Symmetric Group on 3 Letters/Subgroups/Examples/Non-Subgroup

From ProofWiki
Jump to navigation Jump to search

Example of Subset of Symmetric Group on 3 Letters which is not a Subgroup

Let $S_3$ denote the Symmetric Group on $3$ Letters, whose Cayley table is given as:

$\begin{array}{c|cccccc} \circ & e & (123) & (132) & (23) & (13) & (12) \\ \hline e & e & (123) & (132) & (23) & (13) & (12) \\ (123) & (123) & (132) & e & (13) & (12) & (23) \\ (132) & (132) & e & (123) & (12) & (23) & (13) \\ (23) & (23) & (12) & (13) & e & (132) & (123) \\ (13) & (13) & (23) & (12) & (123) & e & (132) \\ (12) & (12) & (13) & (23) & (132) & (123) & e \\ \end{array}$


Consider the subset $H$ of $S_3$:

$H = \set {e, \tuple {12}, \tuple {13}, \tuple {23} }$

Then $H$ is not a subgroup of $S_3$.


Proof

We have:

$\tuple {12} \circ \tuple {13} = \tuple {123}$

But $\tuple {123} \notin H$.

Thus $H$ is not closed under $\circ$.

Hence the result by definition of subgroup.

$\blacksquare$


Sources