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| − | Let <math>|a|=n</math>. Then <math>\phi_a^n(x)=a^nxa^{-n}=x</math>. Does this mean that <math>|\phi_a|=n</math> as well? | + | Let <math>|a|=n</math>. Then <math>\phi_a^n(x)=a^nxa^{-n}=x</math>. Does this mean that <math>|\phi_a|=n</math> as well? Can someone explain what the order of an isomorphism is? |
Revision as of 09:45, 11 February 2009
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?
