I do not understand this Abstract Algebra notation. Can you help me? $$
\langle((a,b),1),(1,(c,d))\rangle
$$
I am doing an independent study of Abstract Algebra. I found the notation that is in the attachment in a Standford webpage, but I do not understand the meaning. It is about the $S_3\times S_3$ group that has $36$ elements. (Not the $S_3$ that has $6$ elements)
I am clear that it uses the symbol of the generator. But why (ab), 1
what does mean? 
If you want to check the link here it is:
(Please go to the second paragraph)
Stanford webpage
 A: $(a,b)$ is the transposition that exchanges $a$ and $b$ and leaves everything else fixed.
You can write e.g. $(1,2)$ as permutation:
$(1,2) \hat= \begin{bmatrix}1 & 2 & 3 \\ 2 & 1 & 3\end{bmatrix} \hat= \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{pmatrix}$
This notation can also be extended, e.g:
$$
\begin{align}
(1,3,2): \quad& 1 \mapsto 3 \\
& 3 \mapsto 2 \\
& 2 \mapsto 1
\end{align}
$$
(Be careful here when you have more than two elements: Some author read this notation the other way around.)
$1$ is the neutral  Element (of $S_3$ here).
So $(x,y) \in S_3 \times S_3$ means $x,y$ are both Elements of $S_3$.
In the first case of your notation $x = (a,b)$ and $y=1$.
$\langle z  \rangle$ is the group that is generated by $z$. If $z$ is just one element, then $\langle z \rangle = \{z^k |k\in \mathbb Z\} = \{\ldots,z^{-1},1,z,z^2,z^3,\ldots\}$ is a cyclic group. In your case 
$\langle u,v \rangle = \{u^i v^j | i,j \in \mathbb Z\} \cup \{v^i u^j | i,j \in \mathbb Z\} $
A: In $S_3\times S_3$, a general element looks like $(\sigma,\tau)$, where both $\sigma$ and $\tau$ are elements of $S_3$. To be strict, $\sigma$ belongs to the "first" $S_3$ in the product $S_3\times S_3$, and $\tau$ belongs to the second one.
The notation $((ab),1)$ therefore stands for the element that corresponds to $(ab)$ in the first $S_3$, together with the identity in the second one.
