A subgroup in Group theory I am currently studying group theory, and its a quite new concept for me. 
When learning about subgroups, I bumped into something. 
In a problem where one wants to show that some set A is a group, is it sufficient enough to show that that set (A) satisfies all axioms of being a subgroup of a group B?
i.e.
If we know for sure that B is a group, 
and we show that some set A satisfies all axioms of being a subgroup of a group B,
can we conclude that A is a group?
 A: It can be done easier. Let $G$ be a group and $U$ be a nonempty subset of $G$.
Then $U$ forms a subgroup of $G$ iff (1) for all $g,h\in U$, $gh\in U$, and (2) for each $g\in U$, $g^{-1}\in U$.
Then $U$ forms a subgroup of $G$ without checking the axioms point by point.
For instance, the unit element $e\in G$ lies in $U$, since $U$ is nonempty and so $U$ has an element $g$. Then $g^{-1}\in U$ by (2) and so by (1) $e= gg^{-1}\in U$.
A: No, as associativity may not be satisfied within the set. Consider, for instance, the structure $(S,*)$, with $S=\{1,a,b\}$, and the following definition table for $*$:
$$
   \begin{matrix} 
    *  &  \bf1 & \bf{a} & \bf{b} \\
      \bf1 & 1 & a & b \\
     \bf{a} & a & 1 & b \\
     \bf{b} & b & a & 1
   \end{matrix} 
$$
This satisfies the conditions for being a subgroup (it is closed under * and inverses, and it is nonempty), except for the fact that it is not in a group. But $(S,*)$ is not a group, because associativity fails: $1=a(ba) \neq (ab)a=a$
A: If $G$ is a group and $H\subseteq G$, then $H$ is a subgroup of $G$ iff it is nonvoid and closed under multiplication and inversion.  You do not need to verify all of the axioms in this case because much is inherited from $G$.
