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This one has been bugging me for a while - I have a number of definitions for a symmetric group, none of which I understand. Clearly, $S_n$ is a group, which I follow. Now, the wikipedia definition (most readily to hand) is:

The symmetric group on a set X is the group whose underlying set is the collection of all bijections from X to X and whose group operation is that of function composition.

Of that, I can see (clearly) we have an underlying set which combined with an operation follows set rules as per the definition of a group. I understand a bijection from X to X implies that $\forall x \in X$, $x \mapsto x$. So, just as in geometric symmetry a set is symmetrical if it becomes itself again after some map.

All well and good, but how do composite functions relate to this? Surely taking the bijective requirement into consideration the only operation permitted is one that effectively leaves the set unchanged?

So, can anyone please either explain how composite functions fit in my definition or (equally fine) provide an example of a function that works so I can try to fit the idea to the definitions I've come across.

Many thanks.

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up vote 8 down vote accepted

I think you misunderstood the definition/use of bijection. I'll use an example:

For $X = \{1, 2, 3\}$, the appropriate symmetric group is $S_3$, and it is the set of all possible permutations of X:

  • $f_1: 1\to 1, 2\to 2, 3\to 3$
  • $f_2: 1\to 1, 2\to 3, 3\to 2$
  • $f_3: 1\to 3, 2\to 2, 3\to 1$
  • $f_4: 1\to 2, 2\to 1, 3\to 3$
  • $f_5: 1\to 2, 2\to 3, 3\to 1$
  • $f_6: 1\to 3, 2\to 1, 3\to 2$

Generally, bijections from a set X to itself are permutations, and intuitively they are "re-orderings" of the members of the set. It then follows that for a set of size n there are exactly n! permutations, and thus $|S_n| = n!$.

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I see, so the group S is in fact made up of member functions and the group operation is composition. Of course! Thank ye very much! – user892 Aug 6 '10 at 23:34
@Ninefingers: The $f_i$ are the bijections themselves. The members of $X$ represent the corners of a triangle, canonically numbered clockwise for the identity triangle. Each of the $f_i$ transform that triangle by flipping, rotation, or identity -- these are all symmetries. Thus, "symmetry group." – Heath Hunnicutt Aug 7 '10 at 2:15

Each member of $S_n$ is a bijection and the composition of two bijections is another bijection. Therefore $S_n$ is closed under composition.

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The original question claims that the author doesn't yet understand the definitions of the symmetric group.

In this case, I think it's best to start with (what I think) is the most accessible definition.

$S_n$, the symmetric group on the $n$ numbers $1,\ldots,n$ is the set of all possible ways to permute the set of numbers $1,\ldots,n$

For example, one possible permutation (element of $S_n$) is

$g = (1,2,\ldots,n) \Rightarrow (2,3,\ldots,n,1)$

This set of permutations forms a group: the multiplication is simply performing one permutation, and then the next. So

$g\cdot g = g^2: (1,2,\ldots,n) \Rightarrow (3,4,\ldots,n,1,2)$

I think you should get comfortable with this definition first. Once you think about it for a while, you'll realize that each permutation can be represented as a function on the set ${1,\ldots,n}$. For example, the permutation $g$ above would be the function that sends $1 \mapsto 2$, $2 \mapsto 3$, and so on (in general, $n \mapsto n + 1$, where $n$ maps back to $1$).

If you think a little further, you'll realize that any function you get from a permutation in such a fashion must be bijective (that is, a permutation can't, say, send all the numbers $1,\ldots,n$ to $1$, since that's clearly not a permutation).

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