Abstract-Algebra-1-Permutation-Groups, guaranteed and verified 100% Pass

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Permutation Groups
Def. A permutation of a set 𝐴 is a function 𝜙: 𝐴 → 𝐴 that’s 1-1 and onto

We can think of a permutation as a rearrangement of the elements of 𝐴.



Ex. Let 𝐴 = {1, 2, 3, 4, 5}. Examples of permutations:

𝜙1 ({1,2, 3, 4, 5}) = {4, 3, 1, 2, 5}
𝜙2 ({1,2, 3, 4, 5}) = {5, 2, 3, 1, 4}
𝜙1 𝜙2
1→4 1→5
2→3 2→2
3→1 3→3
4→2 4→1
5→5 5→4


We can form a new permutation by taking the composition of the above
permutations: 𝜙2 ∘ 𝜙1 ({1, 2, 3, 4, 5}). This is permutation multiplication.


𝜙2 ∘ 𝜙1 i. e. 𝜙2 ∘ 𝜙1
1→4→1 1→1
2→3→3 2→3
3→1→5 3→5
4→2→2 4→2
5→5→4 5→4

, 2


We can write 𝜙1 and 𝜙2 as:
1 2 3 4 5
𝜙1 = ( )
4 3 1 2 5
1 2 3 4 5
𝜙2 = ( ).
5 2 3 1 4


1 2 3 4 5 1 2 3 4 5
Then 𝜙2 ∘ 𝜙1 = ( )( )
5 2 3 1 4 4 3 1 2 5
1 2 3 4 5
=( ).
1 3 5 2 4

Theorem: Let 𝐴 be a nonempty set, and let 𝑆𝐴 be the set of permutations of 𝐴.
Then 𝑆𝐴 is a group under permutation multiplication.


Proof:

0) 𝑆𝐴 is clearly closed under permutation multiplication.
1) Permutation multiplication is just a composition of functions and
composition of functions is associative so this multiplication is as well.
2) The permutation 𝑖(𝑎) = 𝑎 for all 𝑎 ∈ 𝐴 acts as an identity.
3) For any permutation 𝜎, 𝜎 −1 is just the permutation 𝜎 in the opposite
direction that reverses what 𝜎 does.
1 2 3 4 5
For example, if 𝜎 = ( )
2 3 5 4 1
1 2 3 4 5
then 𝜎 −1 = ( ) thus, 𝜎 −1 ∘ 𝜎 = 𝑖 and 𝜎 ∘ 𝜎 −1 = 𝑖.
5 1 2 4 3
Thus 𝑆𝐴 is a group.



We will generally be concerned with 𝑆𝐴 where 𝐴 is a finite set, but that doesn’t
have to be the case.