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Math Help - Union of directed family of subgroups is a group

  1. #1
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    Union of directed family of subgroups is a group

    Show that the union of a nonempty directed family of subgroup of a group G is a subgroup of G.

    Proof so far:

    Let G be a group, and suppose that  \{ S _{a} \} _{a \in A} be a directed family of subgroups of G.

    Then  \bigcup \{ S _{a} \} _{a \in A} is the union.

    Since it is nonempty, the subgroup of {e}, act as the identity of this subgroup.

    Now, suppose that x and y are elements of this union, then there exist i,j in A such that  x \in S_{i} and  y \in S_{j} . And since this family is directed, there exist some k in A such that  S_{i} \subseteq S_{k} and  S_{j} \subseteq S_{k} . Implies that both x,y are in  S_{k} And since the union is consist of all subgroups, that means  S_{k} is a subgroup, implies that xy^{-1} \in S_{k} , thus proves the claim.

    Is this right? Thank you.
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  2. #2
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    Let S = \cup_{i\in I} S_i .
    Here is how to start off.
    If x,y\in S then x\in S_a and y\in S_b.
    Therefore there is c such that a\leq c,b\leq c.
    And thus, S_a \subseteq S_c and S_b\subseteq S_c.
    Thus, x,y\in S_c thus xy\in S_c\implies xy\in S.
    Thus S is closed.

    Continue with the other properties.
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