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Thread: Subgroup of index 2 is normal

  1. #1
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    Subgroup of index 2 is normal

    Prove that every subgroup of index 2 is normal.

    The problem with me here is that I'm not quite sure what subgroup of index 2 means. Does that have anything to do with a family? So like a family of subgroup with an index that contains two elements?

    Thanks.
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  2. #2
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    The index of a subgroup $\displaystyle H$ of a group $\displaystyle G$ is the number of distinct left or right cosets of that subgroup. This is denoted $\displaystyle |G:H|.$

    If $\displaystyle G$ and $\displaystyle H$ are finite, Lagrange’s theorem states that $\displaystyle |G|=|G:H|\times|H|.$ Hence, in the finite case, the index of $\displaystyle H$ in $\displaystyle G$ is the order of $\displaystyle G$ divided by the order of $\displaystyle H$. Be careful though: $\displaystyle G$ and $\displaystyle H$ are not always finite. In the infinite case, the index $\displaystyle |G:H|$ is taken to be the number of cosets of $\displaystyle H$ in $\displaystyle G$ (this can be finite even if $\displaystyle G$ and $\displaystyle H$ are infinite).

    Suppose $\displaystyle H$ is a subgroup of $\displaystyle G$ of index 2. Let $\displaystyle g\in G$.

    If $\displaystyle g\in H$ then $\displaystyle gH=H$ and $\displaystyle Hg=H$, so $\displaystyle gH=Hg.$

    If $\displaystyle g\notin H,$ then $\displaystyle G$ is the disjoint union of the left cosets $\displaystyle H=eH$ and $\displaystyle gH,$ and the disjoint union of the right cosets $\displaystyle H=He$ and $\displaystyle Hg.$ Hence, again $\displaystyle gH=Hg.$

    Since $\displaystyle gH=Hg$ for all $\displaystyle g\in G,$ $\displaystyle H$ is normal in $\displaystyle G.$ The beauty of this result is that it doesn’t assume that $\displaystyle G$ or $\displaystyle H$ is finite, only that the index of $\displaystyle H$ is finite.
    Last edited by JaneBennet; Sep 11th 2008 at 06:52 AM.
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  3. #3
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    index is the number of the left coset of group G,why gH=Hg for every element of G?
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    Index is 2. So there are just two right cosets - H itself and Hx where x doesn't belong to H. every element g in G belongs one of the two cosets (as they are mutually exclusive and exhaistive over G). So Hx = G - H

    As there is one-to-one mapping between left and right cosets, similarly there are two left cosets H and yH. So yH = G - H = Hx

    So for every x,y not in H, Hx = yH.

    If x is in H, Hx = xH = H. This is very trivial prove using the fact H is a sub-group hence closed.
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  5. #5
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    @11235813 - Infact you should try this - Prove that if every right coset of H is a left coset as well, then H is normal. Then prove for index 2 is just a corollary to this result.

    Quote Originally Posted by aman_cc View Post
    Index is 2. So there are just two right cosets - H itself and Hx where x doesn't belong to H. every element g in G belongs one of the two cosets (as they are mutually exclusive and exhaistive over G). So Hx = G - H

    As there is one-to-one mapping between left and right cosets, similarly there are two left cosets H and yH. So yH = G - H = Hx

    So for every x,y not in H, Hx = yH.

    If x is in H, Hx = xH = H. This is very trivial prove using the fact H is a sub-group hence closed.
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  6. #6
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    Thanks!

    Thank You!
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