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### Automorphism on a Finite Group

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Date: 10/12/2001 at 20:14:39
From: Nathan Hancock
Subject: Automorphism on a finite Group

Dear Dr. Math:

This problem, encountered in my Abstract Algebra book, has got me
stumped.

Question: Let G be a finite group, f an automorphism of G such that
f^2 is the identity automorphism of G. Suppose that f(x)=x implies
that x=e (the identity). Prove that G is abelian and f(a)=a^-1 for
all a in G.

My professor offered the hint that we should define h(x)=x^-1(f(x))
and show that it is 1-1 and onto, but even that has thrown me through
a loop. I don't see how h(x) being 1-1 and onto relates to the
problem, or even how to show that.

Thank you.
```

```
Date: 10/12/2001 at 23:00:30
From: Doctor Paul
Subject: Re: Automorphism on a finite Group

This is a nice problem. Here's how to proceed:

Define a function by h(x) = x^-1 * f(x)

We want to show that every element x of G has the form x^-1 * f(x).
To do this, we will show that the mapping defined above by h(x) is a
bijection.

We will show that it is injective. Since G is finite, if we know that
h(x) is injective, it must be surjective as well (this is a property
of finite sets) and then the bijective nature of h(x) will have been
established.

Recall that to show injectivity, we show h(c) = h(b) ==> c = b.

Let c and b be elements of G. Suppose h(c) = h(b). We want to show
that c = b.

h(c) = h(b)

c^-1 * f(c) = b^-1 * f(b)

b*c^-1 = f(b) * f(c)^-1

f is an automorphism (hence homomorphism) so we can write:

b*c^-1 = f(b*c^-1)

but f only maps an element to itself if that element is the identity.

So b*c^-1 = e

which implies

c = b.

Thus injectivity has been shown. Surjectivity follows from the fact
that G is finite. Thus h is a bijection.

So every element of G has the form x^-1 * f(x).

And now you ask: "So what?" What does this have to do with the task at
hand? Actually, a lot.

Now suppose c,b are elements of G. Then cb is in G (since G is a group
and groups are closed under the group operation) as well, so we can
write:

c = d^-1 * f(d)

b = e^-1 * f(e)

cb = g^-1 * f(g)

for some elements d, e, and g in the group G.

cb = g^-1 * f(g)

Now start using the information you know to derive what you want to
show:

( d^-1 * f(d) ) * ( e^-1 * f(e) ) = g^-1 * f(g)

Now take the inverse of both sides and continue to play with things.

Remember that f is a homomorphism so you can use some of the
properties of homomorphims to rewrite products.

[( d^-1 * f(d) ) * ( e^-1 * f(e) )]^-1 = [g^-1 * f(g)]^-1

( e^-1 * f(e) )^-1 * ( d^-1 * f(d) )^-1 = f(g)^-1 * g

( f(e)^-1 * e ) * ( f(d)^-1 * d ) = f(g^-1) * g

( f(e^-1) * e ) * ( f(d^-1) * d ) = f(g^-1) * g

Now recall that f^2 is the identity map so that means f(f(g)) = g.
Similarly, f(f(e)) = e and f(f(d)) = d. I'm going to make that
substition:

( f(e^-1) * f(f(e)) ) * ( f(d^-1) * f(f(d)) ) = f(g^-1) * f(f(g))

and since f is a homomorphism I can combine the two terms on the right
hand side of the equal sign ( this is because f(a)*f(b) = f(ab) )

( f(e^-1) * f(f(e)) ) * ( f(d^-1) * f(f(d)) ) = f( g^-1 * f(g) )

I can make a similar "combining" move on the left hand side of the
equal sign to obtain:

( f( e^-1 * f(e) ) * ( f( d^-1 * f(d) ) = f( g^-1 * f(g) )

but recall that

c = d^-1 * f(d)

b = e^-1 * f(e)

cb = g^-1 * f(g)

making the appropriate substitions, I obtain:

f(b) * f(c) = f(cb)

f(bc) = f(cb)

Since f is an automorphism, f is an isomorphism so f^-1 exists and is
an isomorphism as well. So we'll apply f^-1 on both sides to obtain
the desired result:

bc = cb

Thus G is abelian.

To see that f(a) = a^-1 for all a in G, use the fact that h(x) as
defined above is a bijection to write:

a = d^-1 * f(d) for some d in G.

Then f(a) =

f( d^-1 * f(d) ) =

f(d^-1) * f(f(d)) =

f(d)^-1 * d =

a^-1

Thanks for a fun question. If there's anything about the above
argument that you don't follow, please write back.

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

```
Date: 10/12/2001 at 23:04:13
From: Doctor Paul
Subject: Re: Automorphism on a finite Group

Just a moment ago I answered your question and I realized I left
something out. If you have trouble following the math because of the
way math types up on a computer screen in ASCII format, try viewing

http://www.math.rutgers.edu/~tunnell/math551/sol3.pdf

Look at problem 7 on pages 3 and 4. It isn't worked exactly the same
way as yours, but I think you'll agree that the differences are
trivial.

- Doctor Paul, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Modern Algebra

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