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Thread: 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 $\displaystyle \{ S _{a} \} _{a \in A} $ be a directed family of subgroups of G.

    Then $\displaystyle \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 $\displaystyle x \in S_{i}$ and $\displaystyle y \in S_{j} $. And since this family is directed, there exist some k in A such that $\displaystyle S_{i} \subseteq S_{k} $ and $\displaystyle S_{j} \subseteq S_{k} $. Implies that both x,y are in $\displaystyle S_{k} $ And since the union is consist of all subgroups, that means $\displaystyle S_{k} $ is a subgroup, implies that $\displaystyle 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 $\displaystyle S = \cup_{i\in I} S_i $.
    Here is how to start off.
    If $\displaystyle x,y\in S$ then $\displaystyle x\in S_a$ and $\displaystyle y\in S_b$.
    Therefore there is $\displaystyle c$ such that $\displaystyle a\leq c,b\leq c$.
    And thus, $\displaystyle S_a \subseteq S_c$ and $\displaystyle S_b\subseteq S_c$.
    Thus, $\displaystyle x,y\in S_c$ thus $\displaystyle xy\in S_c\implies xy\in S$.
    Thus $\displaystyle S$ is closed.

    Continue with the other properties.
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