Question

Prove that the center of ${\mathbf{S}}_{\mathrm{n}}(\mathbf{n}\mathbf{>}\mathbf{2})$ is the identity subgroup.

Short answer

It is proved that the center of ${\mathrm{S}}_{\mathrm{n}}(\mathrm{n}>2)$ is the identity subgroup.

Step-by-step solution

Statement of theorem 7.23

Theorem 7.23 states that when $\sigma =(a_1{a}_2\cdots a_k)$ and $\tau =(b_1{b}_2\cdots b_r)$ can be disjoint cycles in $S_n$, then $\sigma \tau =\tau \sigma$.

Show that the center of Sn(n>2) is the identity subgroup

Consider that $\tau \in S_n$as the generic elements of factorization into disjoint cycles $\tau ={\prod }_{\mathrm{i}=1}^{\mathrm{p}}({\mathrm{a}}_1^{\mathrm{i}}\cdots {\mathrm{a}}_{\mathrm{k}}^{\mathrm{i}})$ . Assume that at least one of the $k^i$ would be greater than 2, which might be regarded as $k^1$ without loss of generality, therefore

$$\begin{gathered} \tau \left(a_1^1{a}_2^1\right)=(a_1^1{a}_3^1\cdots a_{k^1}^1)\prod _{i=2}^P(a_1^i\cdots a_{k^i}^i) \\ \left(a_1^1{a}_2^1\right)\tau =(a_1^1{a}_3^1\cdots a_{k^1}^1)\prod _{i=2}^P(a_1^i\cdots a_{k^i}^i) \end{gathered}$$

As a result, $T$ would be not in the center. It is observed that this also demonstrates that transpositions are not in the center.

It is still necessary to demonstrate that the product of more than one transpositions will not be in the center. Assume that in the factorization of role="math" localid="1654328779774" $\tau$, there is all role="math" localid="1654328786940" $k^i$equal 2, then

$\begin{array}{c}\tau \left(a_1^1{a}_2^1{a}_1^2\right)=\left(a_2^1{a}_1^2{a}_3^2\right)\prod _{i=3}^P\left(a_1^i{a}_2^i\right)\\ \left(a_1^1{a}_2^1{a}_1^2\right)\tau =\left(a_1^1{a}_1^2{a}_3^2\right)\prod _{i=3}^P\left(a_1^i{a}_2^i\right)\end{array}$

Hence, it is proved that the center of ${\mathrm{S}}_{\mathrm{n}}(\mathrm{n}>2)$ is the identity subgroup.