This is the task im trying to solve:
This seems so obvious. Axiom 1 where $f(e,{i,j})={i,j}$ is clearly right, since e is just the identity permutation, but i have no idea how to prove it formally. Do i just explain how the identity permutation works?
Same with axiom two. Permutations act transitively, i understand that. I just dont know how to formalize it.
First, you have to prove that $\sigma\in S_5 \wedge \{i,j\}\in X\Rightarrow \{\sigma(i),\sigma(j)\}\in X$, and this is true because by injectivity:
$$\sigma\in S_5\wedge \{i,j\}\in X \Rightarrow \sigma(i)\ne\sigma(j)\Rightarrow \{\sigma(i),\sigma(j)\}\in X \tag 1$$
Then, you have to prove that $\iota_{S_5}\cdot \{i,j\}=\{i,j\}, \forall\{i,j\}\in X$, and this is true because:
$$\iota_{S_5}\cdot \{i,j\}=\{\iota_{S_5}(i),\iota_{S_5}(j)\}=\{i,j\}, \forall\{i,j\}\in X \tag 2$$
Finally, you have to prove that $(\sigma\tau)\cdot\{i,j\}=\sigma\cdot(\tau\cdot\{i,j\}),\forall \sigma,\tau\in S_5, \forall \{i,j\}\in X$, and this is true because:
\begin{alignat}{1} (\sigma\tau)\cdot\{i,j\}&=\{(\sigma\tau)(i),(\sigma\tau)(j)\} \\ &=\{\sigma(\tau(i)),\sigma(\tau(j))\} \\ &=\sigma\cdot\{\tau(i),\tau(j)\} \\ &=\sigma\cdot(\tau\cdot\{i,j\})\\ \tag 3 \end{alignat}
So, perhaps just $(1)$ deserved some attention.