# Thread: S3 multiplication table

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

2. ## 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.

3. ## Re: S3 multiplication table

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

Originally Posted by topsquark
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.

Originally Posted by topsquark
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

4. ## 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.

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# multiplication table symmetric group s3

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