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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? Can someone explain what the order of an isomorphism is? | 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? | ||
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| + | 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. | ||
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| + | --K. Brumbaugh | ||
Revision as of 16:55, 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?
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
