Automorphism on a Finite GroupDate: 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. 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 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 this .pdf file (you need Adobe's Acrobat Reader which you can download free from http://www.adobe.com/ ). 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/ |
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