1.4 Permutations
1.
a.
c.
e.
3.
a.
c.
e.
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):
. 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.
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 ...
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