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Definition of normal subgroup

  1. Feb 6, 2007 #1

    quasar987

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    My professor of topology gave us a quick overview of the group theory results we will be needing later and among the things he said, is that a normal subgroup of a group G is a subgroup H such that for all x in G, xHx^{-1}=H.

    Is this correct? The wiki article seems to indicate that equality btw xHx^{-1} and H is unnecessary, but rather that inclusion is sufficient.
     
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  3. Feb 7, 2007 #2

    matt grime

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    They are easily shown to be equivalent.
     
  4. Feb 7, 2007 #3

    quasar987

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    Mmmh... So for all x in G, [itex]xHx^{-1}\subset H \Leftrightarrow xHx^{-1}= H[/itex].

    For a fixed x, suppose there exists h' in H such that there are no h in H with h'=xhx^{-1}. This is nonsense since h=x^{-1}h'x does the trick.
     
    Last edited: Feb 7, 2007
  5. Feb 7, 2007 #4

    StatusX

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    But then how do you know h is in H? And I don't see what this has to do with your original question.
     
  6. Feb 7, 2007 #5

    quasar987

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    My original question was (essentially) Are the two definitions of normal subgroup "H is is normal subgroup of G if for all x in G, [itex]xHx^{-1}\subset H[/itex]" and "H is is normal subgroup of G if for all x in G, [itex]xHx^{-1}= H[/itex]" equivalent?

    Then matt grime said "Yes". Then I tried to prove this assertion, so I said the following: "To show they are equivalent, I only need to show that for all x in G, [itex]xHx^{-1}\subset H \Rightarrow xHx^{-1}= H[/itex] since the converse is immediate."

    So let's proceed by contradiction by supposing that for a certain fixed x in G, there exists an h' in H such that there are no h in H with h' = xhx^{-1}. This is automatically seen to be an absurdity since the element h of G defined by h=x^{-1}h'x is such that h' = xhx^{-1} and it is in H because by hypothesis, h' is in H and [itex]xHx^{-1}\subset H[/itex], i.e. [itex]h=x^{-1}h'x\in H[/itex].


    I had to write it all in large to convince myself fully that it is correct, and still seems so to me. Do you still have an objection StatusX?
     
    Last edited: Feb 7, 2007
  7. Feb 7, 2007 #6

    StatusX

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    That's clearer than your last post, but still a little hard to follow, with an unnecessary proof by contradiction. The proof is easy:

    Assume for all x in G, xHx^-1 is a subset of H. Then for all h in H, we can define h'=x^-1hx, which is in H, and so h=xh'x^-1, ie, H is a subset of xHx^-1.
     
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