Note that, in the table of example 5.4.1, s4∘s5=s1 (compute from right to left, id est s5 then s4).
(a) s1∘s1=I, thus there are only two elements in <s1>.
(b) s5∘s5=I, its order is 2.
(c) s3∘s3=s4, s4∘s4=s3, s3∘s4=s4∘s3=I. If we just keep I, s3 and s4, we have the following sub-table:
| ∘ | I | s3 | s4 |
|---|---|---|---|
| I | I | s3 | s4 |
| s3 | s3 | s4 | I |
| s4 | s4 | I | s3 |
It’s the table of C3.
(d) According to the table above, the order of s3 is 3.
(e) I=s1∘s1, s1=s1, s2=s2.
s3=s2∘s1, s4=s1∘s2, s5=s2∘s1∘s2.