prove $[E:K\cap L]=[E:K][E:L]$

Question

Let $E/K,E/L$ be finite Galois extensions with Galois groups $G,H$. If $G\cap H= \left\{1\right\}$, show $G\cdot H = \left\{ g h \mid g\in G,h\in H \right\}$ is a group iff $[E:K\cap L]=[E:K][E:L]$.

But isn't the set $G\cdot H$ always a group? The condition $G\cap H= \left\{1\right\}$ implies $KL=E$ and then there's a theorem saying $[KL:L]=[K:L\cap K]$ which is different...

What do you mean $;G\cap H=0;$ ? Do you mean you know the Galois groups are abelian and then, for some reason, you're writing the operations additively and zero is the unit element? But then you take the product $;GH;$ ...? Now, for the question to make sense you must be able to embed both groups in a bigger group, otherwise $;GH;$ isn't even defined, and once you've done this: no, product of subgroups isn't a subgroup in general. — DonAntonio, Jul 20 '16 at 08:49
I fixed the 0 to {1}. Okay, the product isn't a subgroup cause they group aren't abelian. What do I do here? — user355165, Jul 20 '16 at 08:54
Nop, not being abelian is not a reason for the product of two subgroups not being a subgroup. If $;G;$ is any group and $;H,K\le G;$ , then $;HK\le G\iff HK=KH;$ . This is a basic exercise in basic group theory. Thus, for example, it is enough to require that one of the subgroups is normal for the product to be a subgroup (but not necessary). Anyway: in what bigger group are you going to embed your groups $;G,H;$ so that the product $;GH;$ is even defined, to begin with? — DonAntonio, Jul 20 '16 at 08:57
Don't be. It is just that your problem cannot be properly attacked, leave alone, solved, without that info. Either you're omitting some relevant data, or this is part of another, bigger question, etc. — DonAntonio, Jul 20 '16 at 09:01
@DonAntonio what are some conditions on the extensions making the equality $[E:K\cap L]=[E:K][E:L]$ true? — user355165, Jul 20 '16 at 09:25
Are you sure about $[KL:L]=[L:L\cap K]$? Because putting that together with $KL=E$ gives you $[E:L]=[L:L\cap K]$, and comparing that to $[E:K\cap L]=[E:K][E:L]$ and $[E:K\cap L]=[E:L][L:K\cap L]$ tells you $[E:L]=[E:K]$, which seems unlikely. — Gerry Myerson, Jul 20 '16 at 09:43
Gerry mentions a good point. I think the theorem should probably be $;[KL:L]=[\color{red}K:K\cap L]$ — DonAntonio, Jul 20 '16 at 09:45
You are both right of course, sorry. I corrected my question. As far as embedding in a larger group, maybe $\mathrm{Aut}_E(K\cap L)$ works? — user355165, Jul 20 '16 at 09:48
@user355165 That's exactly what I'm trying to do in a comment that was too long and thus is becoming now an answer... — DonAntonio, Jul 20 '16 at 09:50
I shall need more time for this as there are some things that look suspicious and/or I don't remember well. — DonAntonio, Jul 20 '16 at 10:22
@DonAntonio I think a few applications of theorem 2.6 here are needed. — Arrow, Jul 20 '16 at 10:30
@Arrow Perhaps so, yet the OP's question is not about two different extensions of one single basis field, but rather about two different subextensions of the same field with different basis fields. This is what confuses me here, though I was thinking perhaps getting into quotient groups we could use somehow that th. 2.6 you say. Thank you. — DonAntonio, Jul 20 '16 at 10:56
@DonAntonio I think along with $[LK:L]=[K:L\cap K]$ and its mirror version the formula $[LK:L\cap K]=\frac{[L:L\cap K][K:L\cap K]}{[L\cap K:L\cap K]}=[L:L\cap K][K:L\cap K]$ resolves everything, so the problem reduces to justifying its use. — Arrow, Jul 20 '16 at 11:10
@DonAntonio do you have any insight to share on this problem? — user355165, Jul 20 '16 at 17:25
$[KL:L\cap K]=[L:L\cap K]\cdot [K:L\cap K]$ if and only if $L$ and $K$ are linearly disjoint over $L\cap K$. This is automatic, when both $L$ and $K$ are Galois over $L\cap K$. — Jyrki Lahtonen, Jul 20 '16 at 19:15

score 2 · Answer 1 · edited Apr 13 '17 at 12:20

There is something wrong with the title of this question. Namely the claim in the title is false without some extra assumptions.

Consider the following example.

Let $E=\Bbb{C}(x)$ be the field of rational functions. Let $G$ be the cyclic group of order two generated by the $\Bbb{C}$-automorphism $\sigma:x\mapsto -x$. We immediately see that the fixed field of $G$ is the subfield $K=\Bbb{C}(x^2)$.

Let $H$ be the cyclic group of order two generated by the $\Bbb{C}$-automorphism $\tau:x\mapsto 1-x$. We easily see that the corresponding fixed field is $L=\Bbb{C}(x^2-x)$.

But in this case it is easy to show that $$ K\cap L=\Bbb{C}. $$ See this thread and my old answer for a proof of that fact. Anyway, here $$ [E:L]=[E:K]=2,\quad [E:K\cap L]=\infty. $$

Note that in this case $\sigma\tau:x\mapsto x+1$, so $\sigma\tau$ is of infinite order. Therefore $GH$, a set with only four elements, most certainly is not a group.

Answering one direction. Assume that $G\cdot H$ is a group, call it $\Omega$. Because $G\cap H=\{1\}$ we know that $|\Omega|=|G|\cdot |H|$, in particular we know that $\Omega$ is finite. By Artin's lemma the fixed field $F=E^\Omega$ has the properties that $[E:F]=|\Omega|$ and $Gal(E/F)=\Omega$. Because $\Omega$ is the smallest group containing both $G$ and $H$, Galois correspondence tells us that $F$ is the largest subfiel contained in both $K$ and $L$. Therefore $F=K\cap L$. Also, $$ [E:K\cap L]=[E:F]=|\Omega|=|G|\cdot |H|=[E:K]\cdot [E:L]. $$

[Edit:]

I think the other direction follows from the following line of thought. Assume that $[E:K\cap L]=|G|\cdot |H|$. Clearly $K\cap L$ is fixed pointwise by all the automorphisms of $E$ that are of the form $gh$. Because $G$ and $H$ intersect trivially, all the automorphisms of this form are distinct. Therefore $|Aut(E/K\cap L)|\ge |G|\cdot |H|$. On the other hand for a finite extension $E/F$ we have $|Aut(E/F)|\le [E:F]$ and here there is equality if and only if $E/F$ is Galois. Therefore we can conclude that $E/(K\cap L)$ is Galois, and $Aut(E/K\cap L)=G\cdot H$. In particular, $G\cdot H$ is a group.

prove $[E:K\cap L]=[E:K][E:L]$

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