Question

Show that the center of S₃is the identity subgroup.

Short answer

Answer

It is proved that Q^** is not a cyclic group.

Step-by-step solution

Step-by-Step Solution Step 1: Condition for an identity subgroup

Step 2: Show that the center of is the identity subgroup

Show that the center of S3 is the identity subgroup

For all non-identity element $a\in S_3$, there is some $b\in S_3$such that $ab\ne ba$.

Let $a=\left(\begin{array}{ccc}1& 2& 3\\ 2& 1& 3\end{array}\right)andb=\left(\begin{array}{ccc}1& 2& 3\\ 1& 3& 2\end{array}\right)$

Find ab as follows:

$$\begin{gathered} \begin{array}{c}ab=\left(\begin{array}{ccc}1& 2& 3\\ 2& 1& 3\end{array}\right)\left(\begin{array}{ccc}1& 2& 3\\ 1& 3& 2\end{array}\right)\\ =\left(\begin{array}{ccc}1& 2& 3\\ 2& 3& 1\end{array}\right)\end{array} \\ Findbaasfollows \\ \begin{array}{c}ba=\left(\begin{array}{ccc}1& 2& 3\\ 1& 3& 2\end{array}\right)\left(\begin{array}{ccc}1& 2& 3\\ 2& 1& 3\end{array}\right)\\ =\left(\begin{array}{ccc}1& 2& 3\\ 3& 1& 2\end{array}\right)\end{array} \end{gathered}$$

It can be observed that $ab\ne ba$.

Similarly, let $a=\left(\begin{array}{ccc}1& 2& 3\\ 3& 2& 1\end{array}\right)andb=\left(\begin{array}{ccc}1& 2& 3\\ 2& 3& 1\end{array}\right)$

Find abas follows:

$$\begin{gathered} \begin{array}{c}ab=\left(\begin{array}{ccc}1& 2& 3\\ 3& 2& 1\end{array}\right)\left(\begin{array}{ccc}1& 2& 3\\ 2& 3& 1\end{array}\right)\\ =\left(\begin{array}{ccc}1& 2& 3\\ 1& 3& 2\end{array}\right)\end{array} \\ Findbaasfollows \\ \begin{array}{c}ba=\left(\begin{array}{ccc}1& 2& 3\\ 2& 3& 1\end{array}\right)\left(\begin{array}{ccc}1& 2& 3\\ 3& 2& 1\end{array}\right)\\ =\left(\begin{array}{ccc}1& 2& 3\\ 2& 1& 3\end{array}\right)\end{array} \end{gathered}$$

It can be observed that$ab\ne ba$ .

Thus, it is proved that the center of S₃is the identity subgroup.