5. Solution 1. We must have
σ1=1, 2, 3 or 4; in each case we find
σ1=
σ3, a contradiction.

Solution 2. Let

. Then we show
στ=(
a b c d) is a cycle, contrary to
στ=(1 2)(3 4):

6. If σk=k, then σ−1k=σ−1(σk)=k. If also τk=k, then (τσ)k=τ(σk)=τk=k.
7.
a. Here

where
a,
b,
c,
d are 2, 3, 4, 5 in some order. Thus there are 4 choices for
a, 3 for
b, 2 for
c, and 1 for
d; and so we have 4·3·2·1=4!=24 choices in all for
σ.
b. Now where a, b, c are ...