Property of subgroups Is the following property correct?
Property:  let $(G;f_G)$ group, $B \subseteq G$ and $B \neq \emptyset $, then $(B;f_G|_B)$ is group iff $\forall a,b \in B$ we have that $a f_G|_B b'\in B $ and $ b $ is symmetry element  of $b'$.
Thanks in advance!!
 A: That statement is true.  To show that a subset of a group is a subgroup, the subset must contain the identity, must contain the inverse of each of its elements, and must be closed under the group operation.
$B$ contains the identity because $bf_G|_Bb' \in B$.  Write $e$ for the identity element.
The product $ef_G|_Bb'$ lies in $B$ for any $b \in B$, so $B$ contains the inverses of each of its elements.
Finally, to show closure, write $(b')' = b$.  Then $af_G|_B(b')' = af_G|_Bb \in B$
(I'm unfamiliar with your notation and vocabulary.  I'm answering as if $f_G$ means the group operation, $af_Gb$ means "the product of $a$ and $b$," and "symmetry element" means "inverse."  Most users of this site would say "If $B \subset G$ and $ab^{-1} \in B$ for all $a, b \in B$, then $B$ is a subgroup of $G$.)
Edit: As egreg and Thomas Anderson point out, the first step relies on the assumption that $B$ is non-empty.  Were $B$ empty, there would be no $b$ to use in the product $bf_G|_Bb'$.
A: The trick is to prove it in steps. First show that, since $B$ is not empty, we can use the condition to show that $1_G\in B$.
Then use that to show that if $b\in B$ then $b'\in B$.
Then use that to show that if $a,b\in B$ then $af_G|_Bb\in B$.
