24 Subgroups Generated by Subsets of a Group

#todo/school : Do problems 1, 2 📅 2025-02-02 ✅ 2025-02-02

1

Question

Prove that if $H$ is a Subgroup of $G$ then $⟨ H ⟩ = H$ (see Subgroup Generator)

Proof

Suppose $H \leq G$ . Here:

⟨ H ⟩ = ⋂_{H \subseteq H_{i} : H_{i} \leq G} H_{i}

by definition. We want to show this equals $H$ itself.
$(\subseteq)$ : Let $h \in ⟨ H ⟩$ . Then $\forall H_{i} \leq G, H \subseteq H_{i} (h \in H_{i})$ . Assume that $h \notin H$ . Then since $H \leq H$ then $h \notin H_{i}$ for some $H_{i}$ (possibly $H$ itself). But then that contradicts $h \in ⟨ H ⟩$ , so then $h \in H$ .
$(\supseteq)$ : Let $h \in H$ . To show that $h \in ⟨ H ⟩$ , let $H \leq H_{i}$ be arbitrary. We want to show that $h \in H_{i}$ , but by being a subgroup we get that for free.

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2

Question

Prove that if $A \subseteq B$ then $⟨ A ⟩ \leq ⟨ B ⟩$ . Give an example where $A \subseteq B$ and $A \neq B$ but $⟨ A ⟩ = ⟨ B ⟩$ . (see Subgroup Generator)

Proof

(Subset): Let $a \in ⟨ A ⟩$ . Then $\forall A \subseteq H, H \leq G (a \in H)$ . Now using $H := ⟨ B ⟩$ then we get that $a \in ⟨ B ⟩$ as desired. It goes to show that obviously that $⟨ B ⟩ \leq G$ since it is a subgroup of any $H \leq G$ , but to show $A \subseteq ⟨ B ⟩$ if we let $h \in A$ then since $A \subseteq B$ then $h \in B$ . If we let $B \subseteq H, H \leq G$ then $h \in H$ and thus $h \in ⟨ B ⟩$ as desired.

The closure and inverse come from the fact that $⟨ A ⟩$ is from how it is an intersection of Subgroups, so it itself must be a subgroup.

Consider the example $⟨ G ⟩ = ⟨ G - {1} ⟩$ . This is an example where these generators equal, but the sets themselves don't.

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