Thread: The set of all permutations of a A with composition operation is a group:How to show?

Let be the subset of consisting of the permutations:









Now show that is a group of permutations.


Now the problem is how do I prove this? Because I know that in order to be a group of permutations I've to show the following properties:

a) The operation of composition of functions qualifies as an operation on the set of all the permutations of

b) This operation is associative.

c) There is a permutation such that and for any permutation of .

d) Finally for every permutation of there is another permutation of such that
and

Do I exhaust all the possible values and show that the above four properties hold true for all combinations?


Is there any easy way to prove that the set of all the above permutations is a group?

Originally Posted by x3bnm

Let be the subset of consisting of the permutations:




Now show that is a group of permutations.
Is there any easy way to prove that the set of all the above permutations is a group?

Well it is it is easy to see that each of those is its own inverse.
Moreover, the set is closed under the operation.

Originally Posted by Plato

Well it is it is easy to see that each of those is its own inverse.
Moreover, the set is closed under the operation.

Thanks Plato for help.

b) is obvious: composition of functions is associative:

if h(x) = y, and g(y) = z and f(z) = w, then:

(fo(goh))(x) = f((goh)(x)) = f(g(h(x))) = f(g(y)) = f(z) = w

((fog)oh)(x) = (fog)(h(x)) = (fog)(y) = f(g(y)) = f(z) = w.

as Plato pointed out, every element (function) is its own inverse, so we have (d).

(c) is redundant: closure (a) & inverses (d) together imply (c). you can safely avoid proving this (it will be true: if a*b is in S whenever a,b are, and a^-1 is in S when a is, then if a is in S, then so is a^-1, so taking b = a^-1, we have:

a*b = a*a^-1 = e is in S).

there is a trick you can use to show closure:

let a = (1 2)(3 4) (this is f in your original post...by (1 2)(3 4) i mean f(1) = 2, f(2) = 1, f(3) = 4, and f(4) = 3).
let b = (1 3)(2 4) (this is g in your original post).

show your last element is a*b.

this means that G = {e,a,b,a*b}

show that a*b = b*a. this implies that G is abelian (why? hint: a and b are generators...why?). this cuts down the number of products you have to verify closure for down from 16 to 10. 4 of those you will already have verified by proving you have inverses (the ones of the form g*g). that leaves just 6 to check. if you happen to notice that ε is the identity function on {1,2,3,4}, this makes 3 of the remaining products to check for closure trivial. that leaves just three products you have to work for:

a*b <---but you'll already have calculated this verifying a*b = b*a
a*(a*b) <---asociativity and inverses make this trivial
b*(a*b) <---use associativity and the fact that b*a = a*b, and what you learned about inverses.

extra credit: show that G is isomorphic to the klein 4-group V (for vierergruppe, german for "four-group"), and also to the direct product: (Z₂xZ₂,+) (where + means addition mod 2, the elements are:

{(0,0),(1,0),(0,1),(1,1)} you can think of this as "a pair of unlinked switches").

Originally Posted by x3bnm

Let be the subset of consisting of the permutations: Is there any easy way to prove that the set of all the above permutations is a group?

My reply was an answer to the above question.
I assumed that you know that is a group.

There is a standard theorm: If is a subset of a group then is a subgroup of if and only if implies that .

but...but...the "two-step" makes a much nicer dance....

Plato thanks a lot for the neat theory of yours. It's heck of a lot of time saver.

Deveno, thanks for the explanation. Very interesting reading and pretty smart notation you showed there for closure. It's very concise. Your explanation makes me want to ask for more of Abstract Algebra.