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Group of inner automorphisms is isomorphic to a quotient

  1. Mar 8, 2017 #1
    1. The problem statement, all variables and given/known data
    Let ##G## be any group. Recall that the center of ##G##, or ##Z(G)## is ##\{ x \in G ~ | ~ xg =
    gx, ~ \forall g \in G\}##. Show that ##G / Z(G)## is isomorphic to ##Inn(G)##, the group of inner automorphisms of ##G## by ##g##.

    2. Relevant equations


    3. The attempt at a solution
    I am not sure where to get started. I know that I am trying to find a particular isomorphism, but not sure how to find what that isomorphism must be, or whether that map will go from ##G / Z(G)## to ##Inn(G)## or the other way around.
     
  2. jcsd
  3. Mar 8, 2017 #2

    fresh_42

    Staff: Mentor

    Can you establish a surjective homomorphism ##G \twoheadrightarrow Inn(G)\;##?
     
  4. Mar 8, 2017 #3
    Define a general element of ##Inn(G)## to be ##\varphi_g(x) = gxg^{-1}##

    Would ##\mu : G \rightarrow Inn(G)## where ##\mu (g) = \varphi_g## be a surjection?
     
  5. Mar 8, 2017 #4

    fresh_42

    Staff: Mentor

    Yes, because every inner automorphism (=conjugation) looks like a ##\mu (g)##, so ##g## is the pre-imange. Now what is the kernel of ##\mu##?
     
  6. Mar 8, 2017 #5
    Scratch that

    ##Ker( \mu ) = Z(G)##
     
  7. Mar 8, 2017 #6

    fresh_42

    Staff: Mentor

    No. ##\{e\} \subseteq \ker \mu## but not necessarily the entire kernel. The kernel is defined as the set of all elements that maps to ##e'## in the codomain. Now what is this ##e'## then and what does it mean, that ##\mu ## maps an element ##g## on it?
     
  8. Mar 8, 2017 #7
    Sorry, I made a mistake and was too slow to correct. I think that ##Ker (\mu ) = Z(G)##

    Will the fundamental homomorphism theorem be used?
     
  9. Mar 8, 2017 #8

    fresh_42

    Staff: Mentor

    Not sure what this theorem is, but sounds right. How do you now, that ##\ker \mu = Z(G)\,##? This is the essential part of the proof, so you should drop a line on it.
     
  10. Mar 8, 2017 #9
    ##\ker \mu = \{x \in G ~ | ~ \mu (x) = \mu (e) \} = \{x \in G ~ | ~ \varphi_x = \varphi_e \} = \{x \in G ~ | ~ \varphi_x (g) = \varphi_e (g), ~ \forall g \in G \} = \{x \in G ~ | ~ x g x^{-1} = g, ~ \forall g \in G \} = \{x \in G ~ | ~ xg = gx, ~ \forall g \in G\} = Z(G)##
     
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