M^{B} = \mathbf{Q}(\sqrt{2}) = K$, and $M^{G} = \mathbf{Q}$.

So the translation of the statement $A$ is normal in $B$ and $B$ is normal in $G$ but $A$ is not no rmal in $G$ into Galois theory (in this case) is that $L$ is Galois over $K$ and $K$ is Galois over $\mathbf{Q}$ but $L$ is not Galois over $\mathbf{Q}$.