rubiks cube

Associativity:
Identity:
Let $g \in G$ . By definition of $G$ , we know there exists $m_{1}, m_{2}, \dots, m_{n} \in M$ such that $g = m_{1} ⋆ m_{2} ⋆ \dots ⋆ m_{n}$ for some $n \in ℕ_{1}$ . By definition of $e$ , $m ⋆ e = e ⋆ m = m$ for all $m \in M$ . So by associativity,
$e ⋆ g & = e ⋆ (m_{1} ⋆ m_{2} ⋆ \dots ⋆ m_{n}) & = (e ⋆ m_{1}) ⋆ m_{2} ⋆ \dots ⋆ m_{n} & = (m_{1} ⋆ e) ⋆ m_{2} ⋆ \dots ⋆ m_{n} & = m_{1} ⋆ (e ⋆ m_{2}) ⋆ \dots ⋆ m_{n} & = m_{1} ⋆ (m_{2} ⋆ e) ⋆ \dots ⋆ m_{n} & = m_{1} ⋆ m_{2} ⋆ \dots (e ⋆ m_{n}) & = m_{1} ⋆ m_{2} ⋆ \dots (m_{n} ⋆ e) & = (m_{1} ⋆ m_{2} ⋆ \dots ⋆ m_{n}) ⋆ e & = g ⋆ e$ Therefore, $e$ is the identity element in $G$ .
Inversion:
Let $m \in M \subset G$ . Then, $m ⋆ m^{3} = m^{4} = e$ and $m^{3} ⋆ m = m^{4} = e$ . So, $m^{- 1} = m^{3}$ .

Let $g \in G$ . By definition of $G$ , we know there exists $m_{1}, m_{2}, \dots, m_{n} \in M$ such that $g = m_{1} ⋆ m_{2} ⋆ \dots ⋆ m_{n}$ for some $n \in ℕ_{1}$ . Now, consider $g^{'} = m_{n}^{3} ⋆ \dots ⋆ m_{2}^{3} ⋆ m_{1}^{3}$ . By associativity, we have the following. $g ⋆ g^{'} & = (m_{1} ⋆ m_{2} ⋆ \dots ⋆ m_{n}) ⋆ (m_{n}^{3} ⋆ \dots ⋆ m_{2}^{3} ⋆ m_{1}^{3}) & = m_{1} ⋆ m_{2} ⋆ \dots ⋆ (m_{n} ⋆ m_{n}^{3}) ⋆ \dots ⋆ m_{2}^{3} ⋆ m_{1}^{3} & = m_{1} ⋆ m_{2} ⋆ \dots ⋆ e ⋆ m_{n - 1}^{3} ⋆ \dots ⋆ m_{2}^{3} ⋆ m_{1}^{3} & = m_{1} ⋆ m_{2} ⋆ \dots ⋆ \dots ⋆ m_{2}^{3} ⋆ m_{1}^{3}$

🏗️ $Θ ρ ϵ η Π α τ π$ 🚧 (under construction)

Rubik's Cube Move Inverter

🏗️ ΘρϵηΠατπ🚧 (under construction)

Rubik's Cube Move Inverter

🏗️ $Θ ρ ϵ η Π α τ π$ 🚧 (under construction)