Let $(G,\ast)$ be a \reftext{def:group-2026a}{group}, written multiplicatively as $ab=a\ast b$, with identity element $e_G$ and inverses $a^{-1}$ as in \ref{thm:group-identity-inverse-uniqueness-2026a}, and let $H\subseteq G$. Then the following hold.

1. $H$ is a \reftext{def:subgroup-2026a}{subgroup} of $(G,\ast)$ if and only if $H$ is nonempty and

2. If $H$ is a subgroup of $(G,\ast)$, then $H$ together with the restriction of $\ast$ to $H$ is itself a group. Its identity element is $e_G$, and for $a\in H$ its inverse in this group is the element $a^{-1}$ computed in $G$.
3. The set $G$ is a subgroup of $(G,\ast)$, and so is $\{e_G\}$; the latter is called the \textbf{trivial subgroup} of $(G,\ast)$.

The nonemptiness hypothesis in claim 1 cannot be omitted: the empty set satisfies the displayed condition vacuously but is not a subgroup, since it does not contain $e_G$.