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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?

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    $\begingroup$ 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, 2014 at 13:30
  • $\begingroup$ $L_1$ is not Galois since its not normal Dustan. $\endgroup$
    – Raxel
    May 30, 2014 at 13:38
  • $\begingroup$ @DustanLevenstein, what's the problem? In that case $\;L_1/\Bbb Q\;$ is not Galois... $\endgroup$
    – DonAntonio
    May 30, 2014 at 13:38
  • $\begingroup$ Thanks, I missed that part of the assertion. $\endgroup$
    – Dustan Levenstein
    May 30, 2014 at 13:39
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    $\begingroup$ 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, 2014 at 14:10

1 Answer 1

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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:

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

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  • $\begingroup$ I'm not familiar with that reversed play symbol. What does it mean in words? $\endgroup$
    – Raxel
    May 30, 2014 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, 2014 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, 2014 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, 2014 at 13:45
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    $\begingroup$ 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, 2014 at 14:07

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