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Thread: 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?


    Thank you
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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 $\displaystyle \forall n \in N $ and $\displaystyle \forall g \in G $, $\displaystyle 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 $\displaystyle \forall g \in G $, $\displaystyle gN = Ng $

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

    Take $\displaystyle x,y \in H $

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

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

    So $\displaystyle 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 $\displaystyle H$ of $\displaystyle G$ such that $\displaystyle ghg^{-1} \in H$ for all $\displaystyle g\in G$ and $\displaystyle h\in H$. Now suppose that $\displaystyle H$ is a subgroup that has this poperty. We will prove that $\displaystyle bH=Hb$ for any $\displaystyle b\in G$, by showing $\displaystyle bH\subseteq Hb \mbox{ and }Hb\subseteq bH$. Say $\displaystyle x\in bH$ then $\displaystyle x=bh_1$ but $\displaystyle bh_1b^{-1}\in H \implies bh_1b^{-1} = h_2 \implies bh_1 = h_2 b$ for some $\displaystyle h_2\in H$. And so $\displaystyle x=bh_1=h_2b\in Hb$. And the other way around is similar.
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