As I learned, a subset $H$ of a group $G$ is a subgroup of $G$ if and only if :
$(1)$ $H$ is closed under the binary operation of $G$.
$(2)$ The identity element $e$ of $G$ is in $H$.
$(3)$ For all $a$ in $H$, $a^{-1}$ belongs in $H$ also.
But isn't second condition redundant? I think it can be derived from $(1)$ and $(3)$, for let $a\in H$, then $a^{-1}\in H$ by $(3)$, and by $(1)$, $aa^{-1}=e$ is an element of $H$, proving $(2)$.
You need the second one for the subset to be non-empty. If you take $H =\emptyset$, then 1) and 3) are fullfilled but $H$ is not a group.
Edit: I prefer to combine 1) and 3) for example and actually rephrase the second condition in terms of the existence of some element:
A subset $H \subset G$ is a subgroup iff $H$ is non-empty and for all $a, b \in H$ we have $ab^{-1} \in H$.