Find a permutation representation

[fk378]

Homework Statement

Let G be the group S_3. Find the permutation representation of S_3. (Note: this gives an isomorphism of S_3 into S_6)

The Attempt at a Solution

Is there only ONE permutation representation, because the question asks for "the" p.r.
I don't know where to start.

 

[e(ho0n3]

For starters, what is a permutation representation?

 

[HallsofIvy]

It can be shown that every finite group is isomorphic to some group of permutation.

There is an easy way to do that: Write out the operation table for the group. Lable the elements "1", "2", ..., n according to how they are written along the top of the of the table. Now each row of the table is a permutation of 123...n Do you see how to assign each element of the group a corresponding permuation of 12...n?

 

[e(ho0n3]

I get it know. This is just an exercise using Cayley's Theorem.

 

[fk378]

It can be shown that every finite group is isomorphic to some group of permutation.

There is an easy way to do that: Write out the operation table for the group. Lable the elements "1", "2", ..., n according to how they are written along the top of the of the table. Now each row of the table is a permutation of 123...n Do you see how to assign each element of the group a corresponding permuation of 12...n?

I've written out the elements of S_3:
{e, (123), (213), (12), (13), (23)}

I'm sorry but I don't quite understand what you're describing. How do I write out the operation table?

 

[HallsofIvy]

The "operation table" is a table with the elements of the group horizontally across the top and vertically along the left side. Where a column and row intersect is the product of the two elements. For example, "Klein's four group" with elements e, a, b, c being the identity, has operation group
$$\begin{array}{ccccc} & e & a & b & c \\e & e & a & b & c \\ a & a & e & c & b \\ b & b & e & e & a \\ c & c & b & a & e\end{array}$$
If we replace "e", "a", "b", and "c" with "1", "2", "3", and "4" respectively that becomes
$$\begin{array}{ccccc} & 1 & 2 & 3 & 4 \\ 1 & 1 & 2 & 3 & 4 \\ 2 & 2 & 1 & 4 & 3 \\ 3 & 3 & 4 & 1 & 2 \\ 4 & 4 & 3 & 2 & 1 \end{array}$$
The row starting with 1 is "1 2 3 4" so "1" takes 1234 into itself, the identity permutation. The row starting with 2 is "2143" so "2" take 1234 into that, the permutation (12)(34). The row starting with 3 is "3412" so "3" takes 1234 into that, the permutation (13)(24). The row starting with 4 is "4321" so "4" takes 1234 into that, the permutation (14)(23).

The permutation representation of the Klein four group is {e, (12)(34), (13)(24), (14)(23)}.

 

[fk378]

I've made the operation table, I don't see how the symmetry will work with 3 elements on top, and 3 elements on the side.

I have that
1 sends 123-->123
this is the identity
2 sends 123-->231
this is (123)
3 sends 123-->312
this is (132)

Did I set up my table correctly?
Also, I don't see how I would get S_6 from this.

Thank you for your help so far.

 

[HallsofIvy]

S_3 does not contain 3 elements. It contains 3!= 6 elements- which is why it can be represented as a subgroup of S_6.

 

[fk378]

S_3 does not contain 3 elements. It contains 3!= 6 elements.

I thought elements of S_3 were the permutations..no? I mean, in any case I did it wrong. So would I need to do it for entries 1 to 6 instead of what I did?

Also, the operation table is ordered sort of like a symmetric matrix, right?