Relation between a finite group and the Galois group for the field extension generated by the character table entries Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)To prove that the sum of the roots of the characteristic polynomial of a square matrix is equal to the trace of the matrixAn explicit calculation of Galois groupLink between representation theory and Galois theory: Trivial representation in field towers.Systematically describing the Galois Group and Intermediate FieldsField extension $mathbb Q(f)/mathbb Q$ and its Galois groupAutomorphism of $mathbbQ(zeta_n)/mathbbQ$Galois group of the extension $E:= mathbbQ(i, sqrt2, sqrt3, sqrt[4]2)$Solving the Character table for $A_4$ and derived algebra characteristics.Constructing semidirect product out of finite fields and Galois groups and the permutation groups they induceQuestion about Galois Extension and fixed field by a Subgroup of the Galois groupWhat is a primitive character of a Galois groups of a finite cyclic extension of local fields?
Is 1 ppb equal to 1 μg/kg?
Check which numbers satisfy the condition [A*B*C = A! + B! + C!]
Should I discuss the type of campaign with my players?
When is phishing education going too far?
If 'B is more likely given A', then 'A is more likely given B'
Is above average number of years spent on PhD considered a red flag in future academia or industry positions?
Should gear shift center itself while in neutral?
List *all* the tuples!
How can I fade player when goes inside or outside of the area?
Why don't the Weasley twins use magic outside of school if the Trace can only find the location of spells cast?
Is there a Spanish version of "dot your i's and cross your t's" that includes the letter 'ñ'?
IndentationError when pasting code in Python 3 interpreter mode
Is there a "higher Segal conjecture"?
What is the longest distance a 13th-level monk can jump while attacking on the same turn?
How does cp -a work
Examples of mediopassive verb constructions
Right-skewed distribution with mean equals to mode?
"Seemed to had" is it correct?
Is it true that "carbohydrates are of no use for the basal metabolic need"?
Using et al. for a last / senior author rather than for a first author
What does the "x" in "x86" represent?
Antler Helmet: Can it work?
Do you forfeit tax refunds/credits if you aren't required to and don't file by April 15?
Models of set theory where not every set can be linearly ordered
Relation between a finite group and the Galois group for the field extension generated by the character table entries
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)To prove that the sum of the roots of the characteristic polynomial of a square matrix is equal to the trace of the matrixAn explicit calculation of Galois groupLink between representation theory and Galois theory: Trivial representation in field towers.Systematically describing the Galois Group and Intermediate FieldsField extension $mathbb Q(f)/mathbb Q$ and its Galois groupAutomorphism of $mathbbQ(zeta_n)/mathbbQ$Galois group of the extension $E:= mathbbQ(i, sqrt2, sqrt3, sqrt[4]2)$Solving the Character table for $A_4$ and derived algebra characteristics.Constructing semidirect product out of finite fields and Galois groups and the permutation groups they induceQuestion about Galois Extension and fixed field by a Subgroup of the Galois groupWhat is a primitive character of a Galois groups of a finite cyclic extension of local fields?
$begingroup$
Let $E$ be the extension of $mathbbQ$ generated by the character table entries of a finite group $G$.
Let $F$ be the Galois closure of $E$.
Examples:
- $G=C_n$, $E=F=mathbbQ(zeta_n)$ and $Gal(F/mathbbQ) simeq (mathbbZ/nmathbbZ)^times simeq Out(G)=Aut(G)$,
- $G=S_n$, $E=F=mathbbQ$ and so $C_1 = Gal(F/mathbbQ) hookrightarrow Out(G)$ and $Aut(G)$,
- $G=A_n$ with $3 le n le 5$, $E=F$ and $Gal(F/mathbbQ) simeq Out(G) simeq C_2 hookrightarrow Aut(G)$,
- $G=A_6$, $E=F=mathbbQ(sqrt5)$ and $Gal(F/mathbbQ) simeq C_2 hookrightarrow Out(G)simeq C_2 times C_2 hookrightarrow Aut(G)$.
Question: Is it true in general that $Gal(F/mathbbQ)$ or $Aut(E/mathbbQ) hookrightarrow Out(G)$ or $Aut(G)$?
If not, is there at least a non-trivial homomorphism from $Gal(F/mathbbQ)$ to $Aut(G)$? Or any relation?
Bonus question: Is there an example with $E subsetneq F$?
group-theory finite-groups representation-theory galois-theory characters
asked Mar 25 at 20:01
Sebastien PalcouxSebastien Palcoux
2,336928
$endgroup$
|
show 7 more comments
$begingroup$
Let $E$ be the extension of $mathbbQ$ generated by the character table entries of a finite group $G$.
Let $F$ be the Galois closure of $E$.
Examples:
- $G=C_n$, $E=F=mathbbQ(zeta_n)$ and $Gal(F/mathbbQ) simeq (mathbbZ/nmathbbZ)^times simeq Out(G)=Aut(G)$,
- $G=S_n$, $E=F=mathbbQ$ and so $C_1 = Gal(F/mathbbQ) hookrightarrow Out(G)$ and $Aut(G)$,
- $G=A_n$ with $3 le n le 5$, $E=F$ and $Gal(F/mathbbQ) simeq Out(G) simeq C_2 hookrightarrow Aut(G)$,
- $G=A_6$, $E=F=mathbbQ(sqrt5)$ and $Gal(F/mathbbQ) simeq C_2 hookrightarrow Out(G)simeq C_2 times C_2 hookrightarrow Aut(G)$.
Question: Is it true in general that $Gal(F/mathbbQ)$ or $Aut(E/mathbbQ) hookrightarrow Out(G)$ or $Aut(G)$?
If not, is there at least a non-trivial homomorphism from $Gal(F/mathbbQ)$ to $Aut(G)$? Or any relation?
Bonus question: Is there an example with $E subsetneq F$?
group-theory finite-groups representation-theory galois-theory characters
asked Mar 25 at 20:01
Sebastien PalcouxSebastien Palcoux
2,336928
$endgroup$
1
$begingroup$
$n=|G|$ an homomorphism $rho : G to GL_m(BbbC)$ and $det(xI-rho(g))= prod_j (x-alpha_g,j)$ then $rho(g)^n = I$ implies the minimal polynomial divides $x^n-1$ so that $alpha_g,j = zeta_n^e_g,j$ and $Tr(rho(g)) = sum_j alpha_g,j in BbbQ(zeta_n)$ and $E/BbbQ$ is abelian
$endgroup$
– reuns
Mar 25 at 21:15
$begingroup$
@reuns: Can we deduce that $Gal(E/mathbbQ)$ is a subgroup of $G$, or something like that?
$endgroup$
– Sebastien Palcoux
Mar 25 at 21:22
$begingroup$
No, it is a quotient (thus a subgroup since we are in finite abelian group) of $BbbZ/nZ^times = Gal(BbbQ(zeta_n)/Q)$ and $BbbZ/e Z^times$ where $e$ is the exponent of $G$. Also $E$ is called the minimal splitting field
$endgroup$
– reuns
Mar 25 at 21:25
$begingroup$
The character table of dihedral group $D_6$ non-abelian with $6$ elements is integer valued so you can't deduce $E$ only from $n$ or $e$
$endgroup$
– reuns
Mar 25 at 21:40
$begingroup$
There are nonisomorphic groups with the same characters so you can't hope to deduce $G$ from $E$ or $F$
$endgroup$
– Max
Mar 25 at 21:45
|
show 7 more comments
$begingroup$
Let $E$ be the extension of $mathbbQ$ generated by the character table entries of a finite group $G$.
Let $F$ be the Galois closure of $E$.
Examples:
- $G=C_n$, $E=F=mathbbQ(zeta_n)$ and $Gal(F/mathbbQ) simeq (mathbbZ/nmathbbZ)^times simeq Out(G)=Aut(G)$,
- $G=S_n$, $E=F=mathbbQ$ and so $C_1 = Gal(F/mathbbQ) hookrightarrow Out(G)$ and $Aut(G)$,
- $G=A_n$ with $3 le n le 5$, $E=F$ and $Gal(F/mathbbQ) simeq Out(G) simeq C_2 hookrightarrow Aut(G)$,
- $G=A_6$, $E=F=mathbbQ(sqrt5)$ and $Gal(F/mathbbQ) simeq C_2 hookrightarrow Out(G)simeq C_2 times C_2 hookrightarrow Aut(G)$.
Question: Is it true in general that $Gal(F/mathbbQ)$ or $Aut(E/mathbbQ) hookrightarrow Out(G)$ or $Aut(G)$?
If not, is there at least a non-trivial homomorphism from $Gal(F/mathbbQ)$ to $Aut(G)$? Or any relation?
Bonus question: Is there an example with $E subsetneq F$?
group-theory finite-groups representation-theory galois-theory characters
asked Mar 25 at 20:01
Sebastien PalcouxSebastien Palcoux
2,336928
$endgroup$
Let $E$ be the extension of $mathbbQ$ generated by the character table entries of a finite group $G$.
Let $F$ be the Galois closure of $E$.
Examples:
- $G=C_n$, $E=F=mathbbQ(zeta_n)$ and $Gal(F/mathbbQ) simeq (mathbbZ/nmathbbZ)^times simeq Out(G)=Aut(G)$,
- $G=S_n$, $E=F=mathbbQ$ and so $C_1 = Gal(F/mathbbQ) hookrightarrow Out(G)$ and $Aut(G)$,
- $G=A_n$ with $3 le n le 5$, $E=F$ and $Gal(F/mathbbQ) simeq Out(G) simeq C_2 hookrightarrow Aut(G)$,
- $G=A_6$, $E=F=mathbbQ(sqrt5)$ and $Gal(F/mathbbQ) simeq C_2 hookrightarrow Out(G)simeq C_2 times C_2 hookrightarrow Aut(G)$.
Question: Is it true in general that $Gal(F/mathbbQ)$ or $Aut(E/mathbbQ) hookrightarrow Out(G)$ or $Aut(G)$?
If not, is there at least a non-trivial homomorphism from $Gal(F/mathbbQ)$ to $Aut(G)$? Or any relation?
Bonus question: Is there an example with $E subsetneq F$?
group-theory finite-groups representation-theory galois-theory characters
group-theory finite-groups representation-theory galois-theory characters
asked Mar 25 at 20:01
Sebastien PalcouxSebastien Palcoux
2,336928
asked Mar 25 at 20:01
Sebastien PalcouxSebastien Palcoux
2,336928
edited Mar 27 at 7:37
Sebastien Palcoux
asked Mar 25 at 20:01
Sebastien PalcouxSebastien Palcoux
2,336928
asked Mar 25 at 20:01
Sebastien PalcouxSebastien Palcoux
2,336928
asked Mar 25 at 20:01
Sebastien PalcouxSebastien Palcoux
2,336928
2,336928
1
$begingroup$
$n=|G|$ an homomorphism $rho : G to GL_m(BbbC)$ and $det(xI-rho(g))= prod_j (x-alpha_g,j)$ then $rho(g)^n = I$ implies the minimal polynomial divides $x^n-1$ so that $alpha_g,j = zeta_n^e_g,j$ and $Tr(rho(g)) = sum_j alpha_g,j in BbbQ(zeta_n)$ and $E/BbbQ$ is abelian
$endgroup$
– reuns
Mar 25 at 21:15
$begingroup$
@reuns: Can we deduce that $Gal(E/mathbbQ)$ is a subgroup of $G$, or something like that?
$endgroup$
– Sebastien Palcoux
Mar 25 at 21:22
$begingroup$
No, it is a quotient (thus a subgroup since we are in finite abelian group) of $BbbZ/nZ^times = Gal(BbbQ(zeta_n)/Q)$ and $BbbZ/e Z^times$ where $e$ is the exponent of $G$. Also $E$ is called the minimal splitting field
$endgroup$
– reuns
Mar 25 at 21:25
$begingroup$
The character table of dihedral group $D_6$ non-abelian with $6$ elements is integer valued so you can't deduce $E$ only from $n$ or $e$
$endgroup$
– reuns
Mar 25 at 21:40
$begingroup$
There are nonisomorphic groups with the same characters so you can't hope to deduce $G$ from $E$ or $F$
$endgroup$
– Max
Mar 25 at 21:45
|
show 7 more comments
1
$begingroup$
$n=|G|$ an homomorphism $rho : G to GL_m(BbbC)$ and $det(xI-rho(g))= prod_j (x-alpha_g,j)$ then $rho(g)^n = I$ implies the minimal polynomial divides $x^n-1$ so that $alpha_g,j = zeta_n^e_g,j$ and $Tr(rho(g)) = sum_j alpha_g,j in BbbQ(zeta_n)$ and $E/BbbQ$ is abelian
$endgroup$
– reuns
Mar 25 at 21:15
$begingroup$
@reuns: Can we deduce that $Gal(E/mathbbQ)$ is a subgroup of $G$, or something like that?
$endgroup$
– Sebastien Palcoux
Mar 25 at 21:22
$begingroup$
No, it is a quotient (thus a subgroup since we are in finite abelian group) of $BbbZ/nZ^times = Gal(BbbQ(zeta_n)/Q)$ and $BbbZ/e Z^times$ where $e$ is the exponent of $G$. Also $E$ is called the minimal splitting field
$endgroup$
– reuns
Mar 25 at 21:25
$begingroup$
The character table of dihedral group $D_6$ non-abelian with $6$ elements is integer valued so you can't deduce $E$ only from $n$ or $e$
$endgroup$
– reuns
Mar 25 at 21:40
$begingroup$
There are nonisomorphic groups with the same characters so you can't hope to deduce $G$ from $E$ or $F$
$endgroup$
– Max
Mar 25 at 21:45
1
1
$begingroup$
$n=|G|$ an homomorphism $rho : G to GL_m(BbbC)$ and $det(xI-rho(g))= prod_j (x-alpha_g,j)$ then $rho(g)^n = I$ implies the minimal polynomial divides $x^n-1$ so that $alpha_g,j = zeta_n^e_g,j$ and $Tr(rho(g)) = sum_j alpha_g,j in BbbQ(zeta_n)$ and $E/BbbQ$ is abelian
$endgroup$
– reuns
Mar 25 at 21:15
$begingroup$
$n=|G|$ an homomorphism $rho : G to GL_m(BbbC)$ and $det(xI-rho(g))= prod_j (x-alpha_g,j)$ then $rho(g)^n = I$ implies the minimal polynomial divides $x^n-1$ so that $alpha_g,j = zeta_n^e_g,j$ and $Tr(rho(g)) = sum_j alpha_g,j in BbbQ(zeta_n)$ and $E/BbbQ$ is abelian
$endgroup$
– reuns
Mar 25 at 21:15
$begingroup$
@reuns: Can we deduce that $Gal(E/mathbbQ)$ is a subgroup of $G$, or something like that?
$endgroup$
– Sebastien Palcoux
Mar 25 at 21:22
$begingroup$
@reuns: Can we deduce that $Gal(E/mathbbQ)$ is a subgroup of $G$, or something like that?
$endgroup$
– Sebastien Palcoux
Mar 25 at 21:22
$begingroup$
No, it is a quotient (thus a subgroup since we are in finite abelian group) of $BbbZ/nZ^times = Gal(BbbQ(zeta_n)/Q)$ and $BbbZ/e Z^times$ where $e$ is the exponent of $G$. Also $E$ is called the minimal splitting field
$endgroup$
– reuns
Mar 25 at 21:25
$begingroup$
No, it is a quotient (thus a subgroup since we are in finite abelian group) of $BbbZ/nZ^times = Gal(BbbQ(zeta_n)/Q)$ and $BbbZ/e Z^times$ where $e$ is the exponent of $G$. Also $E$ is called the minimal splitting field
$endgroup$
– reuns
Mar 25 at 21:25
$begingroup$
The character table of dihedral group $D_6$ non-abelian with $6$ elements is integer valued so you can't deduce $E$ only from $n$ or $e$
$endgroup$
– reuns
Mar 25 at 21:40
$begingroup$
The character table of dihedral group $D_6$ non-abelian with $6$ elements is integer valued so you can't deduce $E$ only from $n$ or $e$
$endgroup$
– reuns
Mar 25 at 21:40
$begingroup$
There are nonisomorphic groups with the same characters so you can't hope to deduce $G$ from $E$ or $F$
$endgroup$
– Max
Mar 25 at 21:45
$begingroup$
There are nonisomorphic groups with the same characters so you can't hope to deduce $G$ from $E$ or $F$
$endgroup$
– Max
Mar 25 at 21:45
|
show 7 more comments
1 Answer
1
active
oldest
votes
$begingroup$
For the Mathieu group $G=M_11$, we have $E=mathbbQ(isqrt2,isqrt11)$ (see the character table here p. 89), whereas $Out(G)=C_1$. It follows that $Gal(F/mathbbQ) not hookrightarrow Out(G)$.
Otherwise the following sentence comes from this page:
The outer automorphism group acts on the character table by permuting
columns (conjugacy classes) and accordingly rows, which gives another
symmetry to the table.
answered Mar 29 at 11:25
Sebastien PalcouxSebastien Palcoux
2,336928
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3162256%2frelation-between-a-finite-group-and-the-galois-group-for-the-field-extension-gen%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For the Mathieu group $G=M_11$, we have $E=mathbbQ(isqrt2,isqrt11)$ (see the character table here p. 89), whereas $Out(G)=C_1$. It follows that $Gal(F/mathbbQ) not hookrightarrow Out(G)$.
Otherwise the following sentence comes from this page:
The outer automorphism group acts on the character table by permuting
columns (conjugacy classes) and accordingly rows, which gives another
symmetry to the table.
answered Mar 29 at 11:25
Sebastien PalcouxSebastien Palcoux
2,336928
$endgroup$
add a comment |
$begingroup$
For the Mathieu group $G=M_11$, we have $E=mathbbQ(isqrt2,isqrt11)$ (see the character table here p. 89), whereas $Out(G)=C_1$. It follows that $Gal(F/mathbbQ) not hookrightarrow Out(G)$.
Otherwise the following sentence comes from this page:
The outer automorphism group acts on the character table by permuting
columns (conjugacy classes) and accordingly rows, which gives another
symmetry to the table.
answered Mar 29 at 11:25
Sebastien PalcouxSebastien Palcoux
2,336928
$endgroup$
add a comment |
$begingroup$
For the Mathieu group $G=M_11$, we have $E=mathbbQ(isqrt2,isqrt11)$ (see the character table here p. 89), whereas $Out(G)=C_1$. It follows that $Gal(F/mathbbQ) not hookrightarrow Out(G)$.
Otherwise the following sentence comes from this page:
The outer automorphism group acts on the character table by permuting
columns (conjugacy classes) and accordingly rows, which gives another
symmetry to the table.
answered Mar 29 at 11:25
Sebastien PalcouxSebastien Palcoux
2,336928
$endgroup$
For the Mathieu group $G=M_11$, we have $E=mathbbQ(isqrt2,isqrt11)$ (see the character table here p. 89), whereas $Out(G)=C_1$. It follows that $Gal(F/mathbbQ) not hookrightarrow Out(G)$.
Otherwise the following sentence comes from this page:
The outer automorphism group acts on the character table by permuting
columns (conjugacy classes) and accordingly rows, which gives another
symmetry to the table.
answered Mar 29 at 11:25
Sebastien PalcouxSebastien Palcoux
2,336928
answered Mar 29 at 11:25
Sebastien PalcouxSebastien Palcoux
2,336928
answered Mar 29 at 11:25
Sebastien PalcouxSebastien Palcoux
2,336928
answered Mar 29 at 11:25
Sebastien PalcouxSebastien Palcoux
2,336928
2,336928
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3162256%2frelation-between-a-finite-group-and-the-galois-group-for-the-field-extension-gen%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
$n=|G|$ an homomorphism $rho : G to GL_m(BbbC)$ and $det(xI-rho(g))= prod_j (x-alpha_g,j)$ then $rho(g)^n = I$ implies the minimal polynomial divides $x^n-1$ so that $alpha_g,j = zeta_n^e_g,j$ and $Tr(rho(g)) = sum_j alpha_g,j in BbbQ(zeta_n)$ and $E/BbbQ$ is abelian
$endgroup$
– reuns
Mar 25 at 21:15
$begingroup$
@reuns: Can we deduce that $Gal(E/mathbbQ)$ is a subgroup of $G$, or something like that?
$endgroup$
– Sebastien Palcoux
Mar 25 at 21:22
$begingroup$
No, it is a quotient (thus a subgroup since we are in finite abelian group) of $BbbZ/nZ^times = Gal(BbbQ(zeta_n)/Q)$ and $BbbZ/e Z^times$ where $e$ is the exponent of $G$. Also $E$ is called the minimal splitting field
$endgroup$
– reuns
Mar 25 at 21:25
$begingroup$
The character table of dihedral group $D_6$ non-abelian with $6$ elements is integer valued so you can't deduce $E$ only from $n$ or $e$
$endgroup$
– reuns
Mar 25 at 21:40
$begingroup$
There are nonisomorphic groups with the same characters so you can't hope to deduce $G$ from $E$ or $F$
$endgroup$
– Max
Mar 25 at 21:45