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Automorphism on a Finite Group

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 

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 

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 

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 

   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.

Start with something you know to be true:

   cb = g^-1 * f(g)

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

   ( 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 

   ( 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 = 


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   

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 
this .pdf file (you need Adobe's Acrobat Reader which you can download 
free from    ).   

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 

- Doctor Paul, The Math Forum   
Associated Topics:
College Modern Algebra

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