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Math Help - Group Theory Problem

  1. #1
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    Group Theory Problem

    this isn't all that hard, but i decided to post it here, because it's interesting.

    Let P be a partition of a group G with the property that for any pair of elements A,B of the partition, the product set AB is contained entirely within another element C of the partition. Let N be the element of P that contains the identity of G. Prove that N is a normal subgroup of G and that P is the set of its cosets.
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  2. #2
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    Re: Group Theory Problem

    Quote Originally Posted by Deveno View Post
    this isn't all that hard, but i decided to post it here, because it's interesting.

    Let P be a partition of a group G with the property that for any pair of elements A,B of the partition, the product set AB is contained entirely within another element C of the partition. Let N be the element of P that contains the identity of G. Prove that N is a normal subgroup of G and that P is the set of its cosets.

    (*) for any A,B \in P, there exists C \in P such that AB \subseteq C.

    note that an obvious result of (*) is that for any Q_1 \in P and g, h \in G, there exists Q_2 \in P such that gQh \subseteq Q_2.

    now, for any a \in N we have a =  a \cdot 1 \in N^2 and so N \subseteq N^2. we also have N^2 \subseteq Q for some Q \in P, by (*). but then N \subseteq Q and hence N=Q. therefore N=N ^2 and so N is closed under multiplication. let a \in N. then Na^{-1} \subseteq Q for some Q \in P and since 1 \in Na^{-1}, we have 1 \in N \cap Q. thus Q=N and so Na^{-1}  \subseteq N. therefore a^{-1} \in N and so N is a subgroup. now let g \in G. then, by (*), there exists Q \in P such that gNg^{-1} \subseteq Q. but then 1 \in Q \cap N, because 1 \in gNg^{-1} \subseteq Q, and thus Q=N. so we have proved that gNg^{-1} \subseteq N, i.e. N is normal.

    finally, let g \in G and choose Q \in P such that g \in Q. then gN \subseteq Q_1 for some Q_1 \in P, by (*). but then g \in Q_1 \cap Q, because g \in Ng \subseteq Q_1, and so Q_1=Q. hence

    g N \subseteq Q.

    let Q_2 \in P be such that N \subseteq g^{-1}Q \subseteq Q_2. then 1 \in Q_2 \cap N, because 1 \in g^{-1}Q \subseteq Q_2, and so Q_2=N. hence gN=Q and we are done.
    Last edited by NonCommAlg; October 9th 2011 at 02:15 PM.
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    Re: Group Theory Problem

    G is not given to be finite, so N = N^2 is insufficient to establish that N is a subgroup.
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    Re: Group Theory Problem

    Quote Originally Posted by Deveno View Post
    G is not given to be finite, so N = N^2 is insufficient to establish that N is a subgroup.
    ok, you're right but that can be easily fixed. see the solution again.
    the problem is nice but the idea of the solution is very simple used over and over again!
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    Re: Group Theory Problem

    what i found "interesting" about it, was not so much its difficulty level, but the idea that if a partition respects the group multiplication, not only must it be a co-refinement (is that even a word?) of the coset partition, but in fact they are equal.

    i'm pretty sure that this is equivalent to saying that every congruence is in fact a coset partition.

    and yeah, once you establish N is one element of the partition, g-translation pretty much wraps up the rest.
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    Re: Group Theory Problem

    Quote Originally Posted by Deveno View Post
    i'm pretty sure that this is equivalent to saying that every congruence is in fact a coset partition.
    Indeed, if I'm not mistaken, the problem is a restatement of the well-known theorem stating the relation between congruences on groups and normal subgroups.

    One of the two common definitions of a congruence on a semigroup is that it's an equivalence relation R such that, if (a R b) and (c R d), then (ac R bd). The hypothesis of the problem is clearly (I think but I'm often wrong) equivalent to the statement that P is a partition given by a congruence.
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