Q:

order of permutation $(2,9,6,3,7)(4,8)$ in $S_9$

In $S_9$ I want to find the order for the permutation $(2,9,6,3,7)(4,8)$
My result is that the order should be 8. 
Wolfram Alpha tells me instead that it is 10. 
Question: Why is the order of the given permutation 10?
According to this post I have done following: 
$$ord \pi := min{\{ k \in \mathbb N | \pi^{k} = id_x\}}$$
so: 
$$ord((2,9,6,3,7)(4,8)) = lcm(n,m)$$
for:
$$(2,9,6,3,7)^n=(1)  \quad \& \quad (4,8)^m=(1)$$
for $(4,8)^n = (1):$
$${   \begin{pmatrix}
        1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
        1 & 2 & 3 & 8 & 5 & 6 & 7 & 4 & 9 \\
        \end{pmatrix} } \cdot {\begin{pmatrix}
        1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
        1 & 2 & 3 & 8 & 5 & 6 & 7 & 4 & 9 \\
        \end{pmatrix} } $$
$$={\begin{pmatrix}
        1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
        1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
        \end{pmatrix} } \quad n = 2$$
for $(2,9,6,3,7)^m = (1):$
$$ {\pi^0: \quad}{ \begin{pmatrix}
        1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
        1 & 9 & 7 & 4 & 5 & 3 & 2 & 8 & 6 \\
        \end{pmatrix}}$$
$$ {\pi^1: \quad}{ \begin{pmatrix}
        1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
        1 & 6 & 2 & 4 & 5 & 7 & 9 & 8 & 3 \\
        \end{pmatrix}}$$
$$ {\pi^2: \quad}{ \begin{pmatrix}
        1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
        1 & 7 & 6 & 4 & 5 & 9 & 3 & 8 & 2 \\
        \end{pmatrix}}$$
$$ {\pi^3: \quad}{ \begin{pmatrix}
        1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
        1 & 3 & 9 & 4 & 5 & 2 & 6 & 8 & 7 \\
        \end{pmatrix}}$$
$$ {\pi^4: \quad}{ \begin{pmatrix}
        1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
        1 & 9 & 7 & 4 & 5 & 3 & 2 & 8 & 6 \\
        \end{pmatrix}} \quad m = 4$$
So according to: 

So the order of σσ is exactly the smallest natural number n such that
  $(4,5)n=(1)(4,5)n=(1) and (2,3,7)n=(1)$$(2,3,7)n$$=(1)$ (think about this
  fact for a moment).

The order should be $n \cdot m = 2 \cdot 4$

A:

Your calculation of the powers of $(2\;9\;6\;3\;7)$ is wrong. $\pi^0$ is always the identity permutation; only $\pi^1$ will be the cycle you have written down.
What you write as $\pi^0$ is actually $\pi$ itself; what you claim is $\pi^1$ is $\pi^2$, and what you claim is $\pi^2$ actually seems to be $\pi^4$; it looks like you're squaring each line to get the next line, instead of multiplying it by the original $(2\;9\;6\;3\;7)$.
Thus, the order of the 5-cycle should be $5$, not $4$.

