Abstract-Algebra-1-Orbits Cycles and the Alternating Groups, guaranteed and verified 100% Pass

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Orbits, Cycles, and the Alternating Groups

Def. Let 𝜎 be a permutation of a set 𝐴. The equivalence classes in 𝐴 determined
by 𝑎~𝑏 if and only if 𝑏 = 𝜎 𝑛 (𝑎), for some 𝑛 ∈ ℤ, are called the orbits of 𝜎.


Ex. Find the orbits of the permutation:
1 2 3 4 5 6 7 8
𝜎=( ).
7 5 6 1 2 8 4 3


Let’s start with 1 and follow where it goes under powers of 𝜎:

𝜎(1) = 7, 𝜎 2 (1) = 𝜎(7) = 4, 𝜎 3 (1) = 𝜎 (4) = 1.
1 goes to 7, which goes to 4, which goes back to 1. We denote this by (1, 7, 4).


Now go to 2 and see where 𝜎 sends it:

𝜎(2) = 5, 𝜎 2 (2) = 𝜎(5) = 2.
So 2 goes to 5, which goes back to 2. We denote this by (2,5).


Now go to 3:

𝜎(3) = 6, 𝜎 2 (3) = 𝜎 (6) = 8, 𝜎 3 (3) = 𝜎(8) = 3.
So 3 goes to 6, which goes to 8, which goes back to 3. We denote this by
(3, 6, 8).

Notice that we already know what 𝜎 does to 4 (and 5 − 8). For example,

𝜎 sends 4 into 1, into 7, into 4. That’s already captured in (1, 7, 4).
So 𝜎 has 3 orbits:

(1, 7, 4), (2, 5), (3, 6, 8).

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Def. A permutation 𝜎 ∈ 𝑆𝑛 is a cycle if it has at most one orbit containing more
than one element. The length of a cycle is the number of elements in its

largest orbit.


1 2 3 4 5
Ex. The permutation ( ) is a cycle because its orbits are
4 5 3 2 1
(1, 4, 2, 5) and (3).

Let’s go back to the first example of the permutation:
1 2 3 4 5 6 7 8
𝜎=( )
7 5 6 1 2 8 4 3
with orbits: (1, 7, 4), (2, 5), (3, 6, 8).


We can associate to each orbit a permutation that is a cycle:
1 2 3 4 5 6 7 8
(1, 7, 4) = ( )
7 2 3 1 5 6 4 8
(2, 5) = (1 2 3 4 5 6 7 8)
1 5 3 4 2 6 7 8
1 2 3 4 5 6 7 8
(3, 6, 8) = ( )
1 2 6 4 5 8 7 3
and:
1 2 3 4 5 6 7 8
𝜎=( ) = (1, 7,4)(2, 5)(3, 6, 8).
7 5 6 1 2 8 4 3


That is, we can write any 𝜎 ∈ 𝑆𝐴 as a product of disjoint cycles (i.e. any integer is
moved by only one cycle).