For Problem V-11:

Starting from the right, $1 \mapsto 2 \mapsto 3, 2 \mapsto 1 \mapsto 2, 3 \mapsto 3 \mapsto 1$, so the answer is $\left(\begin{array}{ccc}1&2&3\\ 3&2&1\end{array}\right)$.

For Problem V-12: $(12)(23) = (123)$ since going right to left, $1 \mapsto 1 \mapsto 2, 2 \mapsto 3 \mapsto 3, 3 \mapsto 2 \mapsto 1$. $(23)(12) = (132)$ since $1 \mapsto 2 \mapsto 3, 3 \mapsto 3 \mapsto 2, 2 \mapsto 1 \mapsto 1$. Therefore they do not commute.

Since the particular symbols do not really matter and since $(ab) = (ba)$, we see that two transpositions that have just one symbol in common do not commute; instead their products are give distinct 3-cycles.

For Problem V-13:

(This problem had a misprint: the last 4 should be 1.)

$1 \mapsto 2 \mapsto 4 \mapsto 7 \mapsto 1$, $3 \mapsto 3$, and $5 \mapsto 6 \mapsto 5$, so the ``cycle decomposition'' is $(1247)(56)$.

For Problem V-14: $$\begin{array}{c\vert cccccc} & \mathbf{1}& (123) & (132) & ... ...(23) & (13) & (12) & (132) & (123) & \mathbf{1}\\ \end{array}$$

Note: For $S_4$ the table would be $24 \times 24$, and some elements would be non-cycles such as $(12)(34)$.