Proving a subgroup of a Galois group is normal












1














"Let $N/K$ be a finite Galois extension with Galois group $G =
Gal(N/K)$.



Let $M$ be an intermediate field of the extension, let
$E = Gal(N/M)$, and let $F = cap_{sigma in G} σEσ^{-1}$
.



Show that $F$ is a normal subgroup of $G$"



Where do I even begin here?



I know for all $sigma in G$, $F subset sigma E sigma ^{-1}$, so $sigma ^{-1} F sigma subset E$, so $F$ is a subgroup of $E$ under conjugation? Is this true?



Now how do I prove $F$ is normal?






group-theory galois-theory galois-extensions







share|cite|improve this question
























  • Have you shown that $F$ is normal in $G$?
    – rogerl
    Nov 27 '18 at 22:08










  • No, thats the bit I am struggling most with
    – Dino
    Nov 28 '18 at 5:10
















1














"Let $N/K$ be a finite Galois extension with Galois group $G =
Gal(N/K)$.



Let $M$ be an intermediate field of the extension, let
$E = Gal(N/M)$, and let $F = cap_{sigma in G} σEσ^{-1}$
.



Show that $F$ is a normal subgroup of $G$"



Where do I even begin here?



I know for all $sigma in G$, $F subset sigma E sigma ^{-1}$, so $sigma ^{-1} F sigma subset E$, so $F$ is a subgroup of $E$ under conjugation? Is this true?



Now how do I prove $F$ is normal?






group-theory galois-theory galois-extensions







share|cite|improve this question
























  • Have you shown that $F$ is normal in $G$?
    – rogerl
    Nov 27 '18 at 22:08










  • No, thats the bit I am struggling most with
    – Dino
    Nov 28 '18 at 5:10














1












1








1







"Let $N/K$ be a finite Galois extension with Galois group $G =
Gal(N/K)$.



Let $M$ be an intermediate field of the extension, let
$E = Gal(N/M)$, and let $F = cap_{sigma in G} σEσ^{-1}$
.



Show that $F$ is a normal subgroup of $G$"



Where do I even begin here?



I know for all $sigma in G$, $F subset sigma E sigma ^{-1}$, so $sigma ^{-1} F sigma subset E$, so $F$ is a subgroup of $E$ under conjugation? Is this true?



Now how do I prove $F$ is normal?






group-theory galois-theory galois-extensions







share|cite|improve this question















"Let $N/K$ be a finite Galois extension with Galois group $G =
Gal(N/K)$.



Let $M$ be an intermediate field of the extension, let
$E = Gal(N/M)$, and let $F = cap_{sigma in G} σEσ^{-1}$
.



Show that $F$ is a normal subgroup of $G$"



Where do I even begin here?



I know for all $sigma in G$, $F subset sigma E sigma ^{-1}$, so $sigma ^{-1} F sigma subset E$, so $F$ is a subgroup of $E$ under conjugation? Is this true?



Now how do I prove $F$ is normal?






group-theory galois-theory galois-extensions




group-theory galois-theory galois-extensions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 28 '18 at 5:02

























asked Nov 27 '18 at 21:50









Dino

836




836












  • Have you shown that $F$ is normal in $G$?
    – rogerl
    Nov 27 '18 at 22:08










  • No, thats the bit I am struggling most with
    – Dino
    Nov 28 '18 at 5:10


















  • Have you shown that $F$ is normal in $G$?
    – rogerl
    Nov 27 '18 at 22:08










  • No, thats the bit I am struggling most with
    – Dino
    Nov 28 '18 at 5:10
















Have you shown that $F$ is normal in $G$?
– rogerl
Nov 27 '18 at 22:08




Have you shown that $F$ is normal in $G$?
– rogerl
Nov 27 '18 at 22:08












No, thats the bit I am struggling most with
– Dino
Nov 28 '18 at 5:10




No, thats the bit I am struggling most with
– Dino
Nov 28 '18 at 5:10










1 Answer
1






active

oldest

votes


















3














This is just an exercise n group theory: take a group $;G;$ , a subgroup $;Hle G;$ and form what's called the core of $;H;$ , merely: the subgroup $;bigcap_{xin G}xHx^{-1};$ . You can read about this in the web, but we can also do as follows:



Let $;X:=Hbackslash G;$ be the set of left cosets of $;H;$ in $;G;$ , and define an action of $;G;$ on $;X;$ as follows: $;gcdot xH:=(gx)H;$ . It's easy to check this is indeed an action. As usual, this action determines a homomorphism $;phi:Gto Sym_Xcong S_n;$ , when $;n=[G:H];$ , by the rule $;phi g(xH):=;(gx)H$ (you only have to convince yourself that the above actually determines a permutation on $;Sym_X;$ for any $;gin G$ ).



Well, it is now a nice exercise to prove:



$$begin{align*}(1);;&kerphi=bigcap_{xin G}xHx^{-1}\{}\(2);; &kerphi;;text{is the maximal normal subgroup of $;G;$ contained in $H$}end{align*}$$



Give a try to the above two claims






share|cite|improve this answer























  • Let's assume we have no idea what the core of H is, so cannot use the other definitions of it. How would go about proving the question?
    – Dino
    Nov 28 '18 at 5:09






  • 1




    @Dino Exactly as proposed above: show it is the kernel of a homomorphism and voila: you have it is normal! To show it is contained in $;H;$ is very easy (apply one element of the kernel to the left coset $;H;$ in $;X;$ ...)
    – DonAntonio
    Nov 28 '18 at 8:21













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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3














This is just an exercise n group theory: take a group $;G;$ , a subgroup $;Hle G;$ and form what's called the core of $;H;$ , merely: the subgroup $;bigcap_{xin G}xHx^{-1};$ . You can read about this in the web, but we can also do as follows:



Let $;X:=Hbackslash G;$ be the set of left cosets of $;H;$ in $;G;$ , and define an action of $;G;$ on $;X;$ as follows: $;gcdot xH:=(gx)H;$ . It's easy to check this is indeed an action. As usual, this action determines a homomorphism $;phi:Gto Sym_Xcong S_n;$ , when $;n=[G:H];$ , by the rule $;phi g(xH):=;(gx)H$ (you only have to convince yourself that the above actually determines a permutation on $;Sym_X;$ for any $;gin G$ ).



Well, it is now a nice exercise to prove:



$$begin{align*}(1);;&kerphi=bigcap_{xin G}xHx^{-1}\{}\(2);; &kerphi;;text{is the maximal normal subgroup of $;G;$ contained in $H$}end{align*}$$



Give a try to the above two claims






share|cite|improve this answer























  • Let's assume we have no idea what the core of H is, so cannot use the other definitions of it. How would go about proving the question?
    – Dino
    Nov 28 '18 at 5:09






  • 1




    @Dino Exactly as proposed above: show it is the kernel of a homomorphism and voila: you have it is normal! To show it is contained in $;H;$ is very easy (apply one element of the kernel to the left coset $;H;$ in $;X;$ ...)
    – DonAntonio
    Nov 28 '18 at 8:21


















3














This is just an exercise n group theory: take a group $;G;$ , a subgroup $;Hle G;$ and form what's called the core of $;H;$ , merely: the subgroup $;bigcap_{xin G}xHx^{-1};$ . You can read about this in the web, but we can also do as follows:



Let $;X:=Hbackslash G;$ be the set of left cosets of $;H;$ in $;G;$ , and define an action of $;G;$ on $;X;$ as follows: $;gcdot xH:=(gx)H;$ . It's easy to check this is indeed an action. As usual, this action determines a homomorphism $;phi:Gto Sym_Xcong S_n;$ , when $;n=[G:H];$ , by the rule $;phi g(xH):=;(gx)H$ (you only have to convince yourself that the above actually determines a permutation on $;Sym_X;$ for any $;gin G$ ).



Well, it is now a nice exercise to prove:



$$begin{align*}(1);;&kerphi=bigcap_{xin G}xHx^{-1}\{}\(2);; &kerphi;;text{is the maximal normal subgroup of $;G;$ contained in $H$}end{align*}$$



Give a try to the above two claims






share|cite|improve this answer























  • Let's assume we have no idea what the core of H is, so cannot use the other definitions of it. How would go about proving the question?
    – Dino
    Nov 28 '18 at 5:09






  • 1




    @Dino Exactly as proposed above: show it is the kernel of a homomorphism and voila: you have it is normal! To show it is contained in $;H;$ is very easy (apply one element of the kernel to the left coset $;H;$ in $;X;$ ...)
    – DonAntonio
    Nov 28 '18 at 8:21
















3












3








3






This is just an exercise n group theory: take a group $;G;$ , a subgroup $;Hle G;$ and form what's called the core of $;H;$ , merely: the subgroup $;bigcap_{xin G}xHx^{-1};$ . You can read about this in the web, but we can also do as follows:



Let $;X:=Hbackslash G;$ be the set of left cosets of $;H;$ in $;G;$ , and define an action of $;G;$ on $;X;$ as follows: $;gcdot xH:=(gx)H;$ . It's easy to check this is indeed an action. As usual, this action determines a homomorphism $;phi:Gto Sym_Xcong S_n;$ , when $;n=[G:H];$ , by the rule $;phi g(xH):=;(gx)H$ (you only have to convince yourself that the above actually determines a permutation on $;Sym_X;$ for any $;gin G$ ).



Well, it is now a nice exercise to prove:



$$begin{align*}(1);;&kerphi=bigcap_{xin G}xHx^{-1}\{}\(2);; &kerphi;;text{is the maximal normal subgroup of $;G;$ contained in $H$}end{align*}$$



Give a try to the above two claims






share|cite|improve this answer














This is just an exercise n group theory: take a group $;G;$ , a subgroup $;Hle G;$ and form what's called the core of $;H;$ , merely: the subgroup $;bigcap_{xin G}xHx^{-1};$ . You can read about this in the web, but we can also do as follows:



Let $;X:=Hbackslash G;$ be the set of left cosets of $;H;$ in $;G;$ , and define an action of $;G;$ on $;X;$ as follows: $;gcdot xH:=(gx)H;$ . It's easy to check this is indeed an action. As usual, this action determines a homomorphism $;phi:Gto Sym_Xcong S_n;$ , when $;n=[G:H];$ , by the rule $;phi g(xH):=;(gx)H$ (you only have to convince yourself that the above actually determines a permutation on $;Sym_X;$ for any $;gin G$ ).



Well, it is now a nice exercise to prove:



$$begin{align*}(1);;&kerphi=bigcap_{xin G}xHx^{-1}\{}\(2);; &kerphi;;text{is the maximal normal subgroup of $;G;$ contained in $H$}end{align*}$$



Give a try to the above two claims







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 28 '18 at 8:21

























answered Nov 27 '18 at 22:10









DonAntonio

177k1491225




177k1491225












  • Let's assume we have no idea what the core of H is, so cannot use the other definitions of it. How would go about proving the question?
    – Dino
    Nov 28 '18 at 5:09






  • 1




    @Dino Exactly as proposed above: show it is the kernel of a homomorphism and voila: you have it is normal! To show it is contained in $;H;$ is very easy (apply one element of the kernel to the left coset $;H;$ in $;X;$ ...)
    – DonAntonio
    Nov 28 '18 at 8:21




















  • Let's assume we have no idea what the core of H is, so cannot use the other definitions of it. How would go about proving the question?
    – Dino
    Nov 28 '18 at 5:09






  • 1




    @Dino Exactly as proposed above: show it is the kernel of a homomorphism and voila: you have it is normal! To show it is contained in $;H;$ is very easy (apply one element of the kernel to the left coset $;H;$ in $;X;$ ...)
    – DonAntonio
    Nov 28 '18 at 8:21


















Let's assume we have no idea what the core of H is, so cannot use the other definitions of it. How would go about proving the question?
– Dino
Nov 28 '18 at 5:09




Let's assume we have no idea what the core of H is, so cannot use the other definitions of it. How would go about proving the question?
– Dino
Nov 28 '18 at 5:09




1




1




@Dino Exactly as proposed above: show it is the kernel of a homomorphism and voila: you have it is normal! To show it is contained in $;H;$ is very easy (apply one element of the kernel to the left coset $;H;$ in $;X;$ ...)
– DonAntonio
Nov 28 '18 at 8:21






@Dino Exactly as proposed above: show it is the kernel of a homomorphism and voila: you have it is normal! To show it is contained in $;H;$ is very easy (apply one element of the kernel to the left coset $;H;$ in $;X;$ ...)
– DonAntonio
Nov 28 '18 at 8:21




















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