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Math Help - 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 H of a group G is the number of distinct left or right cosets of that subgroup. This is denoted |G:H|.

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

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

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

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

    Since gH=Hg for all g\in G, H is normal in G. The beauty of this result is that it doesn’t assume that G or H is finite, only that the index of H is finite.
    Last edited by JaneBennet; September 11th 2008 at 07:52 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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    @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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    Thanks!

    Thank You!
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