Results 1 to 4 of 4

Math Help - S3 multiplication table

  1. #1
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,211
    Thanks
    419
    Awards
    1

    S3 multiplication table

    1) Do generators exist for S3 (the symmetric group)? I thought the symmetric groups all had generators?

    2) When working out products of elements in S3 I get that (231)(213) = (123). This is a product of two 3-cycles and they are not inverses so (123) has to represent a 3-cycle. How do I tell otherwise that this product isn't the identity?

    3) When working out the multiplication table I noted several elements that are equivalent, such as (132) and (213). But I can't make this identification lest the table have multiple instances of (132) in a single row of the table. However this implies that the elements of S3 must contain (132) and (213), which makes no sense if |S3| is only 6.

    Gaaaaaaaaaaaaaaaaah!!!! This was supposed to be an easy exercise!

    -Dan
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Dec 2012
    From
    Athens, OH, USA
    Posts
    709
    Thanks
    294

    Re: S3 multiplication table

    Hi Dan,
    I think you're a little confused about cycles and products of cycles. I hope the attached discussion helps.

    S3 multiplication table-mhfgroups18.png
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,211
    Thanks
    419
    Awards
    1

    Re: S3 multiplication table

    Okay, I'm seeing things a bit better now.

    Quote Originally Posted by topsquark View Post
    2) When working out products of elements in S3 I get that (231)(213) = (123). This is a product of two 3-cycles and they are not inverses so (123) has to represent a 3-cycle. How do I tell otherwise that this product isn't the identity?
    The reason the RHS is coming out to look like the identity is because (132), the inverse of (231), is equivalent to (213). The RHS doesn't just look like the identity...it is the identity.

    Quote Originally Posted by topsquark View Post
    3) When working out the multiplication table I noted several elements that are equivalent, such as (132) and (213). But I can't make this identification lest the table have multiple instances of (132) in a single row of the table. However this implies that the elements of S3 must contain (132) and (213), which makes no sense if |S3| is only 6.
    And now this question is trivial.


    Thanks for the help, johng, it's a lot clearer. Though if I may inquire, does S3 have a set of generators? I still haven't been able to find any.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Dec 2012
    From
    Athens, OH, USA
    Posts
    709
    Thanks
    294

    Re: S3 multiplication table

    Hi again,
    Any group G is said to be finitely generated iff there is a finite set \{g_i\in G:1\leq i\leq n\} such that every element of G is a "word" in the generators g_i. Here a word is a product of k elements of G, x_1^{e_1}x_2^{e_2}\cdots x_k^{e_k} where each x_j is one of the g_i and e_j is an integer (positive, negative or 0); k need not be the same for different group elements.

    Trivially, then any finite group is finitely generated -- just take the generating set to be the entire group. An infinite direct sum is not finitely generated.

    However, I think what you want is a non=trivial generating set for S_3. Let a=(12) and b=(123). Then e=a0, (12)=a1, (13)=(12)(123)=ab, (23)=(12)(123)2=ab2, (123)=b and (132)=(132)(132)=b2.

    S_3 is said to have presentation by generators and relations: \{a,b:a^2=b^3=1 \text{ and } aba=b^2\}. As above S_3=\{e,a,b,ab,ab^2,b,b^2\}. The group table is then computable from the relations. For example, (ab^2)(ab)=(ab^2a)(b)=(aba)(aba)(b)=b^2b^2b=b^2; the 2nd equality since a2=1. Oops, I automatically used symbol 1 for the identity; 1 and e are synonyms as group elements.

    Typically, this is the way finite groups are described; i.e. via generators and relations. You need the concept of free groups to see that any finite group has a presentation.

    One last example: Let n be a positive integer greater than or equal to 2. The dihedral group D_n=\{x,y:y^n=x^2=e, xyx=y^{-1}\}. D_n is the group of symmetries of a regular n-gon. As you can see, D_3\cong S_3.

    You might try and convince yourself that D_n=\{1,y,y^2\cdots ,y^{n-1},yx,y^2x,\cdots y^{n-1}x\} and then work out the product of any two elements.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: May 6th 2012, 08:17 PM
  2. Multiplication Table
    Posted in the Number Theory Forum
    Replies: 10
    Last Post: March 19th 2009, 03:20 PM
  3. Adjacency Matrix / Multiplication Table
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: March 1st 2009, 06:50 AM
  4. Help with a multiplication table...
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: April 12th 2008, 02:27 PM
  5. Group multiplication table
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: February 8th 2008, 04:38 PM

Search Tags


/mathhelpforum @mathhelpforum