Conjugation is Automorphism

  • #1
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Homework Statement


Let ##G## be a group and let ##G## act on itself by left conjugation, so each ##g \in G## maps ##G## to ##G## by ##x \mapsto gxg^{-1}##. For fixed ##g \in G##, prove that conjunction by ##G## is an automomorphism of G. Deduce that ##x## and ##gxg^{-1}## have the same order for all ##x## in ##G## and that for any subset ##A## of ##G##, ##|A| = |gAg^{-1}|##.

Homework Equations




The Attempt at a Solution


Here is how I approached the problem. Let ##\sigma_g = gxg^{-1}##. I will take it as given that ##\sigma_g## is a bijection from ##G## to ##G##, as I have already proved this in the general case before and don't feel like doing it again. Thus, we must only show that ##\sigma_g## is a homomorphism. Let ##x_1, x_2 \in G##. Then ##\sigma_g (x_1x_2) = g x_1 x_2 g^{-1} = g x_1 g^{-1} g x_2 g^{-1} = \sigma_g (x_1) \sigma_g(x_2)##. So ##\sigma_g## is an automorphism.

We can say that ##|x| = |gxg^{-1}|## because isomorphisms preserve order and we can say that ##|A| = |gAg^{-1}|## because ##\sigma_g## is injective and so the elements in ##A## will be mapped to distinct elements.
 

Answers and Replies

  • #2
14,359
11,673

Homework Statement


Let ##G## be a group and let ##G## act on itself by left conjugation, so each ##g \in G## maps ##G## to ##G## by ##x \mapsto gxg^{-1}##. For fixed ##g \in G##, prove that conjunction by ##G## is an automomorphism of G. Deduce that ##x## and ##gxg^{-1}## have the same order for all ##x## in ##G## and that for any subset ##A## of ##G##, ##|A| = |gAg^{-1}|##.

Homework Equations




The Attempt at a Solution


Here is how I approached the problem. Let ##\sigma_g = gxg^{-1}##.
##\sigma_g(x) = gxg^{-1}##
I will take it as given that ##\sigma_g## is a bijection from ##G## to ##G##, as I have already proved this in the general case before and don't feel like doing it again. Thus, we must only show that ##\sigma_g## is a homomorphism. Let ##x_1, x_2 \in G##. Then ##\sigma_g (x_1x_2) = g x_1 x_2 g^{-1} = g x_1 g^{-1} g x_2 g^{-1} = \sigma_g (x_1) \sigma_g(x_2)##. So ##\sigma_g## is an automorphism.

We can say that ##|x| = |gxg^{-1}|## because isomorphisms preserve order ...
You can also just write the same line as before again: ##(gxg^{-1})^n=gx^ng^{-1}## and one equals ##e=1## if and only if the other does, too.
... and we can say that ##|A| = |gAg^{-1}|## because ##\sigma_g## is injective and so the elements in ##A## will be mapped to distinct elements.
You also need surjectivity as ##G## might not be finite, in which case (finite case) injectivity implies surjectivity and vice versa. Otherwise ##A## could be larger.
 
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  • #3
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Correction:
in which case (finite case) injectivity implies surjectivity
is wrong. I confused this with vector spaces and dimensions. If we know nothing but that ##A## is a set, then even in a finite case, a proper subset is an injective embedding which is not surjective.
 
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