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Normal subgroups

  1. Apr 26, 2008 #1
    Let G be a group and H a subgroup of G. We define the following:

    [tex]N_{G}(H) = \{g \in G \,\,|\,\, g^{-1}hg \in H,\, for\, all\,\, h\in H\}[/tex]

    Show that [tex]N_{G}(H)[/tex] is a subgroup of G.


    I've shown that for all [itex]x,\, y[/itex] of [itex]N_{G}(H)[/itex], [itex]xy[/itex] is an element of [itex]N_{G}(H)[/itex], but how do I show that [itex]x^{-1}[/itex] is an element of [itex]N_{G}(H)[/itex] ?
  2. jcsd
  3. Apr 26, 2008 #2
    H is a subgroup, hence H has inverses so for the element x in H, and the element g-1hg consider the element g-1x-1g which is in NG(H) since x-1 is in H.
  4. Apr 27, 2008 #3


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    I think d leet mean "and the element g-1xg".

    What is the product (g-1xg)(g-1x-1g)?
  5. Apr 27, 2008 #4
    It's the identity.

    But, aren't we supposed to show that (g-1)-1x(g-1) = gxg-1 is in H?
  6. Apr 27, 2008 #5
    yup that's right, think about what halls just said though and it answers your question, you know (g^-1xg)(g^-1x^-1g) = e. This tells you what?
  7. Apr 27, 2008 #6
    It tells me the inverse of g-1xg is g-1x-1g (which must also be contained in H), but how does that show that g-1 is in NG(H)? (i.e. do all elements of NG(H) have inverses?).
  8. Apr 27, 2008 #7


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    ?? Was that a typo. You are not trying to prove that g-1 is in NG(H), you are trying to prove x-1 is.

    And the fact that g-1x-1g is in H shows that x-1 is in NG(H).
  9. Apr 27, 2008 #8
    Now I'm getting confused lol. According to the definition of NG(H) from my original post, we want to show that for all g in NG(H), g-1 is also in NG(H). In other words we have to show that for any g in NG(H) and all x in H, (g-1)-1x(g-1) is H.

    Am I correct in saying this?
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