Week 5 HW, Chapter 6, Problem 35, MA453, Spring 2008, Prof. Walther
Problem Statement
Could somebody please state the problem?
Discussion
Let a belong to a group G and let |a| be finite. Let $ \phi_a $ be the automorphism of G given by $ \phi_a (x) = axa^{-1} $. Show that |$ \phi_a $| divides |a|. Exhibit an element a from a group for which 1<|$ \phi_a $|<|a|.
By the properties of isomorphisms, we know that $ \scriptstyle\phi_\alpha\phi_\beta\ =\ \phi_{\alpha\beta} $. Inductively then we know that $ \scriptstyle(\phi_\alpha)^k\ =\ \phi_{\alpha^k} $. Let $ \scriptstyle\mid a\mid\ =\ n $. Then $ \scriptstyle a^n\ =\ e $, and $ \scriptstyle\phi_{a^n}\ =\ \phi_e $. So, $ \scriptstyle(\phi_a)^n\ =\ \phi_{a^n}\ =\ \phi_e $, and $ \scriptstyle\mid\phi_a\mid\ \textstyle\mid\scriptstyle\ n $.
- --Nick Rupley 12:15, 11 February 2009 (UTC)
Let $ |a|=n $. Then $ \phi_a^n(x)=a^nxa^{-n}=x $. Does this mean that $ |\phi_a|=n $ as well? Can someone explain what the order of an isomorphism is?
The order of the morphism is the number of iterations of the morphism it takes to return back to the original value. So in this problem, the order is n, because when a^n = a^(-n) = e, you get x back.
--K. Brumbaugh
For the part where we have to exhibit an element that has the property, how do you know what to start with? I wasn't clear about it. Just wondering.
--Jrendall 17:33, 14 February 2009 (UTC) --Jrendall 00:33, 19 February 2009 (UTC)
