Don't we have $1\to 3$?

Anyway, here's a slightly different way to find it.

Let's start with $1$.

$1 \to 3$ for $(1\,3\,\ldots)$.

$3\to 1 \to 2$ for $(1\,3\,2\,\ldots)$.

$2\to 4$ for $(1\,3\,2\,4\,\ldots)$.

$4\to 9$ for $(1\,3\,2\,4\,9\,\ldots)$.

$9\to 1$, which completes the cycle $(1\,3\,2\,4\,9)$.

Now we start a new cycle with the first missing number, which is $6$.

$6\to 8$ for $(6\,8\,\ldots)$.

$8\to 12$ for $(6\,8\,12\,\ldots)$.

$12\to 6$, which completes the cycle $(6\,8\,12)$.

So $\pi_1=(1\,3\,2\,4\,9)(6\,8\,12)$.

And yes, that is the same decomposition that you have, although the numbers are in a different order.