Past exam of the mathematics course of the University of Cambridge / 2020 / ia / Paper 2 / 1E (Groups)

A transposition exchanges exactly two letters and fixes the rest. A permutation cycle $(a_1{a}_2\dots ,a_k)$ sends each displayed letter to the next, sends $a_k$ to $a_1$, and fixes all other letters.

After imposing $\sigma (1)=2$, the remaining $n-1$ domain elements may be sent bijectively to the remaining $n-1$ values in arbitrary order. Hence the number is

There are $\lfloor n/2\rfloor$ even letters. The stated condition and injectivity force the even letters to permute among themselves, which can be done in $\lfloor n/2\rfloor !$ ways. The remaining odd letters must then permute among themselves. Thus the number is

Choose the support of the four-cycle and then cyclically order it. A chosen set of four letters has $(4-1)!=6$ distinct cycles, so the answer is

The product of two transpositions has three possible forms. Equal transpositions produce the identity, transpositions meeting in one letter produce a three-cycle, and disjoint transpositions produce two disjoint two-cycles. There are two three-cycles on each chosen triple and three pairings of each chosen four-set. Therefore the number of distinct permutations representable this way is