We verify the group axioms, in the following order (for convenience): === G0: Closure === Let $a, b \in S$. Then from the definition of $\ast$, we have $a \ast b \in S$, and hence also $a \circ b = a \ast \paren {e \ast b} \in S$. This proves closure of $\circ$. {{qed|lemma}} === G2: Identity === We assert that $e$ is the identity with respect to $\circ$. We verify this as follows (let $a \in S$): {{begin-eqn}} {{eqn | l = a \circ e       | r = a \ast \paren {e \ast e}       | c = Definition of $\circ$ }} {{eqn | r = a \ast e       | c = Property $(1)$ of $\ast$ }} {{eqn | l = e \circ a       | r = e \ast \paren {e \ast a}       | c = Definition of $\circ$ }} {{eqn | r = \paren {a \ast a} \ast \paren {e \ast a}       | c = Property $(1)$ of $\ast$ }} {{eqn | r = a \ast e       | c = Property $(2)$ of $\ast$ }} {{end-eqn}} It follows that $a \circ e = e \circ a = a \ast e$. It remains to prove the following identity: : $(3): \quad a \ast e = a$ By property $(1)$, it suffices to prove $a \ast \paren {a \ast e} = e$: {{begin-eqn}} {{eqn | l = a \ast \paren {a \ast e}       | r = \paren {a \ast e} \ast \paren {\paren {a \ast e} \ast e}       | c = Property $(2)$ of $\ast$ }} {{eqn | r = \paren {a \ast e} \ast \paren {\paren {a \ast e} \ast \paren {e \ast e} }       | c = Property $(1)$ of $\ast$ }} {{eqn | r = \paren {a \ast e} \ast \paren {a \ast e}       | c = Property $(2)$ of $\ast$ }} {{eqn | r = e       | c = Property $(1)$ of $\ast$ }} {{end-eqn}} So indeed $e$ is the identity for $\circ$. {{qed|lemma}} === G3: Inverses === For any element $a \in S$, we claim that $e \ast a$ is the inverse for $a$. This is verified as follows: {{begin-eqn}} {{eqn | l = \paren {e \ast a} \circ a       | r = \paren {e \ast a} \ast \paren {e \ast a}       | c = Definition of $\circ$ }} {{eqn | r = e       | c = Property $(1)$ of $\ast$ }} {{eqn | l = a \circ \paren {e \ast a}       | r = a \ast \paren {e \ast \paren {e \ast a} }       | c = Definition of $\circ$ }} {{eqn | r = a \ast \paren {e \circ a}       | c = Definition of $\circ$ }} {{eqn | r = a \ast a       | c = $e$ is the identity element for $\circ$ }} {{eqn | r = e       | c = Property $(1)$ of $\ast$ }} {{end-eqn}} We conclude that $e \ast a$ is indeed the inverse for $a$. {{qed|lemma}} === G1: Associativity === Lastly, we prove that $\circ$ is associative, i.e.: :$a \circ \paren {b \circ c} = \paren {a \circ b} \circ c$ To this end, we observe the following property of $\ast$: {{begin-eqn}} {{eqn | n = 4       | l = e \ast \paren {a \ast b}       | r = \paren {b \ast b} \ast \paren {a \ast b}       | c = Property $(1)$ of $\ast$ }} {{eqn | r = b \ast a       | c = Property $(2)$ of $\ast$ }} {{end-eqn}} Subsequently, we compute: {{begin-eqn}} {{eqn | l = \paren {a \circ b} \circ c       | r = \paren {a \ast \paren {e \ast b} } \ast \paren {e \ast c}       | c = Definition of $\circ$ }} {{eqn | r = \paren {a \ast \paren {e \ast b} } \ast \paren {\paren {e \ast c} \ast e}       | c = Property $(3)$ of $\ast$ }} {{eqn | r = \paren {a \ast \paren {e \ast b} } \ast \paren {\paren {\paren {e \ast c} \ast b} \ast \paren {e \ast b} }       | c = Property $(2)$ of $\ast$ }} {{eqn | r = a \ast \paren {\paren {e \ast c} \ast b}       | c = Property $(2)$ of $\ast$ }} {{eqn | r = a \ast \paren {e \ast \paren {b \ast \paren {e \ast c} } }       | c = Property $(4)$ of $\ast$ }} {{eqn | r = a \circ \paren {b \circ c}       | c = Definition of $\circ$ }} {{end-eqn}} This somewhat cumbersome calculation shows that we indeed have associativity. {{qed|lemma}} It follows that $\struct {S, \circ}$ is indeed a group. {{qed}}