Algebra. Permutations with composition proof. Set $X := \lbrace 1,2,3 \rbrace$. Denote the idenity $\text{id}_X$ of $X$ by 1. Define $1\phi :=1$, $2\phi :=3$, $3\phi = 2$, $1\psi =2$, $2\psi = 1$, $3\psi = 3$. Prove that $\phi \psi \phi= \psi \phi \psi$, abbreviate $\delta := \phi \psi \phi$, and finish the following table by computing all composites.
$$
\begin{array}{l||c|c|c|c|c|c}
 & 1 & \phi & \psi & \phi \psi & \psi \phi & \delta \\
\hline \hline
1 & 1 & \phi & \psi &  & & \\
\hline
\phi & \phi & 1 & \phi \psi &  & & \\
\hline
\psi & \psi & \psi \phi & & & & \\
\hline
\phi \psi  & & & & & & \\
\hline
\psi \phi & & & & & & \\
\hline
\delta & & & & & &
\end{array}
$$
Quick question how do I show they are equal? Do I just do it the old fashion way by choosing an element in $X$ and just computing 6 times or is there a more formal way to go about this?
For the table I can not get it to show up, but how would I evaluate things like $\phi^2$ or $\phi \psi$ ?
 A: The objects of interest here are the functions acting on the set $X$, which can be combined by composition
$$
a \alpha \beta = \beta(\alpha(a)) \quad a \in X \quad  \alpha, \beta \in X^X = \{ \varphi \,\left|\, \varphi : X \to X \right. \}
$$
the combined composition function is what is meant by the short hand product
$$
\alpha \beta = \beta \circ \alpha
$$
To get the first foot on the ground one looks how a function acts on the elements of the set. One finds that certain compositions of functions result in the same mapping of set elements like some others and thust one identifies them as the same composition. Like $\phi^2=1$ (the identity function).
Using tuples to act on several elements at once to avoid writing down three individual equations:
$$
(1,2,3) \varphi =: \left( \varphi(1), \varphi(2), \varphi(3) \right) = (x, y, z) \iff \\
\varphi(1) = x \wedge \varphi(2) = y \wedge \varphi(3) = z
$$
one gets:
$$
(1,2,3) \phi = (1,3,2) \\
(1,2,3) \psi = (2,1,3) \\
(1,2,3) \phi \psi = (1,3,2) \psi = (2,3,1) \\
(1,2,3) \psi \phi = (2,1,3) \phi = (3,1,2) \\
(1,2,3) \phi \psi \phi = (2,3,1) \phi = (3,2,1) \\
(1,2,3) \psi \phi \psi = (3,1,2) \psi = (3,2,1) \\
\implies \phi \psi \phi = \psi \phi \psi =: \delta
$$
The functions act as permutations on $X$, there are $3!=6$ different possible. 
From here on one should mostly need the associative law of function composition
$$
(\alpha \beta) \gamma = 
\gamma \circ (\beta \circ \alpha) =
(\gamma \circ \beta) \circ \alpha =
\alpha (\beta \gamma) 
$$
together with the already discovered identities:
$$
\delta^2 = \delta \delta = 
(\phi \psi \phi)(\phi \psi \phi) = 
\phi (\psi (\phi \phi) \psi) \phi = 
\phi(\psi \psi) \phi =
\phi \phi =
1 \\
\delta \phi = (\phi \psi \phi) \phi = \phi \psi \\
\phi \delta = \phi (\phi \psi \phi) = \psi \phi \\
\delta \psi = (\phi \psi \phi) \psi = \phi (\psi \phi \psi) = 
\phi \delta  = \psi \phi \\
\psi \delta = \psi (\psi \phi \psi) = \phi \psi
$$
Putting all this in the table gives:
$$
\begin{array}{l||c|c|c|c|c|c}
 & 1 & \phi & \psi & \phi \psi & \psi \phi & \delta \\
\hline \hline
1 & 1 & \phi & \psi & \phi \psi & \psi \phi & \delta \\
\hline
\phi & \phi & 1 & \phi \psi & \psi & \delta & \psi \phi \\
\hline
\psi & \psi & \psi \phi & 1 & \delta & \phi & \phi \psi \\
\hline
\phi \psi  & \phi \psi & \delta & \phi & \psi \phi & 1 & \psi \\
\hline
\psi \phi & \psi \phi & \psi & \delta & 1 & \phi \psi & \phi \\
\hline
\delta & \delta & \phi \psi & \psi \phi & \phi & \psi & 1
\end{array}
$$
A: $\phi^2$ can be calculated by taking the compositum of the function $\phi$ with itself. this means that $$1(\phi^2) = (1\phi)\phi = 1\phi = 1\\
2(\phi^2) = (2\phi)\phi = 3\phi = 2\\
3(\phi^2) = (3\phi)\phi = 2\phi = 3,$$
which proves that $\phi^2=1$  (which is also written in your table).
