Existence of a normal subgroup with $lvertoperatorname{Aut}{(H)}rvert>vertoperatorname{Aut}{(G)}rvert$












33












$begingroup$


Does there exist a finite group $G$ and a normal subgroup $H$ of $G$ such that $lvertoperatorname{Aut}{(H)}rvert>lvertoperatorname{Aut}{(G)}rvert$?










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$endgroup$












  • $begingroup$
    @theo: It's ok. No problem. When I saw the question, it was like this: Does there $text{existsa}$ and I added space in between.
    $endgroup$
    – user9413
    Jul 2 '11 at 3:54








  • 1




    $begingroup$
    Yes, I got that. You wrote it the way it is now and I wanted to produce the correct "Does there exist a", but instead of deleting the s I deleted the blank. Never mind.
    $endgroup$
    – t.b.
    Jul 2 '11 at 3:57












  • $begingroup$
    I think this was an IMC 2008 problem :)
    $endgroup$
    – Beni Bogosel
    Aug 12 '11 at 5:44










  • $begingroup$
    @Beni: Yes it is. See answer of damiano
    $endgroup$
    – user9413
    Aug 12 '11 at 5:49


















33












$begingroup$


Does there exist a finite group $G$ and a normal subgroup $H$ of $G$ such that $lvertoperatorname{Aut}{(H)}rvert>lvertoperatorname{Aut}{(G)}rvert$?










share|cite|improve this question











$endgroup$












  • $begingroup$
    @theo: It's ok. No problem. When I saw the question, it was like this: Does there $text{existsa}$ and I added space in between.
    $endgroup$
    – user9413
    Jul 2 '11 at 3:54








  • 1




    $begingroup$
    Yes, I got that. You wrote it the way it is now and I wanted to produce the correct "Does there exist a", but instead of deleting the s I deleted the blank. Never mind.
    $endgroup$
    – t.b.
    Jul 2 '11 at 3:57












  • $begingroup$
    I think this was an IMC 2008 problem :)
    $endgroup$
    – Beni Bogosel
    Aug 12 '11 at 5:44










  • $begingroup$
    @Beni: Yes it is. See answer of damiano
    $endgroup$
    – user9413
    Aug 12 '11 at 5:49
















33












33








33


17



$begingroup$


Does there exist a finite group $G$ and a normal subgroup $H$ of $G$ such that $lvertoperatorname{Aut}{(H)}rvert>lvertoperatorname{Aut}{(G)}rvert$?










share|cite|improve this question











$endgroup$




Does there exist a finite group $G$ and a normal subgroup $H$ of $G$ such that $lvertoperatorname{Aut}{(H)}rvert>lvertoperatorname{Aut}{(G)}rvert$?







group-theory finite-groups






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share|cite|improve this question













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share|cite|improve this question








edited Dec 27 '18 at 20:55









the_fox

2,89021537




2,89021537










asked Aug 11 '10 at 8:41







user9413



















  • $begingroup$
    @theo: It's ok. No problem. When I saw the question, it was like this: Does there $text{existsa}$ and I added space in between.
    $endgroup$
    – user9413
    Jul 2 '11 at 3:54








  • 1




    $begingroup$
    Yes, I got that. You wrote it the way it is now and I wanted to produce the correct "Does there exist a", but instead of deleting the s I deleted the blank. Never mind.
    $endgroup$
    – t.b.
    Jul 2 '11 at 3:57












  • $begingroup$
    I think this was an IMC 2008 problem :)
    $endgroup$
    – Beni Bogosel
    Aug 12 '11 at 5:44










  • $begingroup$
    @Beni: Yes it is. See answer of damiano
    $endgroup$
    – user9413
    Aug 12 '11 at 5:49




















  • $begingroup$
    @theo: It's ok. No problem. When I saw the question, it was like this: Does there $text{existsa}$ and I added space in between.
    $endgroup$
    – user9413
    Jul 2 '11 at 3:54








  • 1




    $begingroup$
    Yes, I got that. You wrote it the way it is now and I wanted to produce the correct "Does there exist a", but instead of deleting the s I deleted the blank. Never mind.
    $endgroup$
    – t.b.
    Jul 2 '11 at 3:57












  • $begingroup$
    I think this was an IMC 2008 problem :)
    $endgroup$
    – Beni Bogosel
    Aug 12 '11 at 5:44










  • $begingroup$
    @Beni: Yes it is. See answer of damiano
    $endgroup$
    – user9413
    Aug 12 '11 at 5:49


















$begingroup$
@theo: It's ok. No problem. When I saw the question, it was like this: Does there $text{existsa}$ and I added space in between.
$endgroup$
– user9413
Jul 2 '11 at 3:54






$begingroup$
@theo: It's ok. No problem. When I saw the question, it was like this: Does there $text{existsa}$ and I added space in between.
$endgroup$
– user9413
Jul 2 '11 at 3:54






1




1




$begingroup$
Yes, I got that. You wrote it the way it is now and I wanted to produce the correct "Does there exist a", but instead of deleting the s I deleted the blank. Never mind.
$endgroup$
– t.b.
Jul 2 '11 at 3:57






$begingroup$
Yes, I got that. You wrote it the way it is now and I wanted to produce the correct "Does there exist a", but instead of deleting the s I deleted the blank. Never mind.
$endgroup$
– t.b.
Jul 2 '11 at 3:57














$begingroup$
I think this was an IMC 2008 problem :)
$endgroup$
– Beni Bogosel
Aug 12 '11 at 5:44




$begingroup$
I think this was an IMC 2008 problem :)
$endgroup$
– Beni Bogosel
Aug 12 '11 at 5:44












$begingroup$
@Beni: Yes it is. See answer of damiano
$endgroup$
– user9413
Aug 12 '11 at 5:49






$begingroup$
@Beni: Yes it is. See answer of damiano
$endgroup$
– user9413
Aug 12 '11 at 5:49












6 Answers
6






active

oldest

votes


















23












$begingroup$

A version of Robin Chapman's answer that might be easier to verify:



The additive group $H$ of a field of order $2^{n}$ has two nice types of automorphism: multiplication by a scalar, and the Frobenius automorphism. The semi-direct product $G$ is called $AΓL(1,2^{n})$. It has order $(2^{n})⋅(2^{n}−1)⋅(n) ≤ 2^{3n}$, and for $n ≥ 2$ it is its own automorphism group.



It has the additive group $H$ of the field as a normal subgroup. However, as just an additive group, $H$ has automorphism group all n×n matrices over the field with 2 elements, which has size $(2^{n}−1)⋅(2^{n}−2)⋅⋅⋅(2^{n}−2^{n−1}) ≥ 2^{(n−1)^{2}}$.



For large $n$, $|Aut(G)| = (2^{n})⋅(2^{n}−1)⋅(n) ≤ 2^{3n} ≤ 2^{(n−1)^{2}} ≤ (2^{n}−1)⋅(2^{n}−2)⋅⋅⋅(2^{n}−2^{n−1}) = |Aut(H)|$



In particular:



For $n = 4$, $G = AΓL(1,16)$, $Aut(G) = G$ has order $960$, and $Aut(H) = GL(4,2)$ has order $20160$.



For $n = 10$, $G = AΓL(1,1024)$, $Aut(G) = G$ has order $10475520$, and $Aut(H) = GL(10,2)$ has order $366440137299948128422802227200$.



In other words, $Aut(H)$ can be enormously bigger than $Aut(G)$.



This is reasonably important in finite group theory:



$G$ is a very rigid group with lots of structure. Because it contains all the "automorphisms" of the field, the group itself determines the field. An automorphism of the group will have to be an automorphism of the field, and we've already listed them all. Sometimes this expressed by saying the group G determines the geometry of the affine line on which it acts.



$H$ is a very floppy group to which you can do nearly anything. Without the maps encoding scalar multiplication, $H$ no longer remembers the field that defined it. It is just a vector space, and so instead of the (very few) field automorphisms, you are now allowed to use any vector space automorphism. $H$ has lost its structure.



I think $H$ is the canonical example of a horribly structureless group. Groups like $GL$ and $AΓL$ are pretty standard examples of groups with very clear structure. The symmetric group is very similar. Except for a few early cases, a symmetric group is its own automorphism group because it already contains within itself the set of points on which it is acting.






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$endgroup$





















    12












    $begingroup$

    The quasidihedral group of order 16, QD16, has automorphism group of order 16, but it has Q8 (the quaternion group) as a normal subgroup, and | Aut(Q8) | = 24.



    Steve






    share|cite|improve this answer









    $endgroup$





















      8












      $begingroup$

      An elementary Abelian group can have a large automorphism group.
      Let $H$ be a largish elementary Abelian group, and $G$
      be an extension of $H$, say be a semidirect product with
      a cyclic group acting faithfully on $H$. Then it is possible
      for Aut$(G)$ to be smaller than Aut$(H)$.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Any Specific Example. What do you mean by a "Largish Elementary Abelian Group".
        $endgroup$
        – anonymous
        Aug 11 '10 at 8:53












      • $begingroup$
        I'll let you work out the details, but certainly a group of order $2^n$ would be better here than a group of prime $p$ of similar size.
        $endgroup$
        – Robin Chapman
        Aug 11 '10 at 9:00










      • $begingroup$
        I also got an example, please check!
        $endgroup$
        – anonymous
        Aug 29 '10 at 2:06



















      2












      $begingroup$

      Let $F_4 = Z/2[w]/w^2+w+1$ be the field of order 4.



      Let $H = (F_4^+)^n = (Z/2)^{2n}$ for $n geq 2$.



      Then $Aut(H) = GL(2n,F_2)$ which has order
      $(4^n-1)(4^n-2)(4^n-4)...(4^n-2*4^(n-1))$



      Let $G$ be $langle H,a : a^3 = 1, axa^{-1} = wx text{for} x in H rangle$. So $G$ is the semi-direct product of $H$ with $Z/3$, where the action is given by multiplication by $H$. An automorphism of $G$ must fix $H$ ($H$ consists of the set of elements of order 2). It must also send a to one of the 2*4^n elements of order 3. Let us consider only the automorphisms of $G$ which fix a (which is a subgroup of $Aut(G)$ of index $2*4^n$). These are automorphisms of H which commute with multiplication by w. Hence these are automorphisms of $F_4^n$ as an $F_4$ vector space, or elements of $GL(n,F_4)$. Hence
      $|Aut(G)| = 2*4^n* (4^n-1)(4^n-4)...(4^n-4^(n-1))$.



      For $n geq 2$, it is clear that $|Aut(H)| > |Aut(G)|$.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Good! This is a nice variation on the AGL. H is just a Z/2Z vector space, but inside G you have a F4 vector space, with fewer automorphisms. G itself has extra automorphisms, but only 2|H| of them, and so eventually the F4 restriction gives radically fewer possible automorphisms.
        $endgroup$
        – Jack Schmidt
        Aug 29 '10 at 2:39



















      1












      $begingroup$

      Problem 5






      share|cite|improve this answer









      $endgroup$









      • 1




        $begingroup$
        The group G in the solution is a (sneaky) subgroup of AΓL(1,8) that manages to encode both the multiplication and the Frobenius automorphism using only an element of order 3. AΓL(1,8) itself has size 168, and so ties with Aut(H) = GL(3,2). The solution does not mention it, but both G and Aut(G) have size 24 (so the upper bound 56 is pretty loose).
        $endgroup$
        – Jack Schmidt
        Aug 11 '10 at 21:22



















      0












      $begingroup$

      Also, you my find the answer at IMC 2008, Problem 5, day 1 (Already given)



      EDITED: thanks to JackSchmidt






      share|cite|improve this answer











      $endgroup$









      • 1




        $begingroup$
        The Heisenberg example is completely wrong: |Aut(G)| = p*p*|Aut(H)|. The IMC link was already given by damiano.
        $endgroup$
        – Jack Schmidt
        Mar 27 '11 at 14:47











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      6 Answers
      6






      active

      oldest

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      6 Answers
      6






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      23












      $begingroup$

      A version of Robin Chapman's answer that might be easier to verify:



      The additive group $H$ of a field of order $2^{n}$ has two nice types of automorphism: multiplication by a scalar, and the Frobenius automorphism. The semi-direct product $G$ is called $AΓL(1,2^{n})$. It has order $(2^{n})⋅(2^{n}−1)⋅(n) ≤ 2^{3n}$, and for $n ≥ 2$ it is its own automorphism group.



      It has the additive group $H$ of the field as a normal subgroup. However, as just an additive group, $H$ has automorphism group all n×n matrices over the field with 2 elements, which has size $(2^{n}−1)⋅(2^{n}−2)⋅⋅⋅(2^{n}−2^{n−1}) ≥ 2^{(n−1)^{2}}$.



      For large $n$, $|Aut(G)| = (2^{n})⋅(2^{n}−1)⋅(n) ≤ 2^{3n} ≤ 2^{(n−1)^{2}} ≤ (2^{n}−1)⋅(2^{n}−2)⋅⋅⋅(2^{n}−2^{n−1}) = |Aut(H)|$



      In particular:



      For $n = 4$, $G = AΓL(1,16)$, $Aut(G) = G$ has order $960$, and $Aut(H) = GL(4,2)$ has order $20160$.



      For $n = 10$, $G = AΓL(1,1024)$, $Aut(G) = G$ has order $10475520$, and $Aut(H) = GL(10,2)$ has order $366440137299948128422802227200$.



      In other words, $Aut(H)$ can be enormously bigger than $Aut(G)$.



      This is reasonably important in finite group theory:



      $G$ is a very rigid group with lots of structure. Because it contains all the "automorphisms" of the field, the group itself determines the field. An automorphism of the group will have to be an automorphism of the field, and we've already listed them all. Sometimes this expressed by saying the group G determines the geometry of the affine line on which it acts.



      $H$ is a very floppy group to which you can do nearly anything. Without the maps encoding scalar multiplication, $H$ no longer remembers the field that defined it. It is just a vector space, and so instead of the (very few) field automorphisms, you are now allowed to use any vector space automorphism. $H$ has lost its structure.



      I think $H$ is the canonical example of a horribly structureless group. Groups like $GL$ and $AΓL$ are pretty standard examples of groups with very clear structure. The symmetric group is very similar. Except for a few early cases, a symmetric group is its own automorphism group because it already contains within itself the set of points on which it is acting.






      share|cite|improve this answer











      $endgroup$


















        23












        $begingroup$

        A version of Robin Chapman's answer that might be easier to verify:



        The additive group $H$ of a field of order $2^{n}$ has two nice types of automorphism: multiplication by a scalar, and the Frobenius automorphism. The semi-direct product $G$ is called $AΓL(1,2^{n})$. It has order $(2^{n})⋅(2^{n}−1)⋅(n) ≤ 2^{3n}$, and for $n ≥ 2$ it is its own automorphism group.



        It has the additive group $H$ of the field as a normal subgroup. However, as just an additive group, $H$ has automorphism group all n×n matrices over the field with 2 elements, which has size $(2^{n}−1)⋅(2^{n}−2)⋅⋅⋅(2^{n}−2^{n−1}) ≥ 2^{(n−1)^{2}}$.



        For large $n$, $|Aut(G)| = (2^{n})⋅(2^{n}−1)⋅(n) ≤ 2^{3n} ≤ 2^{(n−1)^{2}} ≤ (2^{n}−1)⋅(2^{n}−2)⋅⋅⋅(2^{n}−2^{n−1}) = |Aut(H)|$



        In particular:



        For $n = 4$, $G = AΓL(1,16)$, $Aut(G) = G$ has order $960$, and $Aut(H) = GL(4,2)$ has order $20160$.



        For $n = 10$, $G = AΓL(1,1024)$, $Aut(G) = G$ has order $10475520$, and $Aut(H) = GL(10,2)$ has order $366440137299948128422802227200$.



        In other words, $Aut(H)$ can be enormously bigger than $Aut(G)$.



        This is reasonably important in finite group theory:



        $G$ is a very rigid group with lots of structure. Because it contains all the "automorphisms" of the field, the group itself determines the field. An automorphism of the group will have to be an automorphism of the field, and we've already listed them all. Sometimes this expressed by saying the group G determines the geometry of the affine line on which it acts.



        $H$ is a very floppy group to which you can do nearly anything. Without the maps encoding scalar multiplication, $H$ no longer remembers the field that defined it. It is just a vector space, and so instead of the (very few) field automorphisms, you are now allowed to use any vector space automorphism. $H$ has lost its structure.



        I think $H$ is the canonical example of a horribly structureless group. Groups like $GL$ and $AΓL$ are pretty standard examples of groups with very clear structure. The symmetric group is very similar. Except for a few early cases, a symmetric group is its own automorphism group because it already contains within itself the set of points on which it is acting.






        share|cite|improve this answer











        $endgroup$
















          23












          23








          23





          $begingroup$

          A version of Robin Chapman's answer that might be easier to verify:



          The additive group $H$ of a field of order $2^{n}$ has two nice types of automorphism: multiplication by a scalar, and the Frobenius automorphism. The semi-direct product $G$ is called $AΓL(1,2^{n})$. It has order $(2^{n})⋅(2^{n}−1)⋅(n) ≤ 2^{3n}$, and for $n ≥ 2$ it is its own automorphism group.



          It has the additive group $H$ of the field as a normal subgroup. However, as just an additive group, $H$ has automorphism group all n×n matrices over the field with 2 elements, which has size $(2^{n}−1)⋅(2^{n}−2)⋅⋅⋅(2^{n}−2^{n−1}) ≥ 2^{(n−1)^{2}}$.



          For large $n$, $|Aut(G)| = (2^{n})⋅(2^{n}−1)⋅(n) ≤ 2^{3n} ≤ 2^{(n−1)^{2}} ≤ (2^{n}−1)⋅(2^{n}−2)⋅⋅⋅(2^{n}−2^{n−1}) = |Aut(H)|$



          In particular:



          For $n = 4$, $G = AΓL(1,16)$, $Aut(G) = G$ has order $960$, and $Aut(H) = GL(4,2)$ has order $20160$.



          For $n = 10$, $G = AΓL(1,1024)$, $Aut(G) = G$ has order $10475520$, and $Aut(H) = GL(10,2)$ has order $366440137299948128422802227200$.



          In other words, $Aut(H)$ can be enormously bigger than $Aut(G)$.



          This is reasonably important in finite group theory:



          $G$ is a very rigid group with lots of structure. Because it contains all the "automorphisms" of the field, the group itself determines the field. An automorphism of the group will have to be an automorphism of the field, and we've already listed them all. Sometimes this expressed by saying the group G determines the geometry of the affine line on which it acts.



          $H$ is a very floppy group to which you can do nearly anything. Without the maps encoding scalar multiplication, $H$ no longer remembers the field that defined it. It is just a vector space, and so instead of the (very few) field automorphisms, you are now allowed to use any vector space automorphism. $H$ has lost its structure.



          I think $H$ is the canonical example of a horribly structureless group. Groups like $GL$ and $AΓL$ are pretty standard examples of groups with very clear structure. The symmetric group is very similar. Except for a few early cases, a symmetric group is its own automorphism group because it already contains within itself the set of points on which it is acting.






          share|cite|improve this answer











          $endgroup$



          A version of Robin Chapman's answer that might be easier to verify:



          The additive group $H$ of a field of order $2^{n}$ has two nice types of automorphism: multiplication by a scalar, and the Frobenius automorphism. The semi-direct product $G$ is called $AΓL(1,2^{n})$. It has order $(2^{n})⋅(2^{n}−1)⋅(n) ≤ 2^{3n}$, and for $n ≥ 2$ it is its own automorphism group.



          It has the additive group $H$ of the field as a normal subgroup. However, as just an additive group, $H$ has automorphism group all n×n matrices over the field with 2 elements, which has size $(2^{n}−1)⋅(2^{n}−2)⋅⋅⋅(2^{n}−2^{n−1}) ≥ 2^{(n−1)^{2}}$.



          For large $n$, $|Aut(G)| = (2^{n})⋅(2^{n}−1)⋅(n) ≤ 2^{3n} ≤ 2^{(n−1)^{2}} ≤ (2^{n}−1)⋅(2^{n}−2)⋅⋅⋅(2^{n}−2^{n−1}) = |Aut(H)|$



          In particular:



          For $n = 4$, $G = AΓL(1,16)$, $Aut(G) = G$ has order $960$, and $Aut(H) = GL(4,2)$ has order $20160$.



          For $n = 10$, $G = AΓL(1,1024)$, $Aut(G) = G$ has order $10475520$, and $Aut(H) = GL(10,2)$ has order $366440137299948128422802227200$.



          In other words, $Aut(H)$ can be enormously bigger than $Aut(G)$.



          This is reasonably important in finite group theory:



          $G$ is a very rigid group with lots of structure. Because it contains all the "automorphisms" of the field, the group itself determines the field. An automorphism of the group will have to be an automorphism of the field, and we've already listed them all. Sometimes this expressed by saying the group G determines the geometry of the affine line on which it acts.



          $H$ is a very floppy group to which you can do nearly anything. Without the maps encoding scalar multiplication, $H$ no longer remembers the field that defined it. It is just a vector space, and so instead of the (very few) field automorphisms, you are now allowed to use any vector space automorphism. $H$ has lost its structure.



          I think $H$ is the canonical example of a horribly structureless group. Groups like $GL$ and $AΓL$ are pretty standard examples of groups with very clear structure. The symmetric group is very similar. Except for a few early cases, a symmetric group is its own automorphism group because it already contains within itself the set of points on which it is acting.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 8 '13 at 5:09









          SE Anarchist

          467




          467










          answered Aug 11 '10 at 14:39









          Jack SchmidtJack Schmidt

          43.2k572152




          43.2k572152























              12












              $begingroup$

              The quasidihedral group of order 16, QD16, has automorphism group of order 16, but it has Q8 (the quaternion group) as a normal subgroup, and | Aut(Q8) | = 24.



              Steve






              share|cite|improve this answer









              $endgroup$


















                12












                $begingroup$

                The quasidihedral group of order 16, QD16, has automorphism group of order 16, but it has Q8 (the quaternion group) as a normal subgroup, and | Aut(Q8) | = 24.



                Steve






                share|cite|improve this answer









                $endgroup$
















                  12












                  12








                  12





                  $begingroup$

                  The quasidihedral group of order 16, QD16, has automorphism group of order 16, but it has Q8 (the quaternion group) as a normal subgroup, and | Aut(Q8) | = 24.



                  Steve






                  share|cite|improve this answer









                  $endgroup$



                  The quasidihedral group of order 16, QD16, has automorphism group of order 16, but it has Q8 (the quaternion group) as a normal subgroup, and | Aut(Q8) | = 24.



                  Steve







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Aug 11 '10 at 9:14







                  user641






























                      8












                      $begingroup$

                      An elementary Abelian group can have a large automorphism group.
                      Let $H$ be a largish elementary Abelian group, and $G$
                      be an extension of $H$, say be a semidirect product with
                      a cyclic group acting faithfully on $H$. Then it is possible
                      for Aut$(G)$ to be smaller than Aut$(H)$.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        Any Specific Example. What do you mean by a "Largish Elementary Abelian Group".
                        $endgroup$
                        – anonymous
                        Aug 11 '10 at 8:53












                      • $begingroup$
                        I'll let you work out the details, but certainly a group of order $2^n$ would be better here than a group of prime $p$ of similar size.
                        $endgroup$
                        – Robin Chapman
                        Aug 11 '10 at 9:00










                      • $begingroup$
                        I also got an example, please check!
                        $endgroup$
                        – anonymous
                        Aug 29 '10 at 2:06
















                      8












                      $begingroup$

                      An elementary Abelian group can have a large automorphism group.
                      Let $H$ be a largish elementary Abelian group, and $G$
                      be an extension of $H$, say be a semidirect product with
                      a cyclic group acting faithfully on $H$. Then it is possible
                      for Aut$(G)$ to be smaller than Aut$(H)$.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        Any Specific Example. What do you mean by a "Largish Elementary Abelian Group".
                        $endgroup$
                        – anonymous
                        Aug 11 '10 at 8:53












                      • $begingroup$
                        I'll let you work out the details, but certainly a group of order $2^n$ would be better here than a group of prime $p$ of similar size.
                        $endgroup$
                        – Robin Chapman
                        Aug 11 '10 at 9:00










                      • $begingroup$
                        I also got an example, please check!
                        $endgroup$
                        – anonymous
                        Aug 29 '10 at 2:06














                      8












                      8








                      8





                      $begingroup$

                      An elementary Abelian group can have a large automorphism group.
                      Let $H$ be a largish elementary Abelian group, and $G$
                      be an extension of $H$, say be a semidirect product with
                      a cyclic group acting faithfully on $H$. Then it is possible
                      for Aut$(G)$ to be smaller than Aut$(H)$.






                      share|cite|improve this answer









                      $endgroup$



                      An elementary Abelian group can have a large automorphism group.
                      Let $H$ be a largish elementary Abelian group, and $G$
                      be an extension of $H$, say be a semidirect product with
                      a cyclic group acting faithfully on $H$. Then it is possible
                      for Aut$(G)$ to be smaller than Aut$(H)$.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Aug 11 '10 at 8:48









                      Robin ChapmanRobin Chapman

                      19.4k24673




                      19.4k24673












                      • $begingroup$
                        Any Specific Example. What do you mean by a "Largish Elementary Abelian Group".
                        $endgroup$
                        – anonymous
                        Aug 11 '10 at 8:53












                      • $begingroup$
                        I'll let you work out the details, but certainly a group of order $2^n$ would be better here than a group of prime $p$ of similar size.
                        $endgroup$
                        – Robin Chapman
                        Aug 11 '10 at 9:00










                      • $begingroup$
                        I also got an example, please check!
                        $endgroup$
                        – anonymous
                        Aug 29 '10 at 2:06


















                      • $begingroup$
                        Any Specific Example. What do you mean by a "Largish Elementary Abelian Group".
                        $endgroup$
                        – anonymous
                        Aug 11 '10 at 8:53












                      • $begingroup$
                        I'll let you work out the details, but certainly a group of order $2^n$ would be better here than a group of prime $p$ of similar size.
                        $endgroup$
                        – Robin Chapman
                        Aug 11 '10 at 9:00










                      • $begingroup$
                        I also got an example, please check!
                        $endgroup$
                        – anonymous
                        Aug 29 '10 at 2:06
















                      $begingroup$
                      Any Specific Example. What do you mean by a "Largish Elementary Abelian Group".
                      $endgroup$
                      – anonymous
                      Aug 11 '10 at 8:53






                      $begingroup$
                      Any Specific Example. What do you mean by a "Largish Elementary Abelian Group".
                      $endgroup$
                      – anonymous
                      Aug 11 '10 at 8:53














                      $begingroup$
                      I'll let you work out the details, but certainly a group of order $2^n$ would be better here than a group of prime $p$ of similar size.
                      $endgroup$
                      – Robin Chapman
                      Aug 11 '10 at 9:00




                      $begingroup$
                      I'll let you work out the details, but certainly a group of order $2^n$ would be better here than a group of prime $p$ of similar size.
                      $endgroup$
                      – Robin Chapman
                      Aug 11 '10 at 9:00












                      $begingroup$
                      I also got an example, please check!
                      $endgroup$
                      – anonymous
                      Aug 29 '10 at 2:06




                      $begingroup$
                      I also got an example, please check!
                      $endgroup$
                      – anonymous
                      Aug 29 '10 at 2:06











                      2












                      $begingroup$

                      Let $F_4 = Z/2[w]/w^2+w+1$ be the field of order 4.



                      Let $H = (F_4^+)^n = (Z/2)^{2n}$ for $n geq 2$.



                      Then $Aut(H) = GL(2n,F_2)$ which has order
                      $(4^n-1)(4^n-2)(4^n-4)...(4^n-2*4^(n-1))$



                      Let $G$ be $langle H,a : a^3 = 1, axa^{-1} = wx text{for} x in H rangle$. So $G$ is the semi-direct product of $H$ with $Z/3$, where the action is given by multiplication by $H$. An automorphism of $G$ must fix $H$ ($H$ consists of the set of elements of order 2). It must also send a to one of the 2*4^n elements of order 3. Let us consider only the automorphisms of $G$ which fix a (which is a subgroup of $Aut(G)$ of index $2*4^n$). These are automorphisms of H which commute with multiplication by w. Hence these are automorphisms of $F_4^n$ as an $F_4$ vector space, or elements of $GL(n,F_4)$. Hence
                      $|Aut(G)| = 2*4^n* (4^n-1)(4^n-4)...(4^n-4^(n-1))$.



                      For $n geq 2$, it is clear that $|Aut(H)| > |Aut(G)|$.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        Good! This is a nice variation on the AGL. H is just a Z/2Z vector space, but inside G you have a F4 vector space, with fewer automorphisms. G itself has extra automorphisms, but only 2|H| of them, and so eventually the F4 restriction gives radically fewer possible automorphisms.
                        $endgroup$
                        – Jack Schmidt
                        Aug 29 '10 at 2:39
















                      2












                      $begingroup$

                      Let $F_4 = Z/2[w]/w^2+w+1$ be the field of order 4.



                      Let $H = (F_4^+)^n = (Z/2)^{2n}$ for $n geq 2$.



                      Then $Aut(H) = GL(2n,F_2)$ which has order
                      $(4^n-1)(4^n-2)(4^n-4)...(4^n-2*4^(n-1))$



                      Let $G$ be $langle H,a : a^3 = 1, axa^{-1} = wx text{for} x in H rangle$. So $G$ is the semi-direct product of $H$ with $Z/3$, where the action is given by multiplication by $H$. An automorphism of $G$ must fix $H$ ($H$ consists of the set of elements of order 2). It must also send a to one of the 2*4^n elements of order 3. Let us consider only the automorphisms of $G$ which fix a (which is a subgroup of $Aut(G)$ of index $2*4^n$). These are automorphisms of H which commute with multiplication by w. Hence these are automorphisms of $F_4^n$ as an $F_4$ vector space, or elements of $GL(n,F_4)$. Hence
                      $|Aut(G)| = 2*4^n* (4^n-1)(4^n-4)...(4^n-4^(n-1))$.



                      For $n geq 2$, it is clear that $|Aut(H)| > |Aut(G)|$.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        Good! This is a nice variation on the AGL. H is just a Z/2Z vector space, but inside G you have a F4 vector space, with fewer automorphisms. G itself has extra automorphisms, but only 2|H| of them, and so eventually the F4 restriction gives radically fewer possible automorphisms.
                        $endgroup$
                        – Jack Schmidt
                        Aug 29 '10 at 2:39














                      2












                      2








                      2





                      $begingroup$

                      Let $F_4 = Z/2[w]/w^2+w+1$ be the field of order 4.



                      Let $H = (F_4^+)^n = (Z/2)^{2n}$ for $n geq 2$.



                      Then $Aut(H) = GL(2n,F_2)$ which has order
                      $(4^n-1)(4^n-2)(4^n-4)...(4^n-2*4^(n-1))$



                      Let $G$ be $langle H,a : a^3 = 1, axa^{-1} = wx text{for} x in H rangle$. So $G$ is the semi-direct product of $H$ with $Z/3$, where the action is given by multiplication by $H$. An automorphism of $G$ must fix $H$ ($H$ consists of the set of elements of order 2). It must also send a to one of the 2*4^n elements of order 3. Let us consider only the automorphisms of $G$ which fix a (which is a subgroup of $Aut(G)$ of index $2*4^n$). These are automorphisms of H which commute with multiplication by w. Hence these are automorphisms of $F_4^n$ as an $F_4$ vector space, or elements of $GL(n,F_4)$. Hence
                      $|Aut(G)| = 2*4^n* (4^n-1)(4^n-4)...(4^n-4^(n-1))$.



                      For $n geq 2$, it is clear that $|Aut(H)| > |Aut(G)|$.






                      share|cite|improve this answer









                      $endgroup$



                      Let $F_4 = Z/2[w]/w^2+w+1$ be the field of order 4.



                      Let $H = (F_4^+)^n = (Z/2)^{2n}$ for $n geq 2$.



                      Then $Aut(H) = GL(2n,F_2)$ which has order
                      $(4^n-1)(4^n-2)(4^n-4)...(4^n-2*4^(n-1))$



                      Let $G$ be $langle H,a : a^3 = 1, axa^{-1} = wx text{for} x in H rangle$. So $G$ is the semi-direct product of $H$ with $Z/3$, where the action is given by multiplication by $H$. An automorphism of $G$ must fix $H$ ($H$ consists of the set of elements of order 2). It must also send a to one of the 2*4^n elements of order 3. Let us consider only the automorphisms of $G$ which fix a (which is a subgroup of $Aut(G)$ of index $2*4^n$). These are automorphisms of H which commute with multiplication by w. Hence these are automorphisms of $F_4^n$ as an $F_4$ vector space, or elements of $GL(n,F_4)$. Hence
                      $|Aut(G)| = 2*4^n* (4^n-1)(4^n-4)...(4^n-4^(n-1))$.



                      For $n geq 2$, it is clear that $|Aut(H)| > |Aut(G)|$.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Aug 29 '10 at 2:05







                      anonymous



















                      • $begingroup$
                        Good! This is a nice variation on the AGL. H is just a Z/2Z vector space, but inside G you have a F4 vector space, with fewer automorphisms. G itself has extra automorphisms, but only 2|H| of them, and so eventually the F4 restriction gives radically fewer possible automorphisms.
                        $endgroup$
                        – Jack Schmidt
                        Aug 29 '10 at 2:39


















                      • $begingroup$
                        Good! This is a nice variation on the AGL. H is just a Z/2Z vector space, but inside G you have a F4 vector space, with fewer automorphisms. G itself has extra automorphisms, but only 2|H| of them, and so eventually the F4 restriction gives radically fewer possible automorphisms.
                        $endgroup$
                        – Jack Schmidt
                        Aug 29 '10 at 2:39
















                      $begingroup$
                      Good! This is a nice variation on the AGL. H is just a Z/2Z vector space, but inside G you have a F4 vector space, with fewer automorphisms. G itself has extra automorphisms, but only 2|H| of them, and so eventually the F4 restriction gives radically fewer possible automorphisms.
                      $endgroup$
                      – Jack Schmidt
                      Aug 29 '10 at 2:39




                      $begingroup$
                      Good! This is a nice variation on the AGL. H is just a Z/2Z vector space, but inside G you have a F4 vector space, with fewer automorphisms. G itself has extra automorphisms, but only 2|H| of them, and so eventually the F4 restriction gives radically fewer possible automorphisms.
                      $endgroup$
                      – Jack Schmidt
                      Aug 29 '10 at 2:39











                      1












                      $begingroup$

                      Problem 5






                      share|cite|improve this answer









                      $endgroup$









                      • 1




                        $begingroup$
                        The group G in the solution is a (sneaky) subgroup of AΓL(1,8) that manages to encode both the multiplication and the Frobenius automorphism using only an element of order 3. AΓL(1,8) itself has size 168, and so ties with Aut(H) = GL(3,2). The solution does not mention it, but both G and Aut(G) have size 24 (so the upper bound 56 is pretty loose).
                        $endgroup$
                        – Jack Schmidt
                        Aug 11 '10 at 21:22
















                      1












                      $begingroup$

                      Problem 5






                      share|cite|improve this answer









                      $endgroup$









                      • 1




                        $begingroup$
                        The group G in the solution is a (sneaky) subgroup of AΓL(1,8) that manages to encode both the multiplication and the Frobenius automorphism using only an element of order 3. AΓL(1,8) itself has size 168, and so ties with Aut(H) = GL(3,2). The solution does not mention it, but both G and Aut(G) have size 24 (so the upper bound 56 is pretty loose).
                        $endgroup$
                        – Jack Schmidt
                        Aug 11 '10 at 21:22














                      1












                      1








                      1





                      $begingroup$

                      Problem 5






                      share|cite|improve this answer









                      $endgroup$



                      Problem 5







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Aug 11 '10 at 21:12









                      damianodamiano

                      1,7161010




                      1,7161010








                      • 1




                        $begingroup$
                        The group G in the solution is a (sneaky) subgroup of AΓL(1,8) that manages to encode both the multiplication and the Frobenius automorphism using only an element of order 3. AΓL(1,8) itself has size 168, and so ties with Aut(H) = GL(3,2). The solution does not mention it, but both G and Aut(G) have size 24 (so the upper bound 56 is pretty loose).
                        $endgroup$
                        – Jack Schmidt
                        Aug 11 '10 at 21:22














                      • 1




                        $begingroup$
                        The group G in the solution is a (sneaky) subgroup of AΓL(1,8) that manages to encode both the multiplication and the Frobenius automorphism using only an element of order 3. AΓL(1,8) itself has size 168, and so ties with Aut(H) = GL(3,2). The solution does not mention it, but both G and Aut(G) have size 24 (so the upper bound 56 is pretty loose).
                        $endgroup$
                        – Jack Schmidt
                        Aug 11 '10 at 21:22








                      1




                      1




                      $begingroup$
                      The group G in the solution is a (sneaky) subgroup of AΓL(1,8) that manages to encode both the multiplication and the Frobenius automorphism using only an element of order 3. AΓL(1,8) itself has size 168, and so ties with Aut(H) = GL(3,2). The solution does not mention it, but both G and Aut(G) have size 24 (so the upper bound 56 is pretty loose).
                      $endgroup$
                      – Jack Schmidt
                      Aug 11 '10 at 21:22




                      $begingroup$
                      The group G in the solution is a (sneaky) subgroup of AΓL(1,8) that manages to encode both the multiplication and the Frobenius automorphism using only an element of order 3. AΓL(1,8) itself has size 168, and so ties with Aut(H) = GL(3,2). The solution does not mention it, but both G and Aut(G) have size 24 (so the upper bound 56 is pretty loose).
                      $endgroup$
                      – Jack Schmidt
                      Aug 11 '10 at 21:22











                      0












                      $begingroup$

                      Also, you my find the answer at IMC 2008, Problem 5, day 1 (Already given)



                      EDITED: thanks to JackSchmidt






                      share|cite|improve this answer











                      $endgroup$









                      • 1




                        $begingroup$
                        The Heisenberg example is completely wrong: |Aut(G)| = p*p*|Aut(H)|. The IMC link was already given by damiano.
                        $endgroup$
                        – Jack Schmidt
                        Mar 27 '11 at 14:47
















                      0












                      $begingroup$

                      Also, you my find the answer at IMC 2008, Problem 5, day 1 (Already given)



                      EDITED: thanks to JackSchmidt






                      share|cite|improve this answer











                      $endgroup$









                      • 1




                        $begingroup$
                        The Heisenberg example is completely wrong: |Aut(G)| = p*p*|Aut(H)|. The IMC link was already given by damiano.
                        $endgroup$
                        – Jack Schmidt
                        Mar 27 '11 at 14:47














                      0












                      0








                      0





                      $begingroup$

                      Also, you my find the answer at IMC 2008, Problem 5, day 1 (Already given)



                      EDITED: thanks to JackSchmidt






                      share|cite|improve this answer











                      $endgroup$



                      Also, you my find the answer at IMC 2008, Problem 5, day 1 (Already given)



                      EDITED: thanks to JackSchmidt







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Mar 27 '11 at 18:42

























                      answered Mar 27 '11 at 11:17









                      Ehsan M. KermaniEhsan M. Kermani

                      6,40412348




                      6,40412348








                      • 1




                        $begingroup$
                        The Heisenberg example is completely wrong: |Aut(G)| = p*p*|Aut(H)|. The IMC link was already given by damiano.
                        $endgroup$
                        – Jack Schmidt
                        Mar 27 '11 at 14:47














                      • 1




                        $begingroup$
                        The Heisenberg example is completely wrong: |Aut(G)| = p*p*|Aut(H)|. The IMC link was already given by damiano.
                        $endgroup$
                        – Jack Schmidt
                        Mar 27 '11 at 14:47








                      1




                      1




                      $begingroup$
                      The Heisenberg example is completely wrong: |Aut(G)| = p*p*|Aut(H)|. The IMC link was already given by damiano.
                      $endgroup$
                      – Jack Schmidt
                      Mar 27 '11 at 14:47




                      $begingroup$
                      The Heisenberg example is completely wrong: |Aut(G)| = p*p*|Aut(H)|. The IMC link was already given by damiano.
                      $endgroup$
                      – Jack Schmidt
                      Mar 27 '11 at 14:47


















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