What is the order of $\tau = (5\ 6\ 7\ 8\ 9)(3\ 4\ 5\ 6)(2\ 3\ 4\ )(1\ 2)$? Is $\tau$ an even or an odd permutation? In $S_9$, what is the order of $\tau = (5\ 6\ 7\ 8\ 9)(3\ 4\ 5\ 6)(2\ 3\ 4)(1\ 2)$? Is $\tau$ an even or an odd permutation? 
For the first question: I tried to write $\tau$ as the composition of disjoint permutations but failed miserably. However, I made an explicit definition for the permutation $\tau$:
\begin{eqnarray*}
\tau(1) &=& 4\\
\tau(2) &=& 1\\
\tau(3) &=& 6\\
\tau(4) &=& 2\\
\tau(5) &=& 7\\
\tau(6) &=& 3\\
\tau(7) &=& 8\\
\tau(8) &=& 9\\
\tau(9) &=& 5\\
\end{eqnarray*}
Now I noticed that:
\begin{eqnarray*}
\tau^3(1) &=& 1\\
\tau^3(2) &=& 2\\
\tau^2(3) &=& 3\\
\tau^3(4) &=& 4\\
\tau^4(5) &=& 5\\
\tau^2(6) &=& 6\\
\tau^4(7) &=& 7\\
\tau^4(8) &=& 8\\
\tau^4(9) &=& 9\\
\end{eqnarray*}
where $\tau^n(k)$ means you apply the permutation $\tau$ to $k$, and then apply it to the result etc. for $n$ times. The powers appearing are $2,3,4$ so the order of $\tau$ is the smaller common multiple and since $\gcd(3,4)=1$ this is equal to 12. Therefore the order of $\tau$ is 12. So I found what I think is the correct answer but I do not really like the method and was wondering if anyone knows a better more elegant way to solve this.
For the second part I re-wrote $\tau$ as:
$$\tau = (5\ 6)(6\ 7)(7\ 8)(8\ 9)(3\ 4)(4\ 5)(5\ 6)(2\ 3)(3\ 4)(1\ 2)$$
which is the composion of ten 2-cycles which means $\tau$ has order $(-1)^10=-1$. I am pretty confident about this last part but I just wanted to add it for completion. If anyone knows a nicer way to determine the order of $\tau$ I would be happy to hear it. Thanks!
 A: These are the codes in which you can just test your work computationally:
 > G:=SymmetricGroup(9);
 > g:=(1,4,2)(3,6)(5,7,8,9);
 > Order(g); 
                                              12

A: You found the explicit two-line representation
$$\tau=\binom{123456789}{416273895}\;;$$
from that it’s straightforward to decompose $\tau$ into disjoint cycles. Start with $1$: it goes to $4$, which goes to $2$, which goes to $1$, so you have a cycle $(142)$. That takes care of $1$ and $2$, so move on to $3$: it goes to $6$, which goes right back to $3$, giving you the cycle $(36)$. Continuing in this fashion, you get one more cycle, $(5789)$. Thus,
$$\tau=(142)(36)(5789)\;.$$
And now it’s apparent that the order of $\tau$ is the least common multiple of $2,3$, and $4$, which is (as you discovered the hard way!) $12$.
Your determination that $\tau$ is even is correct, but you can do it even more easily from the disjoint cycle decomposition: odd cycles are even permutations, and vice versa, so here we have a composition of one even and two odd permutations, which is even. Or you can write out
$$\tau=(14)(12)(36)(57)(58)(59)\;.$$
(Those are composed from left to right.)
A: You did a nice job on both parts.  What I would do differently for the first part is instead of writing out the permutation explicitly I would write it in cycle structure; that makes it easier to see the order, and it's a bit faster.
A: There is no need to evaluate the product of the permutations.
The trick for determining whether the permutation is odd or even is to split each permutations into cycles.  You have done well with writing out the cycle structures.  Count the total number of cycles, so we have 10 cycles.  Thus, we have even permutation.
OR
You can also count the order of each permutation.  That is: Determine the number of cycles for each permutation.  For instance, if you have $(5\ 6\ 7\ 8\ 9)$, then there are $5 - 1 = 4$ cycles.  The quick formula for this is to first count the total numbers in the permutation.  Then, subtract that by 1 to get the order of the permutation.
Work out other permutations the same way and add up all the orders, so we have:
$$(5\ 6\ 7\ 8\ 9)(3\ 4\ 5\ 6)(2\ 3\ 4)(1\ 2)$$
$$(5 - 1) + (4 - 1) + (3 - 1) + (2 - 1) = 10$$
Thus, the permutation is even.
