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I have what is probably a very easy question, but it is something that I am finding hard to "see".

Thanks to Jyrki's very helpful answer below then perhaps I can phrase my question better.

Given a tower of fields $M / L / K$, we have the exact sequence

$1 \rightarrow \operatorname{Gal}(M/L) \rightarrow \operatorname{Gal}(M/K) \rightarrow \operatorname{Gal}(L/K) \rightarrow 1$

If we know the groups $\operatorname{Gal}(M/L)$ and $\operatorname{Gal}(L/K)$, how can we recover the whole Galois group?

I am particularly interested in when the exact sequence above is split, then the whole galois group is a semi-direct product

$\operatorname{Gal}(M/K) = \operatorname{Gal}(M/L) \rtimes \operatorname{Gal}(L/K)$

If we have an action of $\operatorname{Gal}(L/K)$ on $\operatorname{Gal}(M/L)$ by conjugation then we can recover the semidirect product. However, as the semi direct product can in general determine different groups, does this determine the group uniquely? Is there necessarily just one way of defining this action?

In the general case (i.e. when the exact sequence does not split) what can we say about the whole Galois group?

Is anyone able to point me in the right direction with this, I am sure that I am just forgetting something basic, and if anyone could give me a rough outline of what I am missing in my understanding I would very much appreciate it.

Extra: I've found a similiar question here, Group actions in towers of Galois extensions but note that I am interested in non-abelian extensions as well as abelian. This question states that if $\sigma \in \operatorname{Gal}(L/K)$ then any two lifts of $\sigma$ are conjugate to each other by an element of $ \operatorname{Gal}(L/K)$ , which re-raises my initial question of why two lifts are conjugate.

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In line 3, I suppose you mean: Given a tower of normal fields. Otherwise there is no galois group at all... –  awllower Mar 11 '13 at 9:53
    
Thanks. Perhaps I should have stated the extensions are indeed Galois. It is worth noting that some texts define the Galois group simply as the automorphism group of the larger field over the smaller field, and with this definition every field extension has a Galois group, but only Galois extensions (normal/separable) have Galois groups that obey the fundamental theorem of Galois theory. –  user65970 Mar 11 '13 at 18:27

3 Answers 3

Please let me know if these examples are of any use. I am just employing some small groups.

You can construct a Galois extension $M/K$ whose Galois group is $S_3$, and then if $L$ is the intermediate field corresponding to $A_3 \cong C_3$ (I am writing $C_n$ for a cyclic group of order $n$), in your sequence you will have that $$ \operatorname{Gal}(L/K) \cong C_2, \qquad \operatorname{Gal}(M/L) \cong C_3. $$ But you can also construct a Galois extension $M/K$ whose Galois group is $C_6$, and then if $L$ is the intermediate field corresponding to $C_3$, in your sequence you will also have that $$ \operatorname{Gal}(L/K) \cong C_2, \qquad \operatorname{Gal}(M/L) \cong C_3. $$ Both exact sequences split, but you have different actions, and different groups $\operatorname{Gal}(M/K)$.

You can do the same for the groups $C_4$ and $V = C_2 \times C_2$. Taking for $L$ a subfield corresponding to a subgroup of order $2$, we have in both cases $$\operatorname{Gal}(L/K) \cong C_2 \cong \operatorname{Gal}(M/L),$$ but $V$ is a split extension, while $C_4$ is not.

Using the dihedral group of order $8$ and the quaternion group of order $8$ one can construct two towers such that $$\operatorname{Gal}(L/K) \cong V \qquad \operatorname{Gal}(M/L) \cong C_2,$$ here both sequences will be nonsplit, but the two groups $\operatorname{Gal}(M/K)$ are non-isomorphic.

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Yes, thanks Andreas. This illustrates very well that in general the semi-direct product does not give a unique group. I think that maybe where I am having trouble is that if the sequence splits, then we have a section from $\operatorname{Gal}(L/K)$ to $\operatorname{Gal}(M/K)$ and this allows us to construct an action by conjugation of $\operatorname{Gal}(L/K)$ on $\operatorname{Gal}(M/L)$, this action will give us a unique semi-direct product, but is this action necessarily unique? –  user65970 Mar 10 '13 at 16:33

Something seems to have gone wrong with the first claim. What is true is the following. If $\overline{\sigma}_1$ and $\overline{\sigma}_2$ are two extensions of $\sigma\in \operatorname{Gal}(L/K)$, then $\overline{\sigma}_2^{-1}\overline{\sigma}_1$ restricted to $L$ is $\sigma^{-1}\sigma$, i.e. the identity, so it is an element of $\operatorname{Gal}(M/L)$. In other words: $$ \overline{\sigma}_1=\overline{\sigma}_2 g $$ for some $g\in \operatorname{Gal}(M/L)$.

Note that the two lifts cannot always be conjugate to each other in the prescribed way. After all, for all we know $\operatorname{Gal}(M/K)$ might be abelian, and hence no two distinct elements in there would be conjugate in $\operatorname{Gal}(M/K)$.

[Edit] I cannot say anything about the other claim. At least not yet [/Edit]

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Thanks, I hadn't made clear that I was assuming the sequence was split. Of course this is not always the case. I've edited the question to this effect. –  user65970 Mar 9 '13 at 19:05
    
As for the conjugacy, perhaps I misunderstood and "conjugacy" was used to just mean "differ" by an element of $\operatorname{Gal}(M/L)$. I'll try again under this assumption. Thanks again –  user65970 Mar 9 '13 at 19:17

In the abstract, the question you ask is known as the group extension problem. The wikipedia page should give you a good starting point for further reading.

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