Let $S=\{1,2,3,4,5,6,7\}$ and let
$\sigma_0=(1)(2)(3)(4)(5)(6)(7)$
$\sigma_1=(1,2,3)(4)(5)(6)(7)$
$\sigma_2=(1,3,2)(4)(5)(6)(7)$
$\sigma_3=(1)(2)(3)(4)(5,6)(7)$
$\sigma_4=(1,2,3)(4)(5,6)(7)$
$\sigma_5=(1,3,2)(4)(5,6)(7)$
Prove that $G=\{σ0,σ1,σ2,σ3,σ4,σ5\}$ form a group of permutations and determine the multiplication table for the group. Is the obtained group commutative?
I don't understand what exactly I have to do here, do I have to create a $6 \times 6$ table and multiply each permutation with each other? What exactly is a group of permutations and a multiplication table?
Is the final table on this page what I have to do?
http://groupprops.subwiki.org/wiki/Determination_of_multiplication_table_of_symmetric_group:S3
For $1$-cylce it can be omitted.
$\sigma_0=1$(identitiy function)
$\sigma_1=(123),\sigma_2=(132),\sigma_3=(56),\sigma_4=(123)(56),\sigma_5=(132)(56)$
The law of composition defined for symmetric group is not multiplication but is composition of mapping.
For example, $\sigma_1\sigma_4=(123)(123)(56)$. Look from right to left, $1\rightarrow2\rightarrow3$, $3\rightarrow1\rightarrow 2$, $2\rightarrow 3\rightarrow1$, $5\rightarrow6$, $6\rightarrow5$.
So $\sigma_1\sigma_4=(132)(56)$. By using this method, you can create a multiplication table and prove that $G$ satisfies axioms for closure, identity and inverse. It is not necessary to prove the axiom for associativity because the composition of mapping is itself associative.
Another interpretation: I write $\alpha=(123),\beta=(56)$. Then $$G=\{1,\alpha,\alpha^{-1},\beta,\alpha\beta,\alpha^{-1}\beta\}$$ Since $\alpha\beta=\beta\alpha$(you can easily verify it or by using the fact that they are disjoint), $G$ is commutative. And in fact $G$ is cyclic group of order $6$.