Let $S$ be a subset of $R$ such that we have associative relation $*$ defined on $S\times S\to S$ with some properties... Let $S$ be a subset of $\\R$ such that we have associative relation $\\*\\$ defined on $S\times S\to S$ with 
$$a*b*a=b \hspace{0.5cm} \forall a,b\in S, \hspace{1.5cm} \exists e \hspace{0.5cm} s.t\hspace{0.5cm} \forall b\in S\hspace{0.5cm}  b*e=e*b=b $$
Then which of the following is true?
$1.(S,*)$ is a group.
$2.(S,*)$ is commutative.
$3.S$ is finite.
My work:I have proved that first two are true.About third option I have problem I need help I do not know how to prove or disprove it.
Thanks.
 A: Although you've done it already, let me start with an answer to 1.
Taking any $a \in S$ and substituting $b=e$ into the first equation we get
$$a * a = a * e * a = e
$$
and therefore every element of $S$ is its own inverse. This proves number 1, that $(S,*)$ is a group. But it proves more: every non-identity element has order $2$.
Next, taking any $a,b \in S$ we get
$$aba^{-1}b^{-1}=abab=(ab)^2=e
$$
and therefore $S$ is commutative. But it proves more: every group in which each nonidentity element has order 2 is commutative.
Next, before attacking 3, we ponder the situation. Could a converse be true? Could it be true that every group in which each nonidentity element has order 2 satisfies the law $a*b*a=b$? Yes indeed, because $a*b*a^{-1}*b^{-1}=a*b*a*b = (ab)^2 = e$.
Now, on to number 3. If 3 is true then (as a fruit of our pondering) it must be that every group in which each non-identity element has order 2 must be finite. To disprove this, one must produce an example of an infinite group in which every nontrivial element has order 2. 
Can you take it from here?
