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The following is taken from A first Course in Abstract Algebra Rings, Groups, and Fields Third Edition by Anderson and Feil.##\\\\##
(Assumed exercise and example) ##\\\\##
22.15 In this problem we consider a particular important example of a group endomorphism. Suppose ##G## is a group and ##g\in G##. Define the function ##\varphi:G\to G## by setting ##\varphi(h)=ghg^{-1}\\\\##
(a) Prove that ##\varphi## is a group isomorphism##\\\\##
(b) What can you say about ##\varphi## if the group is abelian? What if ##g\in Z(g),## the center of ##G.\\\\##
Example 28.4##\\\\##
Let ##G## act on ##G## by conjugation given by ##g(x)=gxg^{-1};## see Exercise 22.15. In that exercise you show that the function ##g(x)## is a bijection (that is, a permutation of ##G##). We claim also that this action is homomorphic: That is, we claim that ##g(h(x))=(gh)(x).\\\\##
Exercise Question:##\\\\##
6. Let ##S## be a subgroup of the group ##G.## We wish to consider ##G## acting on the right cosets of ##S## by right multiplication. That is, ##g(Sh)=Shg.## Show that this does not work in general. That is, ##G## is not a group of permutations of the right cosets of ##S.## What goes wrong here? ##\\\\##
7. We wish to let ##G## act on ##G## by conjugation ##g(x)=g^{-1}xg.## Note this is different from conjugation we used in Example 28.4. Show that this does not work in general. What goes wrong here?##\\\\##
8. We ask the same question as the previous exercise, where we try to have ##G## act on its set of subgroups by the conjugation, ##g(S)=g^{-1}Sg.## What goes wrong here?##\\\\##
(c) Show that the order of ##h## is equal to the order of ##\varphi(h)=ghg^{-1}.\\\\##
Questions: I have some quick questions. For the Exercises Questions 6, 7,8; why can't right action not work in those three exercises? Is it the way $g$ is defined in all three questions or does it have to do with how or what $G$ acts on? I thought there is symmetry in the definition of group actions. Meaning if we know how to define it for right action, it should be the same for left action? Am I missing something?
(Assumed exercise and example) ##\\\\##
22.15 In this problem we consider a particular important example of a group endomorphism. Suppose ##G## is a group and ##g\in G##. Define the function ##\varphi:G\to G## by setting ##\varphi(h)=ghg^{-1}\\\\##
(a) Prove that ##\varphi## is a group isomorphism##\\\\##
(b) What can you say about ##\varphi## if the group is abelian? What if ##g\in Z(g),## the center of ##G.\\\\##
Example 28.4##\\\\##
Let ##G## act on ##G## by conjugation given by ##g(x)=gxg^{-1};## see Exercise 22.15. In that exercise you show that the function ##g(x)## is a bijection (that is, a permutation of ##G##). We claim also that this action is homomorphic: That is, we claim that ##g(h(x))=(gh)(x).\\\\##
Exercise Question:##\\\\##
6. Let ##S## be a subgroup of the group ##G.## We wish to consider ##G## acting on the right cosets of ##S## by right multiplication. That is, ##g(Sh)=Shg.## Show that this does not work in general. That is, ##G## is not a group of permutations of the right cosets of ##S.## What goes wrong here? ##\\\\##
7. We wish to let ##G## act on ##G## by conjugation ##g(x)=g^{-1}xg.## Note this is different from conjugation we used in Example 28.4. Show that this does not work in general. What goes wrong here?##\\\\##
8. We ask the same question as the previous exercise, where we try to have ##G## act on its set of subgroups by the conjugation, ##g(S)=g^{-1}Sg.## What goes wrong here?##\\\\##
(c) Show that the order of ##h## is equal to the order of ##\varphi(h)=ghg^{-1}.\\\\##
Questions: I have some quick questions. For the Exercises Questions 6, 7,8; why can't right action not work in those three exercises? Is it the way $g$ is defined in all three questions or does it have to do with how or what $G$ acts on? I thought there is symmetry in the definition of group actions. Meaning if we know how to define it for right action, it should be the same for left action? Am I missing something?
