Opposite Group is Group

Theorem
Let $\left({G, \circ}\right)$ be a group.

We define a new product $*$ on $G$ by:
 * $\forall a, b \in G: a * b = b \circ a$

Then $\left({G, *}\right)$ is a group, called the opposite group to $G$.

Proof
We need to prove that this $\left({G, *}\right)$ is actually a group.

G0: Closure
$\left({G, *}\right)$ is closed: $b \circ a \in G \Longrightarrow a * b \in G$.

G1: Associativity
$*$ is associative on $G$:

G2: Identity
Let $e$ be the identity of $\left({G, \circ}\right)$:


 * $a * e = e \circ a = a$
 * $e * a = a \circ e = a$

Thus $e$ is the identity of $\left({G, *}\right)$.

G3: Inverses
Let the inverse of $a \in \left({G, \circ}\right)$ be $a^{-1}$:


 * $a * a^{-1} = a^{-1} \circ a = e$
 * $a^{-1} * a = a \circ a^{-1} = a$

Thus $a^{-1}$ is the inverse of $a \in \left({G, *}\right)$

So all the group axioms are satisfied, and $\left({G, *}\right)$ is a group.