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Math Help - Cosets/normal subgroup

  1. #1
    Junior Member hercules's Avatar
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    Cosets/normal subgroup

    Can someone please explain to me normal subgroups.

    if you have H a subgroup of an abelian group G and b is an element of G, then the right coset to which b belongs is same as the left coset to which b belongs. (proving Hb=bH).

    also, is a subgroup of an abelian group, necessarily abelian?


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  2. #2
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    Hello

    Quote Originally Posted by hercules View Post
    Can someone please explain to me normal subgroups.
    Take a group, G. then N is called a normal subgroup of G iff  \forall n \in N and  \forall g \in G ,  gng^{-1} \in N

    You can also say N is a normal subgroup of G if the sets of left and right cosets of N in G coincide, or  \forall g \in G ,  gN = Ng

    also, is a subgroup of an abelian group, necessarily abelian?
    Say H is a subgroup of an abelian group, G.

    Take x,y \in H

    Then since H is a subgroup,  x,y \in G

    Since G is abelian, you can say x*y=y*x

    So  x*y = y*x \in H by closure

    So yes, a subgroup of an abelian group is abelian.
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  3. #3
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    Quote Originally Posted by hercules View Post
    if you have H a subgroup of an abelian group G and b is an element of G, then the right coset to which b belongs is same as the left coset to which b belongs. (proving Hb=bH).
    There are different ways to define normality, the way I perfer (and most algebra books use) is a subgroup H of G such that ghg^{-1} \in H for all g\in G and h\in H. Now suppose that H is a subgroup that has this poperty. We will prove that bH=Hb for any b\in G, by showing bH\subseteq Hb \mbox{ and }Hb\subseteq bH. Say x\in bH then x=bh_1 but bh_1b^{-1}\in H \implies bh_1b^{-1} = h_2 \implies bh_1 = h_2 b for some h_2\in H. And so x=bh_1=h_2b\in Hb. And the other way around is similar.
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