Recall that the notation means that f is a function whose domain (set of inputs) is X and whose outputs lie in the set Y. Note that there may be elements of Y which are not outputs of f.
Definition. Let be a function from a set X to a set Y.
1. f is injective (or one-to-one) if implies for all .
2. f is surjective (or onto) if for all , there is an such that .
3. f is bijective (or a one-to-one correspondence) if it is both injective and surjective.
Informally, a function is injective if different inputs always produce different outputs. A function is surjective if everything in the target set is an output of the function.
Example. ( Injective and surjective functions) Consider the function given by . f is {\it not} injective, because
Nor is f surjective. There is no , for instance, such that .
Note, however, that if is defined by , then g is surjective. ( denotes the set of real numbers greater than or equal to 0.) I just shrunk the target set so that it coincides with the set of outputs of .
Example. ( Injective and surjective functions) Consider the function given by . f is injective: If two outputs are the same, say
That is, the inputs must have been the same.
This is one way to show that a function f is injective: Assume that , and prove that .
However, f is not surjective: There is no such that , i.e. such that , because is always positive.
You may know that there is a graphical test for injectivity for functions . A function is injective if and only if every horizontal line intersects the graph at most once. You can see that this is true for the graph of .
Example. ( Injective and surjective functions) Define by
f is not injective, since and : Different inputs can produce the same output.
f is surjective: You can see from the graph that every y-value is an output of the function. To prove this algebraically, I have to show that every is an output of f.
If , .
If , then , so .
To prove a function is surjective, take an arbitrary output y and find an input that produces it. As in this example, your input may be specified in terms of y, since that is given.
While you can show that a function is bijective by showing that it's injective and surjective, there's a method which is usually easier: Simply produce an inverse function.
Definition. Let be a function from a set X to a set Y. An inverse for f is a function such that:
1. For all , .
2. For all , .
The next result is extremely useful. It asserts that being bijective is the same as having an inverse.
Lemma. Let be a function from a set X to a set Y. f is bijective if and only if f has an inverse .
Proof. ( ) Suppose that f is bijective. I'll construct the inverse function .
Take . Since f is surjective, there is an element such that . Moreover, x is unique: If and , then . But f is injective, so .
Define
I have defined a function . I must show that it is the inverse of f.
Let . By definition of , to compute I must find an element such that . But this is easy --- just take . Thus, .
Going the other way, let . By definition of , to compute I must find an element such that . Then , so
Therefore, really is the inverse of f.
( ) Suppose f has an inverse . I must show f is bijective.
To show that f is surjective, take . Then , so I've found an element of X --- namely --- which f maps to y. Therefore, f is surjective.
To show that f is injective, suppose and . Then
Therefore, f is injective.
Since f is injective and surjective, it's bijective.
This result says that if you want to show a function is bijective, all you have to do is to produce an inverse. In many cases, it's easy to produce an inverse, because an inverse is the function which "undoes" the effect of f.
Example. ( Proving that a function is bijective) Define by . I'll show f is bijective.
The opposite of cubing is taking the cube root, so I'll guess that the inverse is . Check it:
Thus, g is the inverse of f. By the lemma, f is bijective.
Definition. Let A be a set. A permutation of (or {\it on}) A is a bijection .
Proposition. The set of permutations of a set A is a group under function composition.
Proof. First, the composition of bijections is a bijection: The inverse of is . Thus, function composition is a binary operation on the set of bijections from A to A.
Function composition is always associative. The identity map is a permutation of A, and serves as an identity under function composition. Since bijective maps have inverses which are bijections, if is a bijection, so is . Therefore, is a group.
Terminology. is called the symmetric group on A. If S has n elements, you may as well take (since it doesn't matter what you call the elements). The corresponding symmetric group is denoted , the symmetric group on n letters.
I'll use to denote the identity permutation that sends every element to itself.
One way to write a permutation is to show where each element goes. For example, suppose
This means that
The identity permutation in is
Example. ( Computing a product of permutations) Suppose
The product means " first, then ".
Here's how to compute it:
So
Some people prefer to multiply permutations left-to-right: For them, means " first, then ". You should probably choose one method and use it all the time, to avoid confusing yourself. The right-to-left approach I used above is consistent with the fact that permutations are functions: In function notation, means , i.e. "g first, then f".
Example. ( Finding the inverse of a permutation) It's easy to find the inverse of a permutation. For example, if
I can find by simply reading "upside-down".
For example, , so . In this way, I get
I'm going to find a nicer way to represent permutations. The idea is like factoring an integer into a product of primes; in this case, the elementary pieces are called cycles.
Definition. A cycle is a permutation which maps a finite subset by
This cycle will be denoted .
The cycle has length n. A cycle of length n has order n in . For example,
A cycle of length 2 is called a transposition. A transposition is a permutation that swaps two elements and leaves everything else fixed. For example, is the transposition that swaps 3 and 6.
Example. ( Examples of cycles) The cycle in is the permutation
Likewise,
Example. ( The inverse of a cycle) To find the inverse of a cycle, just run the cycle backwards. Thus,
Example. ( Solving a permutation equation) Solve for x:
and . So the equation is
Hence,
Example. ( A permutation which is not a cycle) Not every permutation is a cycle. For example, the permutation
can be written as a product of the cycles and . Note that the cycles and are disjoint --- no element is moved by both of them. In fact, an arbitrary permutation may be written as a product of disjoint cycles. Every permutation may also be written as a product of transpositions.
The last example is a particular case of the following theorem.
Theorem. Every permutation on a finite set can be written as a product of disjoint cycles.
Proof. Induct on the number of elements in the set.
First, prove the result for a set with 1 element. This is easy --- there is only one permutation (the identity), and it is the cycle .
Next, assume that the result is known for sets with fewer than n elements and try to prove the result for a set with n elements. Suppose, then, that a permutation on a set with less than n elements can be written as a product of disjoint cycles. I have to show that a permutation on a set with n elements --- that is, an element --- can be written as a product of disjoint cycles.
Since is a finite group, has finite order. Let m be the order of . Look at
If , is the cycle
Otherwise, , so .
Now restricted to is a permutation on , so by the inductive assumption it can be written as a product of disjoint cycles. Then
Thus, has been expressed as a product of disjoint cycles. This completes the induction step, and establishes the result for all n.
Example. ( Writing a permutation as a product of cycles) The proof actually contains an algorithm for decomposing a permutation into a product of disjoint cycles. Start with an element and follow its "orbit" under the permutation until the orbit closes up. If you've exhausted all the elements, you're done. Otherwise, pick an element which wasn't in the orbit of the first element and follow the new element's orbit. Keep going.
For example,
Here's a picture which shows how I got : 1 goes to 6, which goes to 5, which goes to 4, which goes back to 1.
After finishing a cycle, I start with the next element that hasn't been "used" so far. I keep going until all the elements have been accounted for.
If you have a permutation like in which an element doesn't move --- in this case, 2 --- you don't need to write " ". 2 is simply omitted from the cycle list, since an element which isn't listed doesn't move.
Lemma. Disjoint cycles commute.
Proof. Roughly speaking, if two cycles move different sets of elements, then their effects are independent and it doesn't matter in which order they're applied.
If and are disjoint cycles, say
where .
Then
Definition. A transposition is a permutation which interchanges two elements and leaves everything else fixed. (That is, a transposition is a cycle of length 2.)
Proposition. Every permutation is a product of transpositions.
Proof. It suffices to show that every cycle is a product of transpositions, since every permutation is a product of cycles. Just observe that
Remark. While the decomposition of a permutation into disjoint cycles is unique up to order and representation of the cycles (i.e. ), a permutation may be written as a product of transpositions in infinitely many ways. You can always tack on trivial terms of the form .
Example. ( Writing a permutation as a product of transpositions)
The decomposition of a permutation into a product of transpositions is not unique.
Definition. A permutation is even if it can be written as a product of an even number of transpositions; a permutation is odd if it can be written as a product of an odd number of transpositions.
Lemma. A permutation cannot be written as a product of both an odd and an even number of transpositions.
Proof. Suppose to the contrary that
where m is even and n is odd, and all the 's and 's are transpositions.
Since ,
Note that the identity permutation has been written as a product of an odd ( ) number of transpositions. If this is impossible, I have a contradiction.
It therefore suffices to show that the identity permutation {\it cannot} be written as a product of an odd number of transpositions. I'll give a proof by contradiction.
Suppose m is odd and
Here is a clever idea. Consider a polynomial in n variables. A permutation transforms f into another polynomial by "permuting the variables":
For example, suppose . Suppose . Then
Now consider the polynomial
For example, if ,
Obviously, the identity permutation takes f to itself.
On the other hand, a transposition for takes the factor to . In other words, a factor of -1 is introduced for each transposition. Since contains an odd number of transpositions, it will send f to .
This is a contradiction: If and are the same permutation, they should have the same effect on f. Therefore, the identity cannot be written as a product of an odd number of transpositions. Hence, a permutation cannot be written as a product of both an even and an odd number of transpositions.
Remark. The decomposition
shows that a cycle of length n is even if n is even, and odd if n is odd.
Definition. The alternating group on n letters is the subgroup of consisting of the even permutations.
I should check that really is a subgroup. First, is even, so . Next, if and are even, then is even (decompose into transpositions, and write the product backwards). Therefore, is even (by concatenating decompositions of and into products of transpositions). Hence, .
If , there are an equal number of even and odd permutations. Therefore, . In fact, is a normal subgroup of .
Example. Here is the multiplication table for :
The alternating group on 3 letters is the "rotation subgroup":
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Copyright 2013 by Bruce Ikenaga