Order of a permutation which produces a 5-cycle 
Question: Let $\alpha = \left ( 1\,3\,5\,7\,9 \right )\left ( 2\,4\,6 \right )\left ( 8\,10\right )$.
  If $\alpha^{m}$ is a 5-cycle, what can you say about m?

Attempt:
$\left | \alpha \right |=lcm\left ( 5,3,2 \right )=30$
$\Rightarrow \alpha ^{30}=\varepsilon $
Unfortunately, I am unable to further beyond this point. 
Any hint is appreciated. Thanks in advance.
 A: Since $\alpha$ is expressed as the product (which actually means a composition) of disjoint cycles, that is, cycles that have no elements in common, and we know that disjoint cycles are conmutative, we can assure that $\alpha^m=(13579)^m (246)^m (8,10)^m$. Since we need this product to be a 5-cycle, we must consider the following:


*

*The product $(246)^m (8,10)^m$ is a 2-cycle if $m$ is a multiple of 3 but not a multiple of 2; it is a 3-cycle if $m$ is a multiple of 2 but not a multiple of 3; it is the product of a 3-cycle and a 2-cycle if $m$ is neither a multiple of 2 nor of 3; and it is the identity if $m$ is a multiple of 6. Therefore, it can never be a 5-cycle, and since it is disjoint with $(13579)$, the only way to obtain a 5-cycle as the result for $\alpha^m$ is for these two factors to disappear, that is, for them to equal the identity. We then conclude that $m$ must be a multiple of 6.

*If $m$ is also a multiple of 5, then $(13579)^m$ equals the identity, and since the identity is not a 5-cycle we must exclude this possibility. 
Then, the answer is that $\alpha^m$ will be a 5-cycle if, and only if, $m$ is a multiple of 6 but not a multiple of 5. This can also be expressed by saying that $m$ must be a multiple of 6 but not a multiple of 30.
A: As you wrote, $\;ord(\alpha)=30\;$ and thus it generates a group of order $\;30\;$ in $\;S_{10}\;$ , so if 
$$\;\langle\alpha^m\rangle=5\implies \alpha^{5m}=1\implies 30\,\mid\,5m\implies\;\ldots \;$$
Observe you have more than one option...
