Abstract Algebra: Bijection, Isomorphism, Symmetric Sets

• RJLiberator
In summary: Thanks so much again.In summary, we need to prove that there exists a homomorphism F from the set of bijections of X to the symmetric set S_n, such that for any bijection pi in Bij(X), F(pi) = F(pi') implies that pi = pi', and for any element sigma in S_n, there exists a bijection pi in Bij(X) such that F(pi) = sigma. This can be done by constructing F and showing that it is one-to-one and onto, and satisfies the homomorphism property.
RJLiberator
Gold Member

Homework Statement

Suppose X is a set with n elements. Prove that Bij(X) ≅ S_n.

Homework Equations

S_n = Symmetric set
≅ = isomorphism

Definition: Let G and G2 be groups. G and G2 are called Isomorphic if there exists a bijection ϑ:G->G2 such that for all x,y∈G, ϑ(xy) = ϑ(x)ϑ(y) where the LHS is operation in G and the RHS is operation in G2.

The Attempt at a Solution

So if we have a set X with n elements.
A Bijection simply sends one element to some other unique element.
The symmetric operation just sends one element to a unique other element as well.

So clearly both sides have unique elements.

IF we take ϑ(xy) in the Bij(x) that sends them to ϑ(x)ϑ(y) in the symmetric group

I don't know enough about bijections to prove this tho.

Let ##Z(n)\equiv \{1,...,n\}##. Then ##S_n## is the set of all bijective maps from ##Z(n)## to itself.

##Bij(X)## is the set of all bijective maps from ##X## to itself.

When we say ##X## has ##n## elements, we mean precisely that there exists a bijection ##\psi: X\to Z(n)##.

Given all that, for any bijection ##\pi:X\to X##, we can construct a bijection ##F(\pi):Z(n)\to Z(n)## - by composing the maps ##\psi## and ##\pi## and/or their inverses in a suitable fashion. So construct that. If you're not sure, draw a four-cornered map diagram, with two adjacent corners being ##Z(n)## and two being ##X##.

If there is an isomorphism of the sort you want, the map ##F:Bij(X)\to S_n## will be it.

Then all you need to prove is that ##F## is one-to-one and onto, that is, that:

1. ##F(\pi)=F(\pi')\Rightarrow \pi=\pi'##
2. ##\forall \sigma\in S_n## there exists ##\pi:X\to X## such that ##F(\pi)=\sigma##

Edit: I just realized you used the isomorphism symbol. So we not only need to prove that ##F## is a bijection. We also need to prove it's a homomorphism, that is, that

3. ##F(\pi)\circ F(\pi')=F(\pi\circ\pi')## where ##\circ## denotes function composition.

These three things, plus the initial construction of ##F##, are fairly straightforward. No tricks are needed.

RJLiberator
Thank you for the help. You are making this problem already much more clear to me.

So, what I need to do is show that the composition of the two maps (aka F) are one to one and onto.

I'm not quite sure what you mean by four-cornered map diagram. But I think you simply mean a square with the 4 corners being the points as you described.

I can try to prove one-to-one and onto tomorrow with this understanding.

1. What is a bijection in abstract algebra?

A bijection is a function between two sets that is both one-to-one and onto. This means that every element in the first set is paired with exactly one element in the second set, and every element in the second set has a unique pair in the first set.

2. How is bijection related to isomorphism?

Bijection is one of the properties required for a function to be considered an isomorphism. In addition to being one-to-one and onto, an isomorphism must also preserve the algebraic structure of the sets it is mapping between.

3. What are symmetric sets in abstract algebra?

Symmetric sets are sets where the order of the elements does not matter. This means that for any two elements in the set, their positions can be swapped without changing the overall structure of the set.

4. How is isomorphism used in abstract algebra?

Isomorphism is used to show that two algebraic structures are essentially the same, even if they may look different at first glance. It allows us to study the properties of one structure by using the properties of the other.

5. Can two sets be isomorphic but not bijections?

Yes, it is possible for two sets to be isomorphic but not bijections. This can happen when there is a function between the sets that is one-to-one and onto, but does not preserve the algebraic structure of the sets. In this case, the function would not be considered an isomorphism.

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