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?

  • 1
    $\begingroup$ What do you mean by explain the titled statement in terms of Galois Theory? It seems like you already understand the galois theoretic interpretation of the fact that normality is not transitive. Are you looking for a counter example? $\endgroup$ Mar 15 at 16:57

1 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 $\mathbf{Q}$.


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