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### Automorphism of a Finite Group

```Date: 11/02/2004 at 15:34:40
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

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).

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
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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