- The set \(\{ e , p , q , r , s \}\) forms a group, \(A\), under the operation *
Given that \(e\) is the identity element and that
$$p ^ { * } p = s \quad s ^ { * } s = r \quad p ^ { * } p ^ { * } p = q$$
- show that
- \(p ^ { * } q = r\)
- \(s ^ { * } p = q\)
- Hence complete the Cayley table below.
| * | \(e\) | \(\boldsymbol { p }\) | \(\boldsymbol { q }\) | \(r\) | \(s\) |
| \(e\) | | | | | |
| \(\boldsymbol { p }\) | | | | | |
| \(\boldsymbol { q }\) | | | | | |
| \(\boldsymbol { r }\) | | | | | |
| \(S\) | | | | | |
A spare table can be found on page 11 if you need to rewrite your Cayley table. - Use your table to find \(p ^ { * } q ^ { * } r ^ { * } s\)
A student states that there is a subgroup of \(A\) of order 3
- Comment on the validity of this statement, giving a reason for your answer.
\includegraphics[max width=\textwidth, alt={}, center]{989d779e-c40a-4658-ad98-17a37ab1d9e1-11_2464_74_304_36}
Only use this grid if you need to rewrite the Cayley table.
| * | \(e\) | \(\boldsymbol { p }\) | \(\boldsymbol { q }\) | \(r\) | \(s\) |
| \(e\) | | | | | |
| \(\boldsymbol { p }\) | | | | | |
| \(\boldsymbol { q }\) | | | | | |
| \(\boldsymbol { r }\) | | | | | |
| \(S\) | | | | | |