Galois group of $X^5-5$
$begingroup$
I am trying to calculate this galois group of $x^5-5$ over $mathbb Q$. I know that there is a tower of extensions with groups $C_4$ and $C_5$ respectively, so the group is order 20. I am guessing it isn't $C_{20}$ though. Could I have hint?
galois-theory
asked Dec 4 '18 at 14:08
DeviloDevilo
11718
$endgroup$
add a comment |
$begingroup$
I am trying to calculate this galois group of $x^5-5$ over $mathbb Q$. I know that there is a tower of extensions with groups $C_4$ and $C_5$ respectively, so the group is order 20. I am guessing it isn't $C_{20}$ though. Could I have hint?
galois-theory
asked Dec 4 '18 at 14:08
DeviloDevilo
11718
$endgroup$
$begingroup$
The unique (transitive) subgroup of order $20$ in $S_5$ is the general affine group $AGL(1,mathbb{F}_5)$.
$endgroup$
– user10354138
Dec 4 '18 at 18:27
add a comment |
$begingroup$
I am trying to calculate this galois group of $x^5-5$ over $mathbb Q$. I know that there is a tower of extensions with groups $C_4$ and $C_5$ respectively, so the group is order 20. I am guessing it isn't $C_{20}$ though. Could I have hint?
galois-theory
asked Dec 4 '18 at 14:08
DeviloDevilo
11718
$endgroup$
I am trying to calculate this galois group of $x^5-5$ over $mathbb Q$. I know that there is a tower of extensions with groups $C_4$ and $C_5$ respectively, so the group is order 20. I am guessing it isn't $C_{20}$ though. Could I have hint?
galois-theory
galois-theory
asked Dec 4 '18 at 14:08
DeviloDevilo
11718
asked Dec 4 '18 at 14:08
DeviloDevilo
11718
edited Dec 4 '18 at 16:59
Devilo
asked Dec 4 '18 at 14:08
DeviloDevilo
11718
asked Dec 4 '18 at 14:08
DeviloDevilo
11718
asked Dec 4 '18 at 14:08
DeviloDevilo
11718
11718
$begingroup$
The unique (transitive) subgroup of order $20$ in $S_5$ is the general affine group $AGL(1,mathbb{F}_5)$.
$endgroup$
– user10354138
Dec 4 '18 at 18:27
add a comment |
$begingroup$
The unique (transitive) subgroup of order $20$ in $S_5$ is the general affine group $AGL(1,mathbb{F}_5)$.
$endgroup$
– user10354138
Dec 4 '18 at 18:27
$begingroup$
The unique (transitive) subgroup of order $20$ in $S_5$ is the general affine group $AGL(1,mathbb{F}_5)$.
$endgroup$
– user10354138
Dec 4 '18 at 18:27
$begingroup$
The unique (transitive) subgroup of order $20$ in $S_5$ is the general affine group $AGL(1,mathbb{F}_5)$.
$endgroup$
– user10354138
Dec 4 '18 at 18:27
add a comment |
1 Answer
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$begingroup$
Note right away that your Galois group $G$ cannot be $C_{20}$ because $G$ is not abelian (the extension $Q(sqrt [5] 5)/Q$ is not normal). To determine $G$, allow me a development which could be used in a more general situation.
Here, because 4 and 5 are coprime, an extension $G$ of $C_4$ by $C_5$ is necessarily split, i.e. $G$ is a semi-direct product (this is a classical theorem in group theory). So $G$ admits a subgroup $Hcong C_4$ (here the subgroup fixing $Q(sqrt [5] 5)$), and you can describe $G$ explicitly by following the action on the roots of $X^5-5$ of $H$ and of the subgroup fixing $Q(zeta_5)$.
If you are interested only in the isomorphism type of $G$, without any calculation, you need first to determine the action of $C_4$ on $C_5$. Such an action is determined by some $theta in Hom (C_4, Aut(C_5))cong Hom (C_4, C_4)$ (because $Aut(C_5)$ is cyclic of order 4). Fixing a generator $C$ of $C_4$ and writing $theta (c)=c^k,k=1,2,3$ (exclude $0$ because $G$ is not abelian), you get 3 possible isomorphic types for $G$, which are the dihedral $D_{20}$, the dicyclic $Dic_{20}$ and the general affine $GA(1,5)$, see https://groupprops.subwiki.org/wiki/Groups_of_order_20. Here the Galois action easily determines $k$ and shows that $Gcong GA(1,5)$. See also the transitivity argument given by @user10354138 .
answered Dec 6 '18 at 11:26
nguyen quang donguyen quang do
8,9491724
$endgroup$
add a comment |
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$begingroup$
Note right away that your Galois group $G$ cannot be $C_{20}$ because $G$ is not abelian (the extension $Q(sqrt [5] 5)/Q$ is not normal). To determine $G$, allow me a development which could be used in a more general situation.
Here, because 4 and 5 are coprime, an extension $G$ of $C_4$ by $C_5$ is necessarily split, i.e. $G$ is a semi-direct product (this is a classical theorem in group theory). So $G$ admits a subgroup $Hcong C_4$ (here the subgroup fixing $Q(sqrt [5] 5)$), and you can describe $G$ explicitly by following the action on the roots of $X^5-5$ of $H$ and of the subgroup fixing $Q(zeta_5)$.
If you are interested only in the isomorphism type of $G$, without any calculation, you need first to determine the action of $C_4$ on $C_5$. Such an action is determined by some $theta in Hom (C_4, Aut(C_5))cong Hom (C_4, C_4)$ (because $Aut(C_5)$ is cyclic of order 4). Fixing a generator $C$ of $C_4$ and writing $theta (c)=c^k,k=1,2,3$ (exclude $0$ because $G$ is not abelian), you get 3 possible isomorphic types for $G$, which are the dihedral $D_{20}$, the dicyclic $Dic_{20}$ and the general affine $GA(1,5)$, see https://groupprops.subwiki.org/wiki/Groups_of_order_20. Here the Galois action easily determines $k$ and shows that $Gcong GA(1,5)$. See also the transitivity argument given by @user10354138 .
answered Dec 6 '18 at 11:26
nguyen quang donguyen quang do
8,9491724
$endgroup$
add a comment |
$begingroup$
Note right away that your Galois group $G$ cannot be $C_{20}$ because $G$ is not abelian (the extension $Q(sqrt [5] 5)/Q$ is not normal). To determine $G$, allow me a development which could be used in a more general situation.
Here, because 4 and 5 are coprime, an extension $G$ of $C_4$ by $C_5$ is necessarily split, i.e. $G$ is a semi-direct product (this is a classical theorem in group theory). So $G$ admits a subgroup $Hcong C_4$ (here the subgroup fixing $Q(sqrt [5] 5)$), and you can describe $G$ explicitly by following the action on the roots of $X^5-5$ of $H$ and of the subgroup fixing $Q(zeta_5)$.
If you are interested only in the isomorphism type of $G$, without any calculation, you need first to determine the action of $C_4$ on $C_5$. Such an action is determined by some $theta in Hom (C_4, Aut(C_5))cong Hom (C_4, C_4)$ (because $Aut(C_5)$ is cyclic of order 4). Fixing a generator $C$ of $C_4$ and writing $theta (c)=c^k,k=1,2,3$ (exclude $0$ because $G$ is not abelian), you get 3 possible isomorphic types for $G$, which are the dihedral $D_{20}$, the dicyclic $Dic_{20}$ and the general affine $GA(1,5)$, see https://groupprops.subwiki.org/wiki/Groups_of_order_20. Here the Galois action easily determines $k$ and shows that $Gcong GA(1,5)$. See also the transitivity argument given by @user10354138 .
answered Dec 6 '18 at 11:26
nguyen quang donguyen quang do
8,9491724
$endgroup$
add a comment |
$begingroup$
Note right away that your Galois group $G$ cannot be $C_{20}$ because $G$ is not abelian (the extension $Q(sqrt [5] 5)/Q$ is not normal). To determine $G$, allow me a development which could be used in a more general situation.
Here, because 4 and 5 are coprime, an extension $G$ of $C_4$ by $C_5$ is necessarily split, i.e. $G$ is a semi-direct product (this is a classical theorem in group theory). So $G$ admits a subgroup $Hcong C_4$ (here the subgroup fixing $Q(sqrt [5] 5)$), and you can describe $G$ explicitly by following the action on the roots of $X^5-5$ of $H$ and of the subgroup fixing $Q(zeta_5)$.
If you are interested only in the isomorphism type of $G$, without any calculation, you need first to determine the action of $C_4$ on $C_5$. Such an action is determined by some $theta in Hom (C_4, Aut(C_5))cong Hom (C_4, C_4)$ (because $Aut(C_5)$ is cyclic of order 4). Fixing a generator $C$ of $C_4$ and writing $theta (c)=c^k,k=1,2,3$ (exclude $0$ because $G$ is not abelian), you get 3 possible isomorphic types for $G$, which are the dihedral $D_{20}$, the dicyclic $Dic_{20}$ and the general affine $GA(1,5)$, see https://groupprops.subwiki.org/wiki/Groups_of_order_20. Here the Galois action easily determines $k$ and shows that $Gcong GA(1,5)$. See also the transitivity argument given by @user10354138 .
answered Dec 6 '18 at 11:26
nguyen quang donguyen quang do
8,9491724
$endgroup$
Note right away that your Galois group $G$ cannot be $C_{20}$ because $G$ is not abelian (the extension $Q(sqrt [5] 5)/Q$ is not normal). To determine $G$, allow me a development which could be used in a more general situation.
Here, because 4 and 5 are coprime, an extension $G$ of $C_4$ by $C_5$ is necessarily split, i.e. $G$ is a semi-direct product (this is a classical theorem in group theory). So $G$ admits a subgroup $Hcong C_4$ (here the subgroup fixing $Q(sqrt [5] 5)$), and you can describe $G$ explicitly by following the action on the roots of $X^5-5$ of $H$ and of the subgroup fixing $Q(zeta_5)$.
If you are interested only in the isomorphism type of $G$, without any calculation, you need first to determine the action of $C_4$ on $C_5$. Such an action is determined by some $theta in Hom (C_4, Aut(C_5))cong Hom (C_4, C_4)$ (because $Aut(C_5)$ is cyclic of order 4). Fixing a generator $C$ of $C_4$ and writing $theta (c)=c^k,k=1,2,3$ (exclude $0$ because $G$ is not abelian), you get 3 possible isomorphic types for $G$, which are the dihedral $D_{20}$, the dicyclic $Dic_{20}$ and the general affine $GA(1,5)$, see https://groupprops.subwiki.org/wiki/Groups_of_order_20. Here the Galois action easily determines $k$ and shows that $Gcong GA(1,5)$. See also the transitivity argument given by @user10354138 .
answered Dec 6 '18 at 11:26
nguyen quang donguyen quang do
8,9491724
edited Dec 6 '18 at 13:04
answered Dec 6 '18 at 11:26
nguyen quang donguyen quang do
8,9491724
answered Dec 6 '18 at 11:26
nguyen quang donguyen quang do
8,9491724
answered Dec 6 '18 at 11:26
nguyen quang donguyen quang do
8,9491724
8,9491724
add a comment |
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$begingroup$
The unique (transitive) subgroup of order $20$ in $S_5$ is the general affine group $AGL(1,mathbb{F}_5)$.
$endgroup$
– user10354138
Dec 4 '18 at 18:27