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Structure of the outer automorphism group of $D_n(q)$

2

2

$begingroup$

From the ATLAS, I know that the outer automorphism group of the Chevalley group $D_n(q)$, $q=p^f$ for some prime $p$ and some $n$ even and $n>4$, is a semidirect product of three groups, $(C_d times C_d) rtimes (C_f times C_g)$, where $d=(2,q-1)$ (the "diagonal" automorphisms), $f$ is such that $q=p^f$ (the "field" automorphisms) and $g=2$ (the graph automorphisms), so
$$operatornameOut(D_n(q))= (C_2 times C_2) rtimes (C_f times C_2)$$

What I want to know is: when $f=3k$ for some $k in mathbbN$, does $C_f$ act on $C_2 times C_2$? Equivalently, do the field automorphisms and the diagonal automorphisms commute?

I am also interested in the $n=4$ case, when
$$operatornameOut(D_4(q))= (C_2 times C_2) rtimes (C_f times S_3)$$
and I ask the same question for $C_f$, but also for $C_3 leq S_3$.

group-theory finite-groups simple-groups automorphism-group

share|cite|improve this question

edited Apr 2 at 21:23

AnalysisStudent0414

asked Apr 2 at 16:38

AnalysisStudent0414AnalysisStudent0414

4,423928

$endgroup$

add a comment | 

2

2

$begingroup$

From the ATLAS, I know that the outer automorphism group of the Chevalley group $D_n(q)$, $q=p^f$ for some prime $p$ and some $n$ even and $n>4$, is a semidirect product of three groups, $(C_d times C_d) rtimes (C_f times C_g)$, where $d=(2,q-1)$ (the "diagonal" automorphisms), $f$ is such that $q=p^f$ (the "field" automorphisms) and $g=2$ (the graph automorphisms), so
$$operatornameOut(D_n(q))= (C_2 times C_2) rtimes (C_f times C_2)$$

What I want to know is: when $f=3k$ for some $k in mathbbN$, does $C_f$ act on $C_2 times C_2$? Equivalently, do the field automorphisms and the diagonal automorphisms commute?

I am also interested in the $n=4$ case, when
$$operatornameOut(D_4(q))= (C_2 times C_2) rtimes (C_f times S_3)$$
and I ask the same question for $C_f$, but also for $C_3 leq S_3$.

group-theory finite-groups simple-groups automorphism-group

share|cite|improve this question

edited Apr 2 at 21:23

AnalysisStudent0414

asked Apr 2 at 16:38

AnalysisStudent0414AnalysisStudent0414

4,423928

$endgroup$

add a comment | 

$begingroup$

From the ATLAS, I know that the outer automorphism group of the Chevalley group $D_n(q)$, $q=p^f$ for some prime $p$ and some $n$ even and $n>4$, is a semidirect product of three groups, $(C_d times C_d) rtimes (C_f times C_g)$, where $d=(2,q-1)$ (the "diagonal" automorphisms), $f$ is such that $q=p^f$ (the "field" automorphisms) and $g=2$ (the graph automorphisms), so
$$operatornameOut(D_n(q))= (C_2 times C_2) rtimes (C_f times C_2)$$

What I want to know is: when $f=3k$ for some $k in mathbbN$, does $C_f$ act on $C_2 times C_2$? Equivalently, do the field automorphisms and the diagonal automorphisms commute?

I am also interested in the $n=4$ case, when
$$operatornameOut(D_4(q))= (C_2 times C_2) rtimes (C_f times S_3)$$
and I ask the same question for $C_f$, but also for $C_3 leq S_3$.

group-theory finite-groups simple-groups automorphism-group

share|cite|improve this question

edited Apr 2 at 21:23

AnalysisStudent0414

asked Apr 2 at 16:38

AnalysisStudent0414AnalysisStudent0414

4,423928

$endgroup$

share|cite|improve this question

edited Apr 2 at 21:23

AnalysisStudent0414

asked Apr 2 at 16:38

AnalysisStudent0414AnalysisStudent0414

4,423928

share|cite|improve this question

edited Apr 2 at 21:23

AnalysisStudent0414

asked Apr 2 at 16:38

AnalysisStudent0414AnalysisStudent0414

4,423928

asked Apr 2 at 16:38

AnalysisStudent0414AnalysisStudent0414

4,423928

asked Apr 2 at 16:38

AnalysisStudent0414AnalysisStudent0414

4,423928

1 Answer
1

active

oldest

votes

1

$begingroup$

The subgroup $(C_d times C_d) rtimes C_g$ is dihedral of order $8$, and since $C_f$ commutes with $C_g$, it follows that $C_f$ must commute with $C_d times C_d$, because $rm Aut(C_2 times C_2) cong S_3$, not $C_6$.

On the other hand, when $n=4$, the $S_3$ subgroup acts faithfully on $C_d times C_d$, and the subgroup $(C_d times C_d) rtimes S_3$ is isomorphic to $S_4$. This is the only situation in which the outer automorphism group of a finite simple group has derived length $3$.

share|cite|improve this answer

answered Apr 2 at 21:16

Derek HoltDerek Holt

54.7k53574

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Any insight on why $(C_d times C_d) rtimes C_g$ is dihedral? I was having trouble making these three different types of automorphism "interact"
  $endgroup$
  – AnalysisStudent0414
  Apr 2 at 21:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I don't have any magical insight about this. You can study the automorphisms of the orthogonal matrix groups $Omega^+_2n(q)$, and work it out, but it's not particularly easy. IIRC, one of the diagonal automorphisms lies in $rm SO^+_2n(q)$, the graph automorphism lies in $rm GO^+_2n(q)$, and the another diagonal automorphism has the effect of multiplying the from by a non-square in the field.
  $endgroup$
  – Derek Holt
  Apr 2 at 21:29
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Will do! (I think that I need to look at these actions in detail anyway, since this copy of $S_4$ could mess up something I am working on). Thank you very much
  $endgroup$
  – AnalysisStudent0414
  Apr 2 at 21:35
  

  
  
  
  

add a comment | 

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1

$begingroup$

The subgroup $(C_d times C_d) rtimes C_g$ is dihedral of order $8$, and since $C_f$ commutes with $C_g$, it follows that $C_f$ must commute with $C_d times C_d$, because $rm Aut(C_2 times C_2) cong S_3$, not $C_6$.

On the other hand, when $n=4$, the $S_3$ subgroup acts faithfully on $C_d times C_d$, and the subgroup $(C_d times C_d) rtimes S_3$ is isomorphic to $S_4$. This is the only situation in which the outer automorphism group of a finite simple group has derived length $3$.

share|cite|improve this answer

answered Apr 2 at 21:16

Derek HoltDerek Holt

54.7k53574

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Any insight on why $(C_d times C_d) rtimes C_g$ is dihedral? I was having trouble making these three different types of automorphism "interact"
  $endgroup$
  – AnalysisStudent0414
  Apr 2 at 21:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I don't have any magical insight about this. You can study the automorphisms of the orthogonal matrix groups $Omega^+_2n(q)$, and work it out, but it's not particularly easy. IIRC, one of the diagonal automorphisms lies in $rm SO^+_2n(q)$, the graph automorphism lies in $rm GO^+_2n(q)$, and the another diagonal automorphism has the effect of multiplying the from by a non-square in the field.
  $endgroup$
  – Derek Holt
  Apr 2 at 21:29
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Will do! (I think that I need to look at these actions in detail anyway, since this copy of $S_4$ could mess up something I am working on). Thank you very much
  $endgroup$
  – AnalysisStudent0414
  Apr 2 at 21:35
  

  
  
  
  

add a comment | 

1

$begingroup$

The subgroup $(C_d times C_d) rtimes C_g$ is dihedral of order $8$, and since $C_f$ commutes with $C_g$, it follows that $C_f$ must commute with $C_d times C_d$, because $rm Aut(C_2 times C_2) cong S_3$, not $C_6$.

On the other hand, when $n=4$, the $S_3$ subgroup acts faithfully on $C_d times C_d$, and the subgroup $(C_d times C_d) rtimes S_3$ is isomorphic to $S_4$. This is the only situation in which the outer automorphism group of a finite simple group has derived length $3$.

share|cite|improve this answer

answered Apr 2 at 21:16

Derek HoltDerek Holt

54.7k53574

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Any insight on why $(C_d times C_d) rtimes C_g$ is dihedral? I was having trouble making these three different types of automorphism "interact"
  $endgroup$
  – AnalysisStudent0414
  Apr 2 at 21:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I don't have any magical insight about this. You can study the automorphisms of the orthogonal matrix groups $Omega^+_2n(q)$, and work it out, but it's not particularly easy. IIRC, one of the diagonal automorphisms lies in $rm SO^+_2n(q)$, the graph automorphism lies in $rm GO^+_2n(q)$, and the another diagonal automorphism has the effect of multiplying the from by a non-square in the field.
  $endgroup$
  – Derek Holt
  Apr 2 at 21:29
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Will do! (I think that I need to look at these actions in detail anyway, since this copy of $S_4$ could mess up something I am working on). Thank you very much
  $endgroup$
  – AnalysisStudent0414
  Apr 2 at 21:35
  

  
  
  
  

add a comment | 

$begingroup$

The subgroup $(C_d times C_d) rtimes C_g$ is dihedral of order $8$, and since $C_f$ commutes with $C_g$, it follows that $C_f$ must commute with $C_d times C_d$, because $rm Aut(C_2 times C_2) cong S_3$, not $C_6$.

On the other hand, when $n=4$, the $S_3$ subgroup acts faithfully on $C_d times C_d$, and the subgroup $(C_d times C_d) rtimes S_3$ is isomorphic to $S_4$. This is the only situation in which the outer automorphism group of a finite simple group has derived length $3$.

share|cite|improve this answer

answered Apr 2 at 21:16

Derek HoltDerek Holt

54.7k53574

$endgroup$

share|cite|improve this answer

answered Apr 2 at 21:16

Derek HoltDerek Holt

54.7k53574

answered Apr 2 at 21:16

Derek HoltDerek Holt

54.7k53574

answered Apr 2 at 21:16

Derek HoltDerek Holt

54.7k53574