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.