1
$\begingroup$

I have the following exercise:

$L$ is a finite Galois extension of $K$ with Galois group $G$ (that is $G=Gal(L/K)$). Suppose $L_1$ and $L_2$ are subextensions and $G_1$ and $G_2$ are the respective subgroups of $G$. Show that $G$ is a direct product of $G_1$ and $G_2$ iff $L_1$ and $L_2$ are Galois extensions of $K$ s.t. $L_1 L_2 =L$ and $L_1 \cap L_2 =K$

What I am just wondering is the respective subgroups $G_1$ and $G_2$. If $G_1$ was Galois extension is $G_1=\operatorname{Gal}(L_1 /K)$ or is $G_1 =\operatorname{Gal}(L/L_1)$? I guessed it was the latter based on theorems for example the fundamental theorem of Galois extensions in the finite case but based on this. And is true that $G_1$ and $G_2$ are in general Galois extensions?

$\endgroup$
  • 1
    $\begingroup$ The latter, yes. And $L/L_i$ is always a Galois extension. It's $L_i/K$ that may not be a Galois extension. $\endgroup$ – Dustan Levenstein May 30 '14 at 13:30
  • $\begingroup$ $L_1$ is not Galois since its not normal Dustan. $\endgroup$ – Raxel May 30 '14 at 13:38
  • $\begingroup$ @DustanLevenstein, what's the problem? In that case $\;L_1/\Bbb Q\;$ is not Galois... $\endgroup$ – DonAntonio May 30 '14 at 13:38
  • $\begingroup$ Thanks, I missed that part of the assertion. $\endgroup$ – Dustan Levenstein May 30 '14 at 13:39
  • 1
    $\begingroup$ Yes. There's no conflict with the result you linked to, because when $G$ is the direct product of $G_1$ and $G_2$, we have $G_1 \simeq G/G_2 = \operatorname{Gal}(L_2/K)$ and $G_2 \simeq G/G_1 = \operatorname{Gal}(L_1/K)$. $\endgroup$ – Dustan Levenstein May 30 '14 at 14:10
1
$\begingroup$

Hints (and you try to connect the dots): If $\;G\;$ is a group and $\;N_1\,,\,N_2\le G\;$ , then:

$$\begin{align*}\bullet&\;\;G=N_1\times N_2\iff\begin{cases}N_1,N_2\lhd G\\{}\\N_1N_2=G\\{}\\N_1\cap N_2=1\end{cases}\end{align*}$$

Now, using the Galois correspondence and theorems around this, and using your notation and your assumptions:

$$\begin{align*}\bullet&\;\;G=G_1G_2\iff L_1L_2=L\\{}\\\bullet&\;\;G_i\lhd G\iff L_i/K\;\;\text{is a normal extension}\iff L_i/K\;\;\text{is Galois}\\{}\\\bullet&\;\;G_1\cap G_2=1\iff L_1\cap L_2=K\;\text{(hint: what extension fits to the trivial sbgp.?)}\end{align*}$$

$\endgroup$
  • $\begingroup$ I'm not familiar with that reversed play symbol. What does it mean in words? $\endgroup$ – Raxel May 30 '14 at 13:40
  • $\begingroup$ @Raxel?? Do you mean $\;\lhd\;$ ? But that is the international symbol for normal subgroup: you must know basic group theory before engaging into Galois theory! $\endgroup$ – DonAntonio May 30 '14 at 13:41
  • $\begingroup$ Well my lecture notes usually uses just words for it and Gallian does not use this symbol. $\endgroup$ – Raxel May 30 '14 at 13:42
  • $\begingroup$ I don't know what Gallian book you're talking about, @Raxel, but if it is "Contemporary Abstract Algebra" he does use that symbol: 2nd edition, page 145, chapter 10. I really don't know any book in any language and after 1960 that doesn't use that symbol...not to mention that in a class at 2nd-3rd undergraduate level one usually knows already several books on some given subject. $\endgroup$ – DonAntonio May 30 '14 at 13:45
  • 1
    $\begingroup$ Well, now you know one rather important symbol used, and it appears in each and every book in abstract algebra/ group theory as far as I recall, and it makes things way shorter and, imo, also simpler. Anyway, the answer's there and it doesn't change. $\endgroup$ – DonAntonio May 30 '14 at 14:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.