The subgroup generator function is inclusion monotone

Let G be a group, with A \subseteq B \subseteq G. Prove that \langle A \rangle \leq \langle B \rangle. Give an example where A \subseteq B with A \neq B but \langle A \rangle = \langle B \rangle.


Let \mathcal{A} = \{ H \leq G \ |\ A \subseteq H \} and \mathcal{B} = \{ H \leq G \ |\ B \subseteq H \}. Since A \subseteq B, we have A \subseteq H whenever B \subseteq H; thus \mathcal{B} \subseteq \mathcal{A}. By definition, we have \langle A \rangle = \cap \mathcal{A} and \langle B \rangle = \cap \mathcal{B}. We know from set theory that \cap \mathcal{A} \subseteq \cap \mathcal{B}, so that \langle A \rangle \subseteq \langle B \rangle.

Now since \langle A \rangle is itself a subgroup of G, we have \langle A \rangle \leq \langle B \rangle.

Now suppose G = \langle x \rangle is cyclic. Then \{x\} \subsetneq G, but we have \langle x \rangle = \langle G \rangle.

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