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Automorphism of a Finite Group

Date: 11/02/2004 at 15:34:40
From: Vlado
Subject: automorphism of finite group

If some automorphism T sends more than three quarters of elements into 
their inverses, prove that T(x) = x^(-1) for all x in G, where G is 
finite. 

I'm not sure how to use the "three quarters" in the proof.  I 
divided group G into two parts regarding the criterium that T sends one 
element to its inverse.



Date: 11/04/2004 at 02:48:20
From: Doctor Jacques
Subject: Re: automorphism of finite group

Hi Vlado,

We will make use of a few simple facts:

If a subgroup G of H contains more than half of the elements of G, 
then H = G.  This follows from Lagrange's theorem.

A consequence of this is that any subset S of G that contains more 
than half of the elements is a generating set, since the subgroup 
generated by S is then equal to G, as a consequence of the previous 
proposition.  This means that any element of G can be written as a 
product of elements of S; in particular, if we know the value of an 
automorphism for all elements of S, then the automorphism is uniquely 
defined on G.

Another useful fact is that the function f : G -> G defined by
f(x) = x^(-1) is an automorphism if and only if G is Abelian (please 
write back if you want to discuss this further).

Given the hypotheses, our plan of attack will be:

* Prove that G is Abelian.

* As G is Abelian, there is an automorphism U such that U(x) = x^(-1)
  for all x.

* As T and U agree on a generating set of G, they are equal.

Consider the set S = {x | T(x) = x^(-1)}.  We know that S contains at 
least 3n/4 elements (n is the order of G).  We could try to use the 
same kind of reasoning used to prove that if S = G, then G is Abelian, 
but the problem is that we do not know that S is closed under 
multiplication: if x and y are in S, we are not sure that xy is also 
in S.

To cope with that problem, let us take x as a fixed element in S 
and consider the set:

  xS = {xs | s in S}

and let K(x) be the intersection of S and xS.

Now, let z be any element of K.  As K is a subset of S, T(z) = z^(-1). 
As K is a subset of xS, there is y in S such that z = xy and
T(y) = y^(-1). If we write y = x^(-1)z, this becomes:

  T(x^(-1)z) = (x^(-1)z)^(-1)
             = z^(-1)x

Consider now:

  u = T(xx^(-1)z)

We may write:

  u = T(z) = z^(-1)     [1]

On the other hand, as T is an automorphism, we may also write:

  u = T(x(x^(-1)z))
    = T(x)T(x^(-1)z)
    = x^(-1)z^(-1)x      [2]

And, by equating [1] and [2], we find easily that xz = zx, where z 
is any element of K(x).

Now, by hypothesis, S contains more than 3n/4 elements, and xS 
contains exactly as many elements as S.  If both sets were disjoint, 
this would give more than 3n/2 elements, whereas G only contains n 
elements.  This means that the intersection K(x) contains more than n/2 
elements.

The centralizer of x,  C(x) is a subgroup of G, and, as x commutes
with all elements of K(x), C(x) includes K(x). As we saw that K(x) 
(and therefore C(x)) contains more than half the elements of G, we 
conclude that C(x) = G.
 
The means that x commutes with all elements of G, i.e., x belongs to 
Z(G), the center of G. As x represents an arbitrary element of S, this
means that Z(G) contains S. Using the same argument again, we see that
Z(G) is a subgroup that contains more than half of the elements of G, 
and we can conclude that Z(G) = G, i.e., G is Abelian.

As G is Abelian, there is an automorphism:

  U : G -> G

defined by U(x) = x^(-1) for all x in G. Note that U(x) = T(x) for 
all x in S (by definition).  Now, S contains more than n/2 elements, 
and is therefore a generating set of G (any element of G can be 
written as the product of elements of S).  As the automorphisms U and 
T take the same value on elements of S, they are equal, and
T(x) = U(x) = x^(-1) for all x in G.

Note that the bound 3n/4 is, in some sense, the best possible.  In the 
dihedral group D4 (of order 8), represented by the permutations:

  a = (1,2,3,4)
  b = (1,3)

The automorphism T(x) = bxb^(-1) exchanges the elements (1,2)(3,4) 
and (1,3)(2,4) (which are obviously not inverses to each other), but 
sends each of the 6 other elements to its inverse (which is in many 
cases equal to the element itself).

Does this help?  Write back if you'd like to talk about this some 
more, or if you have any other questions.

- Doctor Jacques, The Math Forum
  http://mathforum.org/dr.math/ 



Date: 11/04/2004 at 07:34:55
From: Vlado
Subject: Thank you (automorphism of finite group)

Thank you very much.  In the meantime, I continued to search for a
solution and found some rough proof.  Your proof is much nicer and it
helps me a lot.  Thank you again.
Associated Topics:
College Modern Algebra

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