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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

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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 $6$.

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  • $\begingroup$ I completed the table and can see why it is commutative, but do you know what the first part of the question is asking, "Prove that $G=\{σ0,σ1,σ2,σ3,σ4,σ5\}$ form a group of permutations?" Does creating the table answer the fact that they form a group of permutations? $\endgroup$ – idknuttin Nov 13 '16 at 1:26
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    $\begingroup$ @idknuttin Nope, but you can use the multiplication table to verify this fact. Since the table consists only of the elements of $G$, $G$ is closed. Next, you can notice that $\sigma_0$ is the identity of the group. Lastly, you can find that $\sigma_1^{-1}=\sigma_2$, $\sigma_2^{-1}=\sigma_1$,$\sigma_3^{-1}=\sigma_3$,$\sigma_4^{-1}=\sigma_5$,$\sigma_5^{-1}=\sigma_4$ $\endgroup$ – Alan Wang Nov 13 '16 at 4:39

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