# Quotient group of S3

• May 29th 2008, 09:36 AM
topsquark
Quotient group of S3
It's a good thing I'm doing this review. I'm running into all kinds of stuff that I had skipped over.

I am constructing a quotient group of $\displaystyle S_3$:
Let
$\displaystyle e = \left ( \begin{matrix} 1 & 2 & 3 \\ 1 & 2 & 3 \end{matrix} \right )$

$\displaystyle a = \left ( \begin{matrix} 1 & 2 & 3 \\ 1 & 3 & 2 \end{matrix} \right )$

$\displaystyle b = \left ( \begin{matrix} 1 & 2 & 3 \\ 3 & 1 & 2 \end{matrix} \right )$

$\displaystyle c = \left ( \begin{matrix} 1 & 2 & 3 \\ 3 & 2 & 1 \end{matrix} \right )$

$\displaystyle d = \left ( \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{matrix} \right )$

$\displaystyle f = \left ( \begin{matrix} 1 & 2 & 3 \\ 2 & 1 & 3 \end{matrix} \right )$

I constructed the multiplication table as usual. Now I take the subgroup $\displaystyle E = \{ e, a \}$ and construct $\displaystyle S_3 / E$. I get the cosets
$\displaystyle E = \{ e, a \}$
$\displaystyle A = \{ b, c \}$
$\displaystyle B = \{ d, f \}$

Now I want to construct the multiplication table for $\displaystyle S_3 / E$, but when I apply the definition $\displaystyle (Ex)^{-1} = Ex^{-1}$ I get an apparent problem:
$\displaystyle (Eb)^{-1} = Eb^{-1} = Ed = \{ d, f \} = B$
but
$\displaystyle (Ec)^{-1} = Ec^{-1} = Ec = \{ b, c \} = A$

I would have expected both answers to be B, the inverse of A under the group $\displaystyle S_3 / E$. What did I do wrong?

Thanks!
-Dan
• May 29th 2008, 11:20 AM
TheEmptySet
Not a Normal subgroup
I am reviewing some Algebra as well. I THINK that this is problem

E is a subgroup of $\displaystyle S_3$ but it is not a normal subgroup

If A is a subgroup of G. Then A is a normal subgroup if $\displaystyle xA=Ax$ for all $\displaystyle x \in G$

Note that this is a Set equality.

For you $\displaystyle cE \ne Ec$ so E isn't normal

Then the defintion of a Quoteint Group is

If H is a normal subgroup of G, the group G/H that consists of the cosets of H in G is called the quotient groups.

Conclusion: I think that E must be Normal not just a subgroup to construct a quotient group.

I will keep looking.

(Thinking)
• May 29th 2008, 11:31 AM
ThePerfectHacker
When dealing with $\displaystyle S_3$ it is extremely helpful to let $\displaystyle \alpha = (1,2,3)$ and $\displaystyle \beta = (1,2)$. Then $\displaystyle e = \alpha^3 = \beta^2$ and $\displaystyle \alpha = d$, $\displaystyle \alpha^2 = b$, $\displaystyle \beta = f$, $\displaystyle \alpha \beta = c$, $\displaystyle \alpha^2 \beta = a$. And another useful property that $\displaystyle \beta \alpha = \alpha^2 \beta$.

Thus, $\displaystyle S_3 = \{ e , \alpha, \alpha^2, \beta , \alpha\beta ,\alpha^2 \beta\}$ with the properties that: $\displaystyle \alpha^3 = \beta^2 = 1$ and $\displaystyle \beta \alpha = \alpha^2 \beta$.

With that above you can easily multiply elements. For example, $\displaystyle (\alpha \beta)(\alpha^2 \beta) = \alpha (\beta \alpha^2) \beta = \alpha (\beta \alpha) \alpha \beta$.
Which becomes $\displaystyle \alpha (\alpha^2 \beta) \alpha \beta = \beta \alpha \beta = \alpha^2 \beta^2 = \alpha^2$.

Also the subgroup diagram is given below.

You wish to form $\displaystyle S_3/E$ where $\displaystyle E= \{e ,a\} = \{ e, \alpha^2 \beta \}$ but in order to do that you need to know if $\displaystyle E$ is a normal subgroup. Let us see if the left cosets agree with the right coset if we use $\displaystyle \alpha$. Note $\displaystyle \alpha E = \{ \alpha , \beta\}$ while $\displaystyle E\alpha = \{ \alpha, \alpha \beta \}$. Which is a problem.
• May 29th 2008, 12:58 PM
topsquark
Once again I missed that tiny little word "normal." As a lame defense I will mention that my Physics classes apparently used normal subgroups all the time, but never mention them as being that. :)

-Dan
• May 29th 2008, 02:53 PM
ThePerfectHacker
Quote:

Originally Posted by topsquark
Once again I missed that tiny little word "normal." As a lame defense I will mention that my Physics classes apparently used normal subgroups all the time, but never mention them as being that. :)

Maybe that is because in Physics classes you use abelian groups all the time?
(Which automatically makes them normal). (Thinking)
• May 29th 2008, 03:12 PM
topsquark
Quote:

Originally Posted by ThePerfectHacker
Maybe that is because in Physics classes you use abelian groups all the time?
(Which automatically makes them normal). (Thinking)

Actually most of my work where it might have mattered is with simple Lie groups. (Perhaps I should say "the simpler Lie groups." For all I know there is a specific definition for "simple" in reference to Lie groups.) Mostly it has been with the differential structure of the group rather than global, so this is probably why it hasn't been an issue. But for the record much of my work has been with U(1) and SU(2) (with only a little SO(3) and SU(3) thrown in).

Thanks again for the response. :)

-Dan