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Math Help - Cayley table and subgroup

  1. #1
    Senior Member sfspitfire23's Avatar
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    Cayley table and subgroup

    If I need to prove that something is a subgroup and I say that a,b are elements in the subgroup, will constructing a Cayley table with a,b,1 show that the subgroup is closed?
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  2. #2
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    Quote Originally Posted by sfspitfire23 View Post
    If I need to prove that something is a subgroup and I say that a,b are elements in the subgroup, will constructing a Cayley table with a,b,1 show that the subgroup is closed?
    Yes. If H is a subgroup of a group (G, *), then H should be closed with respect to '*'. It should also satisfy the group axioms with respect to '*' of course.
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    Senior Member sfspitfire23's Avatar
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    so it will also show associativity right?
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    Quote Originally Posted by sfspitfire23 View Post
    so it will also show associativity right?
    To show that H is a subgroup of (G, *), it suffice to show that for all a and b in H (as a subset of G), a * b^{-1} \in H. This basically checks the closure of an operation '*' and its existence of an inverse in H. If H satifies this subgroup condition, it means H also satisfies the remaining group axioms (identity, associativity). You can think that "associativity" w.r.t '*' is simply inherited from G.
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