Editing Order (group theory) (section)

==Example==
The [[symmetric group]] S<sub>3</sub> has the following [[Cayley table|multiplication table]].
:{| class="wikitable"
|-
!   •
! ''e'' || ''s'' || ''t'' || ''u'' || ''v'' || ''w''
|-
! ''e''
| <span style="color:#009246">''e''</span> || ''s'' || ''t'' || ''u'' || ''v'' || ''w''
|-
! ''s''
| ''s'' || <span style="color:#009246">''e''</span> || ''v'' || ''w'' || ''t'' || ''u''
|-
! ''t''
| ''t'' || ''u'' || <span style="color:#009246">''e''</span> || ''s'' || ''w'' || ''v''
|-
! ''u''
| ''u'' || ''t'' || ''w'' || <span style="color:#009246">''v''</span> || ''e'' || ''s''
|-
! ''v''
| ''v'' || ''w'' || ''s'' || ''e'' || <span style="color:#009246">''u''</span> || ''t''
|-
! ''w''
| ''w'' || ''v'' || ''u'' || ''t'' || ''s'' || <span style="color:#009246">''e''</span>
|}

This group has six elements, so {{math|1=ord(S<sub>3</sub>)&nbsp;= 6}}. By definition, the order of the identity, {{math|''e''}}, is one, since {{math|1=''e'' <sup>1</sup> = ''e''}}. Each of {{math|''s''}}, {{math|''t''}}, and {{math|''w''}} squares to {{math|''e''}}, so these group elements have order two: {{math|1={{!}}''s''{{!}} = {{!}}''t''{{!}} = {{!}}''w''{{!}} = 2}}. Finally, {{math|''u''}} and {{math|''v''}} have order 3, since {{math|1=''u''<sup>3</sup>&nbsp;= ''vu''&nbsp;= ''e''}}, and {{math|1=''v''<sup>3</sup>&nbsp;= ''uv''&nbsp;= ''e''}}.