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Math Help - Question about normal subgroup

  1. #1
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    Question about normal subgroup

    Suppose G is a group, and H is some subset of G.
    Is it true that if \forall g\in G and \forall h\in H: ghg^{-1}\in H, then H is a normal subgroup?
    Or that H has to be a subgroup (not just any subset) for this statment to be true?

    I'm asking because I need to prove it, and I not 100 percent sure my teacher didn't make a mistake here...

    Thanks in advance.
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  2. #2
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    Re: Question about normal subgroup

    Hi Stormey,
    Suppose G is abelian and H is any subset of G. Isn't it true that ghg-1 is in H for all g in G and all h in H?

    Perhaps your teacher wants you to prove that <H>, the subgroup generated by H, is a normal subgroup.
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  3. #3
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    Re: Question about normal subgroup

    Quote Originally Posted by johng View Post
    Hi Stormey,
    Suppose G is abelian and H is any subset of G. Isn't it true that ghg-1 is in H for all g in G and all h in H?

    Perhaps your teacher wants you to prove that <H>, the subgroup generated by H, is a normal subgroup.
    But it didn't say there that G is abelian, and if G isn't abelian, then this claim isn't true.
    So either G is abelian, or H is a subgroup, for this claim to be true, isn't it?
    Last edited by Stormey; November 29th 2013 at 01:19 AM.
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    Re: Question about normal subgroup

    Stormey,

    One counterexample is always enough to refute a statement. However, your statement is also false for non-abelian groups.

    Example: Let G=S3 be the symmetric group on {1,2,3}, x the 3 cycle (123) and H = {x,x-1}. I leave it to you to prove that ghg-1 is in H for all g in G and h in H. "Clearly", H is not a subgroup of G.
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    Re: Question about normal subgroup

    Excellent.
    This is what I was looking for.
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    Re: Question about normal subgroup

    *edit

    well, I found some counter example to this too:

    if H=\left \{(1 \hspace{3} 2 \hspace{3} 3),(3 \hspace{3} 2 \hspace{3} 1) \right \}, and let's say g=(1 \hspace{3} 2), then:
    ghg^{-1}=(1 \hspace{3} 2)\circ (1 \hspace{3} 2 \hspace{3} 3)\circ (1 \hspace{3} 2)\notin H
    Last edited by Stormey; November 30th 2013 at 01:52 AM.
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    Re: Question about normal subgroup

    Quote Originally Posted by Stormey View Post
    *edit

    well, I found some counter example to this too:

    if H=\left \{(1 \hspace{3} 2 \hspace{3} 3),(3 \hspace{3} 2 \hspace{3} 1) \right \}, and let's say g=(1 \hspace{3} 2), then:
    ghg^{-1}=(1 \hspace{3} 2)\circ (1 \hspace{3} 2 \hspace{3} 3)\circ (1 \hspace{3} 2)\notin H
    Actually (1 ~ 2) \circ (1~2~3) \circ (1~2) = (1~2) \circ (1~3~2) = (1~3~2) which is equivalent to (3 2 1).

    -Dan
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    Re: Question about normal subgroup

    Oh... I also got (1 3 2), but didn't relize it's equivalent to (3 2 1).
    Thanks Dan!
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    Re: Question about normal subgroup

    Quote Originally Posted by Stormey View Post
    Oh... I also got (1 3 2), but didn't relize it's equivalent to (3 2 1).
    Thanks Dan!
    No problem. Someone had to point that out to me about two weeks ago.

    -Dan
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