Does the permutation group S_8 contain elements of order 14?

[i_a_n]

Does the permutation group $S_8$ contain elements of order $14$?My answer: If $\sigma =\alpha \beta$
where $\alpha$ and $\beta$ are disjoint cycles, then
$|\sigma|=lcm(|\alpha|, |\beta|)$ .
Therefore the only possible disjoint cycle decompositions for a permutation $\sigma \in S_8$ with $|\sigma| =14$ is $(7,2)$. Since $7+2\neq 8$ so there is no element of order 14 in $S_8$.

Is my answer right?

 

[I like Serena]

Re: Does the permutation group $S_8$ contain elements of order 14 and the number of those element?

Does the permutation group $S_8$ contain elements of order $14$?My answer: If $\sigma =\alpha \beta$
where $\alpha$ and $\beta$ are disjoint cycles, then
$|\sigma|=lcm(|\alpha|, |\beta|)$ .
Therefore the only possible disjoint cycle decompositions for a permutation $\sigma \in S_8$ with $|\sigma| =14$ is $(7,2)$. Since $7+2\neq 8$ so there is no element of order 14 in $S_8$.

Is my answer right?I feel I've only considered the possibility that
[FONT=MathJax_Math]σ[/FONT] is product of two disjoint cycles but I don't know how to count the number of elements of order 14 in a single cycle. So what's the right answer?

Your answer is right... with a couple of tweaks.

Since 14 has the prime factorization $2 \cdot 7$ there are only two permutation types possible for an element of order 14.
Either it must be a 14-cycle, or it has to be the combination of a 2-cycle and a disjoint 7-cycle.

It is not so much that $7+2\neq 8$, but it is not possible because $7+2 > 8$.
And of course 14 is also greater than 8.
To clarify, $S_8$ does have an element of order 12: a 3-cycle combined with a disjoint 4-cycle.
This works because $3+4 \le 8$ and because 3 and 4 are relatively prime.