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Does there exists a finite group $G$ and a normal subgroup $H$ of $G$ such that $|\operatorname{Aut}{(H)}|>|\operatorname{Aut}{(G)}|$?

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@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. –  user9413 Jul 2 '11 at 3:54
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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. –  t.b. Jul 2 '11 at 3:57
    
I think this was an IMC 2008 problem :) –  Beni Bogosel Aug 12 '11 at 5:44
    
@Beni: Yes it is. See answer of damiano –  user9413 Aug 12 '11 at 5:49

6 Answers 6

up vote 18 down vote accepted

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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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

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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$(H)$.

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Any Specific Example. What do you mean by a "Largish Elementary Abelian Group". –  anonymous Aug 11 '10 at 8:53
    
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. –  Robin Chapman Aug 11 '10 at 9:00
    
Thank you for the information. –  anonymous Aug 11 '10 at 10:33
    
I also got an example, please check! –  anonymous Aug 29 '10 at 2:06

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 $|Aut(H)| > |Aut(G)|$.

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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. –  Jack Schmidt Aug 29 '10 at 2:39

Problem 5

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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). –  Jack Schmidt Aug 11 '10 at 21:22

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

EDITED: thanks to JackSchmidt

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

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