An arbitrary intersection of subgroups is a subgroup

Let G be a group.

  1. Prove that if H and K are subgroups of G, then so is H \cap K.
  2. Prove that if \{H_i\}_{i \in I} is a family of subgroups of G then so is \bigcap_{i \in I} H_i.

  1. Note that H \cap K is not empty since 1 \in H \cap K. Now suppose x, y \in H \cap K. Then since H and K are subgroups, we have xy^{-1} \in H and xy^{-1} \in K by the subgroup criterion; thus xy^{-1} \in H \cap K. By the subgroup criterion, H \cap K is a subgroup of G.
  2. Note that \bigcap_{i \in I} H_i is not empty since 1 \in H_i for each i \in I. Now let x,y \in \bigcap_{i \in I} H_i. Then x,y \in H_i for each i \in I, and by the subgroup criterion, xy^{-1} \in H_i for each i \in I. Thus xy^{-1} \in \bigcap_{i \in I} H_i. By the subgroup criterion, \bigcap_{i \in I} H_i is a subgroup of G.
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