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