Let $(G,\ast)$ be a group. For every subset $A\subseteq G$ define $$\langle A\rangle:=\bigcap_{A\subset S\subset G}S$$ as the set intersection of all subgroups $S$ of $G$ containing $A.$

Obviously, $A\subset \langle A\rangle$ and $\langle A\rangle \subset S$ for every subgroup $S$ of $G.$ Therefore, $\langle A\rangle$ is the *smallest* subgroup of $G$ containing $A.$ We call $A$ the **generating set** of $\langle A\rangle$ and $\langle A\rangle$ the **group generated by** $A.$ If $A$ is finite, we write $\langle a_1,\ldots,a_n\rangle.$

| | | | | created: 2019-07-27 21:23:28 | modified: 2019-07-28 10:28:34 | by: *bookofproofs* | references: [677]

[677] **Modler, Florian; Kreh, Martin**: “Tutorium Algebra”, Springer Spektrum, 2013