Subgroup

Let $G$ be a group (Group and abelian group) with identity $e$. A subset $H\subseteq G$ is a subgroup of $G$, written $H\le G$, when

• (S1) $e\in H$;
• (S2) $H$ is closed under the operation: $x,y\in H$ implies $xy\in H$ (Binary operation on a set; associativity, commutativity, and a subset closed under the operation);
• (S3) $H$ is closed under inverses: $x\in H$ implies $x^{-1}\in H$.

By (S2) the operation of $G$ restricts to a binary operation on $H$; it is associative there because it is associative on $G$, the element $e$ of (S1) is a two-sided identity for it (Left identity, right identity, and two-sided identity for a binary operation), and (S3) supplies for each $x\in H$ a two-sided inverse lying in $H$ (Left inverse, right inverse, and invertible element of a monoid). So a subgroup, with the restricted operation, is itself a group, and its identity and its inverses are those of $G$.

Every group $G$ has the two trivial subgroups $\{e\}$ and $G$ itself; a subgroup $H$ with $H\mathrm{\ne }G$ is called proper.

Group and abelian group

Binary operation on a set; associativity, commutativity, and a subset closed under the operation

Left identity, right identity, and two-sided identity for a binary operation

Left inverse, right inverse, and invertible element of a monoid

• The definition is stated so that no comparison of structures is needed. A subgroup is a subset satisfying three closure conditions, and the group structure it carries is inherited rather than chosen. The converse reading, that a subset which happens to be a group under the restricted operation is a subgroup in the above sense, is a small theorem rather than a tautology, because a priori such a subset could carry a different identity; it is part of One-step subgroup test: a nonempty $H\subseteq G$ is a subgroup iff $g{h}^{-1}\in H$ for all $g,h\in H$; the identity and the inverses of $H$ are then those of $G$, and cancellation in $G$ is what rules that out.

• Conditions (S1)–(S3) are not independent as stated: if $H$ is nonempty and satisfies (S2) and (S3) then it satisfies (S1). The economical single test is One-step subgroup test: a nonempty $H\subseteq G$ is a subgroup iff $g{h}^{-1}\in H$ for all $g,h\in H$; the identity and the inverses of $H$ are then those of $G$.

• Intersections of subgroups are subgroups (The intersection of a nonempty family of subgroups of $G$ is a subgroup of $G$), which is what makes "the smallest subgroup containing a given subset" meaningful (The subgroup $⟨S⟩$ generated by a subset, the cyclic subgroup $⟨g⟩$, and cyclic groups). Unions of subgroups are almost never subgroups.