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Help with proof regarding normal subgroups and cosets

  1. Nov 3, 2011 #1
    1. The problem statement, all variables and given/known data

    I think I've got this one about figured out, I just wanted someone to check it over. (For this problem, (a-1) is a inverse, (b-1) b inverse, etc.)

    "Let G be a group, H a subgroup of G.

    Then, H is normal in G iff every left coset of H is equal to some right coset of H"

    2. Relevant equations

    H normal in G implies aH(a-1) = H for all a in G

    3. The attempt at a solution

    Let a,b be elts of G, and H a normal sbgp of G.

    (=>):

    By normal condition, aH(a-1) = H for all a in G

    -> aH = Ha (right multiply by a)

    H normal in G implies the left coset of H by a coincides with the right coset of H by a.

    Implies that if H is normal, every left coset of H is equal to some right coset of H.

    (<=):

    Let the left coset of H by a be equal to some right coset of H by b.

    So, aH = Hb

    -> aH(b-1) = H (right multiply by (b-1) )

    H = aH(b-1) implies (by taking inverses) that H = bH(a-1)

    Finally, H = HH = (aH(b-1))(bH(a-1)) = aH((b-1)b)H(a-1) = aH(e)H(a-1) = aHH(a-1) = aH(a-1)

    Then we have H = aH(a-1)

    Similar argument shows that H = bH(b-1)

    So, H is normal.

    In conclusion, if every left coset of H is equal to some right coset of H, then H is normal in G.

    I think that handles both directions (<=) & (=>).

    To me, this seems like a pretty pertinent question regarding normal sbgps so I'd like to make sure I hit it on the head.

    Thanks!
     
  2. jcsd
  3. Nov 3, 2011 #2

    Deveno

    User Avatar
    Science Advisor

    you can use the "SUP" tags to produce inverses, like so:

    x-1

    (it's the icon under "advanced" that looks like x2).

    your "=>" part is fine, you exhibit some right coset that aH is equal to, for every a in G, namely Ha.

    for your "<=" part, note that while a is arbitrary, b in fact depends on a, that is, given aH, b is some element such that aH = Hb (it might be the case that there is only one such b, for example).

    so you only need to show that aHa-1 = H, because this will be true for ANY a in G, whereas bHb-1 = H only holds for those b for which aH = Hb (for a specific a).

    that is, the subsequent proof that bHb-1 = H is unecessary, and considerably less general than aHa-1 = H (we have no prior conditions on a, but we do for b).

    a general warning on "set multiplication": be aware that you cannot always use facts about AB to prove facts about ab. for example, if G is abelian, AB = BA, but it is often the case that AB = BA, but G is not abelian. so it's a good idea to check that your statements about sets like:

    (aHb-1)(bHa-1) = H still make sense at "the element level":

    (ahb-1)(bh'a-1) should yield an element of H.

    being able to treat certain subsets of G (cosets of H) like elements, is a special property of normal subgroups, and statements like HH = H, only apply to certain subsets of G. what you have written is indeed true, just be aware that thinking of sets as things you can multiply, CAN get you into trouble (you can't always use the "group rules" of G to calculate these).
     
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